1. No category

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3.1 Exercises, Sample Solutions
7. Answers may vary.
a) A direction vector for the line is:
AB = [6 − 3, 1 − (−2)]
= [3, 3]
Use A(3, −2) to obtain the parametric equations
x = 3 + 3t and y = −2 + 3t.
b) Let t = −1 to obtain (0, −5).
Let t = –2 to obtain (–3, −8).
Let t = 2 to obtain (9, 4).
c)
8. Answers may vary.
2−x y+1
a)
= 3
−4
2−x
x−2 y+1
by −1 to obtain 4 = 3 .
−4
The line passes through (2, −1) and has direction vector [4, 3]. Therefore, its parametric
equations are x = 2 + 4t, y = −1 + 3t.
Let t = −1 to obtain (−2, −4).
Let t = 0 to obtain (2, −1).
Let t = 1 to obtain (6, 2).
Multiply the numerator and denominator of
b)
x+3 3−y
2 = 5
3−y
x+3 y −3
5 by −1 to obtain 2 = −5 .
The line passes through (−3, 3) and has direction vector [2, −5]. Therefore, its parametric
equations are x = −3 + 2t, y = 3 − 5t.
Let t = −1 to obtain (−5, 8).
Let t = 0 to obtain (−3, 3)
Let t = 1 to obtain (−1, −2).
Multiply the numerator and denominator of
9. Answers may vary.
a) A vector equation is [x, y] = [4, 1] + t[−3, 1].
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The corresponding parametric equations are x = 4 − 3t, y = 1 + t.
x − 4 y −1
The corresponding symmetric equation is
= 1 .
−3
b) A direction vector for the line is:
RS = [4 − (−6), −2 − 2]
= [10, −4]
A simpler direction vector is [5, −2].
Use the point R(−6, 2).
A vector equation is [x, y] = [−6, 2] + t[5, −2].
The corresponding parametric equations are x = −6 + 5t, y = 2 − 2t.
x+6 y−2
The corresponding symmetric equation is 5 =
.
−2
G
c) A line parallel to the y-axis is in the direction j = [0, 1].
A vector equation is [x, y] = [2, −3] + t[0, 1].
The corresponding parametric equations are x = 2, y = −3 + t.
Since the x-component of the direction vector is 0, the equation of the line cannot be
written in symmetric form.
10. Answers may vary for parts b and c.
a) 2x − 3y + 12 = 0
3y = 2x + 12
2
y=3x+4
b) Two points on the line 2x − 3y + 12 = 0 are A(0, 4) and B(−6, 0).
Therefore, a direction vector for the line is:
AB = [−6 − 0, 0 − 4]
= [−6, −4]
A simpler direction vector is [3, 2]. Use the point A(0, 4) to obtain the
parametric equations x = 3t, y = 4 + 2t.
x y−4
b) The symmetric equation is 3 = 2 .
11. a) A direction vector for the line is:
AB = [5 − 2, 2 − (−4)]
= [3, 6]
A simpler direction vector is [1, 2]. Use the point A(2, −4).
i) The equation of the line in parametric form is x = 2 + t, y = −4 + 2t.
x−2 y+4
ii) The equation of the line in symmetric form is 1 = 2 .
iii) Since the line has direction vector [1, 2], its slope is 2. The equation of the line in
slope y-intercept form is:
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y+4
=2
x−2
y + 4 = 2x − 4
y = 2x − 8
iv) Rearrange the equation in part iii to obtain the general equation 2x − y − 8 = 0.
b) Parametric and symmetric equations can be extended to 3-space, while the slope
y-intercept and general equations cannot. An advantage of each form of equation is:
i) The equation of every line in can be written in parametric form, even if the
direction vector has a zero component.
ii) Symmetric equations do not involve a parameter.
iii) When the equation is written in slope y-intercept form, it is easy to visualize the line.
iv) The equation of every line can be written in general form, even lines with undefined
slope.
12.
Line parallel to the x-axis
Example: Line parallel to the x-axis
through A(3, 2). Hence a direction
vector for the line is [1, 0].
Vector equation
Vector equation
The direction vector has a y-component of 0.
[x y] = [3, 2] + t[1, 0]
Parametric equations
Parametric equations
The equation for y does not involve a parameter. x = 3 + t, y = 2
Symmetric equations
Symmetric equations
None
None
Slope y-intercept equation
Slope y-intercept form
m = 0 so the equation of the line is of the form
y=2
y = b.
General equation
General form
A = 0 so the equation of the line is of the form
y−2=0
By + C = 0.
Line parallel to the y-axis
Example: Line parallel to the y-axis
through A(3, 2). Hence a direction
vector for the line is [0, 1].
Vector equation
Vector equation
The direction vector has a x-component of 0.
[x y] = [3, 2] + t[0, 1]
Parametric equations
Parametric equations
The equation for x does not involve a parameter. x = 3, y = 2 + t
Symmetric equations
Symmetric equations
None
None
Slope y-intercept equation
Slope y-intercept form
None since slope is undefined.
None
General equation
General form
B = 0 so the equation of the line is of the form
x−3=0
Ax + C = 0.
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13. Substitute the coordinates of the point in the parametric equations of the line and solve for t.
If t is the same in each equation, then the point lies on the line.
A(−1, 1): −1 = 2 − 3t
1 = −1 + 2t
−3 = −3t
2 = 2t
t=1
t=1
Therefore, A lies on the line.
B(−4, 3): −4 = 2 − 3t
3 = −1 + 2t
−6 = −3t
4 = 2t
t=2
t=2
Therefore, B lies on the line.
7 = 2 − 3t
5 = −1 + 2t
5 = −3t
6 = 2t
5
t=3
t = −3
Therefore, C does not lie on the line.
C(7, 5):
D(5, −3): 5 = 2 − 3t
−3 = −1 + 2t
3 = −3t
−2 = 2t
t = −1
t=−1
Therefore, D lies on the line.
G
G
14. The direction numbers of the three sets of equations are m1 = [−3, 2], m2 = [9, −6], and
1 G
G
G
G
G
m3 = [3, −2] respectively. By inspection, m1 = −3 m2 and m1 = − m3 , so the lines are either
parallel or identical. The point (1, 3) lies on the first line. Determine whether it also lies on
the second and third lines by substituting (1, 3) into the equations of the lines.
Line 2:
1 = 7 + 9s
3 = −1 − 6s
−6 = 9s
4 = −6s
2
2
s = −3
s = −3
Since s is the same for each coordinate, (1, 3) lies on the second line. Hence
the first and second sets of equations represent the same line.
1 = −2 + 3k
3 = 5 − 2k
3 = 3k
−2 = −2k
k=1
k=1
Since k is the same for each coordinate, (1, 3) also lies on the third line. The
first and third sets of equations represent the same line.
Therefore, all three equations represent the same line.
Line 3:
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G G
m ⋅m
G
G
15. Use cosθ = G1 G 2 to find the angle between the two lines, where m1 and m2 represent the
m1 m2
direction vectors of the lines.
a) By inspection, the direction vectors [2, 1] and [3, −1] are not scalar multiples of each
other. Hence the lines intersect.
At a point of intersection, the x- and y-coordinates of the lines are equal.
7 + 2t = −3 + 3s →
2t − 3s = −10 c
4+t=4−s
→
t+s=0
d
From equation d: s = −t e
Substitute this expression in equation c and solve for t.
2t − 3(−t) = −10
5t = −10
t = −2
Substitute t = −2 in equation e to get s = 2.
To obtain the point of intersection, substitute t = −2 in the equations of L1 or s = 2 in the
equations of L2.
L1: x = 7 + 2(−2)
y=4−2
=3
=2
The lines intersect at the point (3, 2).
The angle of intersection, θ, of the lines is the acute angle between their direction vectors.
cosθ =
=
[2,1] ⋅ [3, − 1]
2 + 12 3 2 + (−1) 2
2(3) + 1(−1)
2
5 10
5
=
50
θ = 45°
The angle of intersection is 45°.
b) By inspection, the direction vectors [3, 4] and [3, −2] are not scalar multiples of each
other. Hence the lines intersect.
The parametric equations of the lines are:
L1: x = −3 + 3t
L2: x = 6 + 3s
y = −1 + 4t
y = 2 − 2s
At a point of intersection, the x- and y-coordinates of the lines are equal.
3t − 3s = 9 c
−3 + 3t = 6 + 3s →
4t + 2s = 3 d
−1 + 4t = 2 − 2s →
2 × c: 6t − 6s = 18
3 × d: 12t + 6s = 9
−4 × c: −12t + 12s = −36
3 × d: 12t + 6s = 9
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18t = 27
3
t=2
Add:
Add:
18s = −27
3
s = −2
3
3
To obtain the point of intersection, substitute t = 2 in the equations of L1 or s = −2 in the
equations of L2.
3
L1: x = −3 + 3 
2
3
=2
3
y = −1 + 4 
2
=5
3 
The lines intersect at the point  , 5  .
2 
The angle of intersection, θ, of the lines is the acute angle between their direction vectors.
cosθ =
=
=
[3, 4] ⋅ [3, − 2]
3 2 + 4 2 3 2 + (−2) 2
1
25 13
1
5 13
θ Ñ 86.8°
The angle of intersection is approximately 86.8°.
G
G
G
G
16. a) The direction vectors of the lines are m1 = [1, −3] and m2 = [−2, 6]. Since m2 = −2 m1 ,
the lines are either parallel or identical. The point (−5, 2) lies on L1. Determine whether it
also lies on L2 by substituting (−5, 2) into the equations of L2.
−5 = 4 − 2s
2 = 6s
1
2s = 9
s=3
9
s=2
Since s is not the same for each coordinate, (−5, 2) does not lie on L2. Therefore, the lines
are parallel and do not intersect.
G
G
b) By inspection, the direction vectors m1 = [−2, 1] and m2 = [2, 1] are not scalar multiples of
each other. Hence the lines intersect.
At a point of intersection, the x- and y-coordinates of the lines are equal.
6 − 2t = 4 + 2s
→
2t + 2s = 2, or t + s = 1 c
−1 + t = −8 + s
→
t − s = −7 d
c:
t+s=1
c: t + s = 1
d:
t − s = −7
−d: t − s = −7
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Add:
2t = −6
t = −3
Subtract:
2s = 8
s=4
To obtain the point of intersection, substitute t = −3 in the equations of L1 or s = 4 in the
equations of L2.
y = −1 − 3
L1: x = 6 − 2(−3)
= 12
= −4
The lines intersect at the point (12, −4).
17. a) When the line intersects the x-axis, y = 0.
Therefore 0 = 5 − t, or t = 5.
At t = 5, x = 2 + 5
=7
Therefore, the line intersects the x-axis at (7, 0).
Similarly, when line intersects the y-axis, x = 0.
Therefore 0 = 2 + t or t = −2
At t = −2, y = 5 − (−2)
=7
Therefore, the line intersects the y-axis at (0, 7).
b) When the line intersects the x-axis, y = 0.
x+1 0−4
3 = 2
x+1
3 = −2
x + 1 = −6
x = −7
Therefore, the line intersects the x-axis at (−7, 0).
Similarly, when line intersects the y-axis, x = 0.
0+1 y−4
3 = 2
1 y−4
3= 2
2
3=y−4
14
y= 3
 14 
Therefore, the line intersects the y-axis at  0,  .
 3
18. Answers may vary.
G
a) The line x = 4 is parallel to the y-axis, so a direction vector for the line is j = [0, 1].
Use the point (4, 0) to obtain the vector equation [x, y] = [4, 0] + t[0, 1]. The
corresponding parametric equations are x = 4, y = t.
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G
b) The line y = 3 is parallel to the x-axis, so a direction vector for the line is i = [1, 0].
Use the point (0, 3) to obtain the vector equation [x, y] = [0, 3] + t[1, 0]. The
corresponding parametric equations x = t, y = 3.
c) Two points on the line y = 3x − 2 are A(0, −2) and B(1, 1).
Therefore, a direction vector for the line is:
AB = [1 − 0, 1 − (−2)]
= [1, 3]
Use the point A(0, −2) to obtain the vector equation [x, y] = [0, −2] + t[1, 3]. The
corresponding parametric equations x = t, y = −2 + 3t.
d) Two points on the line x + 2y + 4 = 0 are A(0, −2) and B(−4, 0).
Therefore, a direction vector for the line is:
AB = [−4 − 0, 0 − (−2)]
= [−4, 2]
A simpler direction vector is [−2, 1].
Use the point B(−4, 0) to obtain the vector equation [x, y] = [−4, 0] + t[−2, 1] . The
corresponding parametric equations x = −4 − 2t and y = t.
5–x 2–y
19. a) 4 = 3
x–5 y–2
–4 = –3
The line passes through the point (5, 2) and has direction vector [–4, –3]. Therefore, the
corresponding parametric equations are x = 5 – 4t, y = 2 – 3t.
x y−1
b) 2 =
−1
The line passes through the point (0, 1) and has direction vector [2, −1]. Therefore, the
corresponding parametric equations are x = 2t, y = 1 − t.
20. Two points on the line y = 4x + 2 are A(0, 2) and B(1, 6).
Therefore, a direction vector for the line is:
AB = [1 − 0, 6 − 2]
= [1, 4]
Two points on the line y = −x + 3 are C(0, 3) and D(1, 2).
Therefore, a direction vector for the line is:
CD = [1 − 0, 2 − 3]
= [1, −1]
The angle, θ, between the lines is the acute angle between their direction vectors.
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cosθ =
=
=
=
AB ⋅ CD
AB CD
[1, 4] ⋅ [1, − 1]
1 + 4 2 12 + (−1) 2
1− 4
2
17 2
−3
34
θ Ñ 121.0°
Therefore, the angle of intersection is approximately 180° − 121° = 59°.
21. [x, y] = [2, 3] + s[1, −2]
Determine two points on the line. Let s = 0 to obtain (2, 3). Let s = 1 to obtain (3, 1).
1−3
or m = −2. The equation of the line with slope −2 passing
The slope of the line is m =
3−2
through (2, 3) is:
y = −2(x − 2) + 3
= −2x + 4 + 3
= −2x + 7
22. a) The direction vector of the line is [1, 1]. Since this vector has length 12 + 12 = 2 ,
1
1
the direction cosines are cos α =
and cos β =
. Therefore, the direction angles are:
2
2
 1 
 1 
α = cos −1 
β = cos −1 


 2
 2
= 45°
= 45°
b) The direction vector of the line is [3, 4]. Since this vector has length 32 + 42 = 5, the
3
4
direction cosines are cos α = 5 and cos β = 5 . Therefore, the direction angles are:
3
4
α = cos −1  
β = cos −1  
5
5
Ñ 53.1°
Ñ 36.9°
23. Since C lies on the line x = 2 + 2t, y = 8 − t, its coordinates are of the form (2 + 2t, 8 − t).
AB is the hypotenuse so ∠C = 90°. Sides AC and BC are perpendicular and AC ⋅ BC = 0.
AC ⋅ BC = 0
[2 + 2t − 2, 8 − t − 3]⋅[2 + 2t − 9, 8 − t − 2] = 0
[2t, 5 − t]⋅[2t − 7, 6 − t] = 0
2t(2t − 7) + (5 − t)(6 − t) = 0
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4t2 − 14t + 30 − 11t + t2 = 0
5t2 − 25t + 30 = 0
t2 − 5t + 6 = 0
(t − 2)(t − 3) = 0
t = 2 or t = 3
When t = 2, C has coordinates (2 + 2(2), 8 − 2) or (6, 6).
When t = 3, C has coordinates (2 +2(3), 8 − 3) or (8, 5).
24. From exercise 23, C has coordinates (2 + 2t, 8 − t).
a) AC is the hypotenuse so ∠B = 90°. Sides AB and BC are perpendicular and AB ⋅ BC = 0.
AB ⋅ BC = 0
[9 − 2, 2 − 3]⋅[2 + 2t − 9, 8 − t − 2] = 0
[7, −1]⋅[2t − 7, 6 − t] = 0
7(2t − 7) −1(6 − t) = 0
14t − 49 − 6 + t = 0
15t = 55
55
t = 15
11
= 3

11   28 13 
 11 
Therefore, C has coordinates  2 + 2 , 8 −  or  , 
3  3 3
3

b) BC is the hypotenuse so ∠A = 90°. Sides AB and AC are perpendicular and AB ⋅ AC = 0.
AB ⋅ AC = 0
[9 − 2, 2 − 3]⋅[2 + 2t − 2, 8 − t − 3] = 0
[7, −1]⋅[2t, 5 − t] = 0
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7(2t) −1(5 − t) = 0
14t − 5 + t = 0
15t = 5
1
t=3

1   8 23 
1
Therefore, C has coordinates  2 + 2 , 8 −  or  ,  .
3 3 3 
3

G
x
y
25. a) The direction cosines of direction vector v = [x, y] are cos α = G and cos β = G so
v
v
cos2α + cos2β = 1. Since the first direction angle is 60°:
cos2 60° + cos2β = 1
2
1
2
  + cos β = 1
2
1
cos2β = 1 − 4
3
=4
3
=± 2
Since 0° ≤ β ≤ 180°, β = 30° or β = 150°.
The possible second direction angles are 30° or 150°.
1 3 
b) The line with direction angles α = 60° and β = 30° has direction vector  ,  .
2 2 
The vector [1, 3 ] is also a direction vector for the line. Since the line passes through
A(0, 4), its parametric equations are x = t, y = 4 + 3 t.
Similarly, the line with direction angles α = 60° and β = 150° has direction vector
1
3
 ,−
 . The vector [1, – 3 ] is also a direction vector for the line. Since the line
2
2


passes through A(0, 4), its parametric equations are x = t, y = 4 − 3 t.
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c)
26.
Let P be any point on line AB as shown in the diagram above. By the Triangle Law:
OP = OA + AP
= OA + t AB
= OA + t (−OA + OB)
= OA − t OA + t OB
= (1 − t )OA + t OB
G
G G
p = (1 − t )a + tb
G
G
G G
G
The coefficients of a and b add to 1. If we let 1 − t = s, we have p = sa + tb where s + t = 1.
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3.2 Exercises, Sample Solutions
8. Answers may vary.
a) A direction vector for the line is:
AB = [4 − 5, 5 − 1, −1 + 3]
= [−1, 4, 2]
Use the point A(5, 1, −3) to obtain the parametric equations x = 5 − t, y = 1 + 4t,
z = −3 + 2t.
b) Let t = 2 to obtain (3, 9, 1); t = 3 to obtain (2, 13, 3); t = 4 to obtain (1, 17, 5).
9. Answers may vary.
a) The parametric equations of the line are x = 2 + t, y = 3 − 2t, z = −1 − t.
Let t = 0 to obtain (2, 3, −1); t = 1 to obtain (3, 1, −2); t = 2 to obtain (4, −1, −3).
b) The parametric equations of the line are x = 1 + 3t, y = t, z = 1 − 2t.
Let t = 0 to obtain (1, 0, 1); t = 1 to obtain (4, 1, −1); t = 2 to obtain (7, 2, −3).
c) The parametric equations of the lines are x = −3 − 2t, y = 5 + t, z = 2.
Let t = 0 to obtain (−3, 5, 2); t = 1 to obtain (−5, 6, 2); t = 2 to obtain (−7, 7, 2).
d) The parametric equations of the lines are x = −4, y = −2 + 3t, z = 3 + 4t
Let t = 0 to obtain (−4, −2, 3); t = 1 to obtain (−4, 1, 7); t = 2 to obtain (−4, 4, 11).
10. a) Since all points on the line have a z-coordinate of 2, the line is parallel to the xy-plane and
is 2 units above it.
b) Since all points on the line have an x-coordinate of −4, the line is parallel to the yz-plane
and is 4 units behind it.
11. Answers may vary.
a) A vector equation of the line is [x, y, z] = [2, −1, 3] + t[−1, 3, 5].
The corresponding parametric equations are x = 2 − t, y = −1 + 3t, z = 3 + 5t.
x−2 y+1 z−3
The corresponding symmetric equation is
= 3 = 5 .
−1
b) A direction vector for the line is:
AB = [−1 − 4, 0 + 2, 3 − 1]
= [−5, 2, 2]
A vector equation of the line is [x, y, z] = [4, −2, 1] + t[−5, 2, 2].
The corresponding parametric equations are x = 4 − 5t, y = −2 + 2t, z = 1 + 2t.
x−4 y+2 z−1
= 2 = 2 .
The corresponding symmetric equation is
−5
c) A direction vector for the line is:
CD = [5 − 5, 3 + 1, −4 − 0]
= [0, 4, −4]
Use the point C(5, −1, 0).
A vector equation of the line is [x, y, z] = [5, −1, 0] + t[0, 4, −4].
The corresponding parametric equations are x = 5, y = −1 + 4t, z = −4t.
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Since the direction vector has a component of 0, there are no symmetric equations.
y+1 z
.
However, we can write the equation of the line as x = 5, 1 =
−1
G
d) A line parallel to the x-axis is in the direction i = [1, 0, 0].
A vector equation of the line is [x, y, z] = [3, −1, −1] + t[1, 0, 0].
The corresponding parametric equations are x = 3 + t, y = −1, z = −1.
Since the y- and z- components of the direction vector are 0, the equation of the line
cannot be written in symmetric form.
G
e) A line parallel to the y-axis is in the direction j = [0, 1, 0].
A vector equation of the line is [x, y, z] = [−2, 0, 5] + t[0, 1, 0].
The corresponding parametric equations are x = −2, y = t, z = 5.
Since the x- and z- components of the direction vector are 0, the equation of the line
cannot be written in symmetric form.
12. Substitute the coordinates of each point in the symmetric equation. If the ratios are equal, the
point lies on the line.
x + 3 −5 + 3
z−1 7−1
y 2
A(−5, 2, 7):
=
1 =1
3 = 3
−2
−2
=1
=2
=2
The ratios are not equal. A does not lie on the line.
B(3, 0, −1):
x+3 3+3
=
−2
−2
= −3
y 0
1 =1
=0
z − 1 −1 − 1
3 = 3
−2
= 3
The ratios are not equal. B does not lie on the line.
x + 3 −1 + 3
y −1
=
1 = 1
−2
−2
= −1
= −1
The ratios are equal. C lies on the line.
x+3 4+3
y −2
=
D(4, −2, 3):
1 = 1
−2
−2
7
= −2
= −2
The ratios are not equal. D does not lie on the line.
C(−1, −1, −2):
z − 1 −2 − 1
3 = 3
= −1
z−1 3−1
3 = 3
2
=3
G
G
13. The direction numbers of the three sets of equations are m1 = [2, −1, 4], m2 = [−2, 1, −4], and
G
G
G
G 1 G
m3 = [4, −2, 8] respectively. By inspection, m1 = − m2 and m1 = 2 m3 . Thus the lines are
either parallel or coincident. The point (1, 3, 7) lies on the first line. Determine whether it
also lies on the lines represented by the other two sets of equations.
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Set 2
3=6+s
7 = −5 − 4s
1 = −5 − 2s
6 = −2s
s = −3
12 = −4s
s = −3
s = −3
Since s is the same for each coordinate, (1, 3, 7) lies on the second line. Hence the first
and second sets of equations represent the same line.
Set 3
1 = 5 + 4k
3 = 1 − 2k
7 = 15 + 8k
2 = −2k
−8 = 8k
−4 = 4k
k = −1
k = −1
k = −1
Since k is the same for each coordinate, (1, 3, 7) lies on the third line. Hence the first and
third sets of equations represent the same line.
Therefore, all three equations represent the same line.
G G
m1 ⋅ m2
G
G
14. Use cosθ = G G to find the angle between the two lines, where m1 and m2 represent the
m1 m2
direction vectors of the lines.
a)
The parametric equations of L1 and L2 are:
L1: x = −1 + 3t, y = 2 − t, z = 4t
L2: x = −6 + 2s, y = 8 − 5s, z = −1 −3s
At a point of intersection, the x-, y-, and z-coordinates of the lines are equal.
−1 + 3t = −6 + 2s →
3t − 2s = −5 c
2 − t = 8 − 5s
→
t − 5s = −6 d
4t = −1 −3s
→
4t + 3s = −1 e
Use equations c and d.
c: 3t − 2s = −5
−3 × d: −3t + 15s = 18
Add:
13s = 13
s=1
Substitute s = 1 in equation d and solve for t.
t − 5(1) = −6
t = −1
Verify the values of t and s by substituting them in equation e.
LS = 4(−1) + 3(1)
= −1
= RS
The system of equations is consistent so the lines intersect. To obtain the point of
intersection, substitute t = −1 in the equations of L1 or s = 1 in the equations of L2.
L1: x = −1 + 3(−1)
y = 2 − (−1)
z = 4(−1)
= −4
=3
= −4
The lines intersect at the point (−4, 3, −4).
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b) The parametric equations of L1 and L2 are:
L1: x = 1 + 2t, y = −3 + t, z = −3
L2: x = 2 + 3s, y = −1 + 2s, z = s
At a point of intersection, the x-, y-, and z-coordinates of the lines are equal.
→
2t − 3s = 1 c
1 + 2t = 2 + 3s
−3 + t = −1 + 2s
→
t − 2s = 2 d
→
s = −3 e
−3 = s
Use equations d and e. Substitute s = −3 in equation d and solve for t.
t − 2(−3) = 2
t = −4
Verify the values of t and s by substituting them in equation c.
LS = 2(−4) −3(−3)
=1
= RS
The system of equations is consistent so the lines intersect. To obtain the point of
intersection, substitute t = −4 in the equations of L1 or s = −3 in the equations of L2.
L1: x = 1 + 2(−4)
y = −3 − 4
z = −3
= −7
= −7
The lines intersect at the point (−7, −7, −3).
15. a) At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
1 + 2t = 1 + 3s
→
2t − 3s = 0 c
−1 − t = 2 + 2s
→
t + 2s = −3 d
3t = 3 + 4s
→
3t − 4s = 3 e
Use equations c and d. From equation d: t = −2s − 3 f
Substitute this expression for t in equation c.
2(−2s − 3) − 3s = 0
−4s − 6 − 3s = 0
−7s = 6
6
s=−7
6
Substitute s = − 7 in equation f to determine t.
 6
t = −2 −  − 3
 7
9
t = −7
Verify the values of t and s by substituting them in equation e.
 9  6
LS = 3 −  − 4 − 
 7  7
3
= −7
≠ RS
The system of equations is inconsistent so the lines do not intersect.
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b) The parametric equations of L1 and L2 are:
L1: x = −1 + t, y = 1 + 2t, z = 3 − 2t
L2: x = 5 − 2s, y = 3 + s, z = −3 + s
At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
−1 + t = 5 − 2s
→
t + 2s = 6 c
1 + 2t = 3 + s
→
2t − s = 2 d
3 − 2t = −3 + s
→
2t + s = 6 e
Use equations d and e.
d − e: −2s = −4
s=2
Substitute s = 2 in equation d and solve for t.
2t − 2 = 2
2t = 4
t=2
Verify the values of t and s by substituting them in equation c.
LS = 2 + 2(2)
=6
= RS
The system of equations is consistent so the lines intersect. To obtain the point of
intersection, substitute t = 2 in the equations of L1 or s = 2 in the equations of L2.
x = −1 + 2
y = 1 + 2(2)
z = 3 −2(2)
L1:
=1
=5
= −1
The lines intersect at the point (1, 5, −1).
c) The parametric equations of L1 and L2 are:
L1: x = 1 + 2t, y = 3 + 3t, z = 5 + 4t
L2: x = −1 + 2s, y = −4 − s, z = −2 + s
At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
1 + 2t = −1 + 2s
→
2t − 2s = −2, or t − s = −1 c
3 + 3t = −4 − s
→
3t + s = −7 d
5 + 4t = −2 + s
→
4t − s = −7 e
Use equations d and e.
d + e: 7t = −14
t = −2
Substitute t = −2 in equation d and solve for s.
3(−2) + s = −7
−6 + s = −7
s = −1
Verify the values of s and t by substituting them in equation c.
LS = −2 − (−1)
= −1
= RS
The system of equations is consistent so the lines intersect. To obtain the point of
intersection, substitute t = −2 in the equations of L1 or s = −1 in the equations of L2.
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L1:
x = 1 + 2(−2)
y = 3 + 3(−2)
= −3
= −3
The lines intersect at the point (−3, −3, −3).
z = 5 + 4(−2)
= −3
16. Answers may vary.
The line through edge AB passes through A(2, 2, 2) and
B(2, 2, −2). It is parallel to the z-axis so it has direction
G
vector k = [0, 0, 1] and parametric equations x = 2, y = 2,
z = 2 + t.
a) The line through edge AC passesG through A(2, 2, 2) and C(−2, 2, 2). It is parallel to the
x-axis so it has direction vector i = [1, 0, 0] and parametric equations x = 2 + s,
y = 2, z = 2.
At a point of intersection, the x-, y-, and z-coordinates of the two lines must be equal.
Since the y-coordinates are 2 in both lines, this leaves:
2 = 2 + s, or s = 0
2 + t = 2, or t = 0
Substitute t = 0 in the equations of line AB or s = 0 in the equations of line AC to obtain
the point of intersection (2, 2, 2). As expected, these are the coordinates of point A.
b) The line through edge CD is parallel to the line through edge AB. Since the line passes
G
through C(−2, 2, 2) and is in the direction k = [0, 0, 1], its parametric equations are
x = −2, y = 2, z = 2 + s. At a point of intersection, the x-, y- and z-coordinates of the lines
must be equal. This condition is not satisfied by the x-coordinates of the lines: for AB,
x = 2 but for CD, x = −2. Thus the system of equations is inconsistent, and the lines
do not intersect. We know the lines are parallel because their direction vectors are parallel.
c) The line through GC is not parallel to the line through edge AB and doesGnot pass through
points A or B. The line is parallel to the y-axis so it has direction vector j = [0, 1, 0]. Use
C (−2, 2, 2) to obtain the parametric equations x = −2, y = 2 + s, z = 2. At a point of
intersection, the x-, y- and z-coordinates of the lines must be equal. This condition is not
satisfied by x-coordinates of the lines: for AB, x = 2 but for GC, x = −2). Thus the system
of equations is inconsistent, and the lines do not intersect. We know the lines are skew
because their direction vectors are not parallel.
17. a) Answers may vary.
Label the vertices A(2, 2, 2), E(2, −2, −2), D(−2, 2, −2), G(−2, −2, 2).
Line
Direction vector
of line
Parametric
equations
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AD
AD = [−2 − 2, 2 − 2, −2 − 2]
= [−4, 0, −4]
G
Use m = [−1, 0, −1]
x=2−t
y=2
z=2+t
AE
AE = [2 − 2, −2 − 2, −2 − 2]
= [0, −4, −4]
G
Use m = [0, −1, −1]
x=2
y=2−t
z=2−t
AG
AG = [−2 − 2, −2 − 2, 2 − 2]
= [−4, −4, 0]
G
Use m = [−1, −1, 0]
x=2−t
y=2−t
z=2
GD
GD = [−2 − (−2), 2 − (−2), −2 − 2]
= [0, 4, −4]
G
Use m = [0, 1, −1]
x = −2
y = −2 + t
z=2−t
GE
GE = [2 − (−2), −2 − (−2), −2 − 2]
= [4, 0, −4]
G
Use m = [1, 0, −1]
x = −2 + t
y = −2
z=2−t
ED
ED = [−2 − 2, 2 − (−2), −2 − (−2)]
= [−4, 4, 0]
G
Use m = [−1, 1, 0]
x=2−t
y = −2 + t
z = −2
b) Use edges AD and AE.
cosθ =
=
=
AD ⋅ AE
AD AE
[−1, 0,1] ⋅ [0, − 1, − 1]
(−1) 2 + 0 2 + 12 0 2 + (−1) 2 + (−1) 2
[−1, 0,1] ⋅ [0, − 1, − 1]
(−1) 2 + 0 2 + 12 0 2 + (−1) 2 + (−1) 2
−1
=
2 2
−1
= 2
θ = 120°
Use edges EA and ED.
cosθ =
=
EA ⋅ ED
EA ED
[0,1,1] ⋅ [−1,1, 0]
0 2 + 12 + 12 (−1) 2 + 12 + 0 2
1
=
2 2
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1
=2
θ = 60°
Due to symmetry, all other sets of two edges will either have an angle of intersection
of 60° or an angle of intersection of 120°.
18. In each part, the line passes through the point A(−2, 5, 3).
G
a) m = [2, 0, 1]
The line passes through A and is parallel to the xz-plane.
In the parametric equations, the equation for y will not involve a parameter.
There are no symmetric equations.
G
b) m = [0, 4, 1]
The line passes through A and is parallel to the yz-plane.
In the parametric equations, the equation for x will not involve a parameter.
There are no symmetric equations.
G
c) m = [2, 0, 0]
G
= 2i
The line passes through A and is parallel to the x-axis.
In the parametric equations, the equations for y and z will not involve a parameter.
There are no symmetric equations.
19. a) Since points on the xy-plane have a z-coordinate of 0, a line will intersect the xy-plane at
the point where z = 0. Similarly, the line will intersect the xz-plane at the point where
y = 0, and the yz-plane at the point where x = 0.
b)
x−5 y+2 z−1
1 = 3 = −2
x−5 y+2 0−1
The line intersects the xy-plane at z = 0, so 1 = 3 =
−2
x−5 0−1
y+2 0−1
1 = −2
3 = −2
y+2 1
1
x−5=2
3 =2
11
3
x= 2
y+2=2
1
y = −2
 11 1 
The line intersects the xy-plane at  ,− ,0  .
2 2 
.
x−5 0+2 z−1
The line intersects the xz-plane at y = 0, so 1 = 3 =
.
−2
x−5 0+2
z−1 0+2
=
= 3
1
3
−2
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2
x−5=3
17
x= 3
z−1 2
=3
−2
4
z − 1 = −3
1
z = −3
1
 17
The line intersects the xz-plane at  ,0,−  .
3
 3
0−5 y+2 z−1
The line intersects the yz-plane at x = 0, so 1 = 3 =
.
−2
z−1 0−5
0−5 y+2
=
= 1
3
1
−2
y+2
z−1
= −5
−5 = 3
−2
y + 2 = −15
z − 1 = 10
y = −17
z = 11
The line intersects the yz-plane at (0, −17, 11).
c)
x+6 y−2
2 = 3 , z = −2
Since the z-coordinate of the line is never 0, the line does not intersect the xy-plane.
x+6 0−2
The line intersects the xz-plane at y = 0, so 2 = 3 .
x+6 0−2
2 = 3
2
x+6
=
−
3
2
4
x + 6 = −3
22
x=−3
 22

The line intersects the xz-plane at  − , 0, − 2  .

 3
0+6 y−2
The line intersects the yz-plane at x = 0, so 2 = 3 .
0+6 y−2
2 = 3
y−2
3 = 3
y−2=9
y = 11
The line intersects the yz-plane at (0, 11, −2).
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20. If a line is parallel to a coordinate plane, then the direction vector will have one zero
component. One of the parametric equations will not involve a parameter. There will be no
symmetric equations. For example, consider the line through A(2, 5, 1) and direction vector
[1, 2, 0]. The z-component of the direction vector is 0, so the line is parallel to the xy-plane
and has parametric equations x = 2 + t, y = 5 + 2t, z = 1. The line has no symmetric equations,
x−2 y−5
but we can write 1 = 2 , z = 1.
21. If a line is parallel to one of the coordinate axes, then the direction vector will have two zero
components. Two of the parametric equations will not involve a parameter. There will be no
symmetric equations. For example,
G consider the line through A(2, 5, 1) and direction vector
[1, 0, 0]. This direction vector is i so the line is parallel to the x-axis and has parametric
equations x = 2 + t, y = 5, z = 2. The equations cannot be written in symmetric form.
22. a) The parametric equations of L1 and L2 are:
L1: x = 4 + 2t, y = 8 + 3t, z = −1 − 4t
L2: x = 16 − 6s, y = 2 + s, z = −1 + 2s
At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
2t + 6s = 12, or t + 3s = 6 c
4 + 2t = 16 − 6s →
→
3t − s = −6
d
8 + 3t = 2 + s
4t + 2s = 0 , or 2t + s = 0 e
−1 − 4t = −1 + 2s →
Use equations d and e. From equation e: s = −2t f
Substitute this expression for s in equation d and solve for t.
3t − (−2t) = −6
5t = −6
6
t = −5
 6
From equation f: s = −2 − 
 5
12
= 5
Verify the values of t and s by substituting them in equation c.
6  12 
LS = − + 3 
5 5
=6
= RS
The system of equations is consistent so the lines intersect. To obtain the point of
6
12
intersection, substitute t = −5 in the equations of L1 or s = 5 in the equations of L2.
 6
 6
 6
L1: x = 4 + 2 − 
y = 8 + 3 − 
z = − 1 − 4 − 
 5
 5
 5
22
19
8
= 5
= 5
=5
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 8 22 19 
The lines intersect at the point  , ,  .
5 5 5 
b) Answers may vary.
G
The line is perpendicular to both L1 and L2 so its direction vector, m , is the cross
product of the direction vectors of L1 and L2.
3 −4
2
3
G
m = [2, 3, −4] × [−6, 1, 2]
1
2 −6
1
= [6 − (−4), 24 − 4, 2 − (−18)]
= [10, 20, 20]
The vector [1, 2, 2] is also a direction vector of the line. Since the line passes through
 8 22 19 
 , ,  , the point of intersection of L1 and L2, its parametric equations are
5 5 5 
8
22
19
x = 5 + t, y = 5 + 2t, z = 5 + 2t.
23. Answers may vary.
The parametric equations of L1 and L2 are:
L1: x = 4 + 2t, y = 8 + 3t, z = −1 − 4t
L2: x = 7 − 6s, y = 2 + s, z = −1 + 2s
Let L be a line that intersects L1 and L2 at right angles at points
A and B respectively.
Since A lies on L1, its coordinates are of the form
c
(4 + 2t, 8 + 3t, −1 − 4t)
Since B lies on L2, its coordinates are of the form
d
(7 − 6s, 2 + s, −1 + 2s)
Therefore, a direction vector for L is:
AB = [7 − 6s − (4 + 2t), 2 + s − (8 + 3t), −1 + 2s − (−1 − 4t)]
= [3 − 6s − 2t, −6 + s − 3t, 2s + 4t]
Since L intersects L1 and L2 at right angles, AB is perpendicular to their direction vectors.
[3 − 6s − 2t, −6 + s − 3t, 2s + 4t] ⋅ [2, 3, −4] = 0
6 − 12s − 4t − 18 + 3s − 9t − 8s −16t = 0
29t + 17s = −12 e
[3 − 6s − 2t, −6 + s − 3t, 2s + 4t] ⋅ [−6, 1, 2] = 0
−18 + 36s + 12t −6 + s − 3t + 4s + 8t = 0
17t + 41s = 24 f
Solve for s and t.
−41 × e: −1189t − 697s = 492
−17 × e: −493t − 289s = 204
17 × f:
289t + 697s = 408
29 × f: 493t + 1189s = 696
Add: −900t
= 900
Add:
900s = 900
s=1
t = −1
Substitute t = −1 and s = 1 into c and d to obtain the coordinates of A and B.
A has coordinates (4 + 2(−1), 8 + 3(−1), −1 − 4(−1)), or (2, 5, 3).
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B has coordinates (7 − 6(1), 2 + 1, −1 + 2(1)), or (1, 3, 1).
Therefore AB = [1 − 2, 3 − 5, 1 − 3]
= [−1, −2, −2]
The vector [1, 2, 2] is also a direction vector for L. Use point A to obtain the following
parametric equations for L: x = 2 + t, y = 5 + 2t, z = 3 + 2t.
24. Answers may vary.
The direction cosines of the line are:
cos β = cos 45°
cos α = cos 60°
1
1
=2
=
2
cos γ = cos 60°
1
=2
1 1 1
,  . Multiply by 2 2 to obtain a
Therefore, a direction vector for the line is  ,
2 2 2
direction vector whose components are not fractions: [ 2 , 2, 2 ] . The line passes through
x+2 y−1 z−3
(−2, 1, 3) so its symmetric equations are
= 2 =
.
2
2
G
G
25. a) The direction vector v = [x, y, z] has magnitude v = x 2 + y 2 + z 2 . Its direction cosines
are:
cos α =
x
2
2
x +y +z
2
, cos β =
y
2
2
x +y +z
2
, cos γ =
z
2
x + y2 + z2
So cos2α + cos2β + cos2γ = 1. Since the first two direction angles are both 60°:
cos2 60° + cos2 60° + cos2γ = 1
2
2
1 1
2
  +   + cos γ = 1
2
2
   
1 1
cos2γ = 1 − 4 − 4
1
=2
1
2
cos γ = ±
, or ± 2
2
Since 0° ≤ γ ≤ 180°, γ = 45° or γ = 135°.
The possible third direction angles are 45° or 135°.
b) The line with direction angles α = 60°, β = 60°, and γ = 45° has direction vector
1 1 2 
 , ,
 . The vector [ 2 , 2 , 2] is also a direction vector for the line. Since the line
2
2
2


passes through A(0, 0, 4), its parametric equations are x = 2 t, y = 2 t, z = 4 + 2t.
Similarly, the line with direction angles α = 60°, β = 60°, and γ = 135° has direction
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1 1
2
vector  , ,−
 . The vector [ 2 , 2 , −2] is also a direction vector for the line. Since
2
2
2


the line passes through A(0, 0, 4), its parametric equations are x = 2 t, y = 2 t,
z = 4 − 2t.
c)
z
x = 2 t, y = 2 t, z = 4 + 2t
4
y
x
x = 2 t, y = 2 t, z = 4 − 2t
G G G G G
G G G G
G G
G G
G
26. a) Since p = a + tm , p × m = (a + tm) × m . Therefore, p × m = a × m + t (m × m) . But
G G G G
G G G
m × m = 0 , so p × m = a × m .
G G
G
b) Vectors a , p , and m are coplanar.
27. a) A line is always determined by two conditions. For example, it passes
through two given points, or it passes through a given point and has a given
direction vector. Any equation of a line has to satisfy these two conditions.
b) An equation of a line has to satisfy two conditions. There is no inconsistency in having
one vector equation because it combines in one equation the two conditions that the line
passes through a given point and has a given direction vector. There is no inconsistency in
having three parametric equations because these are just a restatement of the vector
equation in component form.
28. a) Use the diagram on page 135.
G G
p − a = AP
G
= tm
( pG − aG ) × mG = tmG × mG
G G
= t ( m × m)
G
=0
G G G G
Therefore, ( p − a ) × m = 0 is the equation of the line passing through the point A and
G
parallel to the vector m .
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G G G G
b) ( p − a ) × m = 0
G
([x, y, z] − [−2, 5, 3]) × [2, 4, 1] = 0
G
[x + 2, y − 5, z − 3] × [2, 4, 1] = 0
y −5
4
z −3
x+2
y −5
1
2
4
G
[y − 5 − 4(z − 3), 2(z − 3) − (x + 2), 4(x + 2) − 2(y − 5)] = 0
[y −4z + 7, − x + 2z − 8, 4x − 2y + 18] = [0, 0, 0]
Since the vectors are equal, their components are equal.
y − 4z + 7 = 0
c
x − 2z + 8 = 0
d
4x − 2y + 18 = 0
e
From equation c: y = −7 + 4z f
From equation d: x = −8 + 2z g
Substitute f and g into e to obtain:
4(−8 + 2z) − 2(−7 + 4z) + 18 = 0
−32 + 8z + 14 − 8z + 18 = 0
0z = 0
This equation is satisfied by all real values of z, so let z = s, where t is any real number.
Substitute s for z in equations f and g to obtain: x = −8 + 2s, and y = −7 + 4s.
These are the parametric equations of a line. The corresponding symmetric equations are:
x+8 y+7 z
2 = 4 =1
Observe that these equations represent the same line as the symmetric equations at the
bottom of page 126.
29. If m1, m2, and m3 are the direction cosines of the line, then m12 + m22 + m32 = 1.
Let P(x, y, z) be any point on the line. The distance from P to A(a1, a2, a3) is:
AP =
( x − a1 ) 2 + ( y − a 2 ) 2 + ( z − a3 ) 2
=
(tm1 ) 2 + (tm 2 ) 2 + (tm3 ) 2
=
t 2 m1 + t 2 m1 + t 2 m3
=
t 2 (m1 + m2 + m3 )
2
2
2
2
2
2
= t 2 (1) , since m12 + m22 + m32 = 1
= |t|
Hence, |t| is the distance from point P(x, y, z) to point A(a1, a2, a3).
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10. Answers may vary.
a) Two vectors in the plane are:
AB = [5 − 1, 1 − 2, 0 + 3] = [4, −1, 3]
AC = [3 − 1, 2 − 2, −6 + 3] = [2, 0, −3]
Use point A(1, 2, −3) to obtain the vector equation:
[x, y, z] = [1, 2, −3] + s[4, −1, 3] + t[2, 0, −3]
b) Two vectors in the plane are:
PQ = [4 − 8, −3 − (−4), 1 − 2]
= [−4, 1, −1]
PR = [−2 − 8, 6 − (−4), 2 − 2]
= [−10, 10, 0]
A simpler direction vector is [−1, 1, 0].
Use point R(−2, 6, 2) to obtain the vector equation:
[x, y, z] = [−2, 6, 2] + s[−4, 1, −1] + t[−1, 1, 0]
11. Answers may vary.
a) Two vectors in the plane are:
AB = [0 − 7, −4 − (−3), 3 − 1]
AC = [1 − 7, −1 − (−3), 0 − 1]
= [−7, −1, 2]
= [−6, 2, −1]
Use point A(7, −3, 1) to obtain the parametric equations:
x = 7 − 7s − 6t, y = −3 − s + 2t, z = 1 + 2s − t
b) Two vectors in the plane are:
PQ = [3 − (−2), −3 − 6, 1 − 1]
PR = [2 − (−2), 5 − 6, −5 − 1]
= [5, −9, 0]
= [4, −1, −6]
Use point P(−2, 6, 1) to obtain the parametric equations:
x = −2 + 5s + 4t, y = 6 − 9s − t, z = 1 − 6t
12. a) The equation of the plane is of the form 4x − 2y + z + D = 0.
Since R(−3, 1, 2) lies on the plane:
4(−3) − 2(1) + 2 + D = 0
D = 12
The equation of the plane is 4x −2y + z + 12 = 0.
b) The equation of the plane is of the form x − y + 4z + D = 0.
Since R(5, 0, −3) lies on the plane:
5 − 0 + 4(−3) + D = 0
D=7
The equation of the plane is x − y + 4z + 7 = 0.
13. The equation of the plane is of the form 3x − y + 2z + D = 0.
a) Since (0, 0, 0) lies on the plane, D = 0.
The equation of the plane is 3x − y + 2z = 0.
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b) Since (1, 2, 3) lies on the plane:
3(1) − 2 + 2(3) + D = 0
D = −7
The equation of the plane is 3x − y + 2z − 7 = 0.
c) Since (−1, 0, 1) lies on the plane:
3(−1) − 0 + 2(1) + D = 0
D=1
The equation of the plane is 3x − y + 2z + 1 = 0.
14. a) The normal vector is [3, 2, 0].
b) Since the coefficient of z is 0, the plane does not have a
z-intercept. Verify this by setting x and y to 0 in the
equation 3x+ 2y + 0z − 12 = 0. This gives 0z − 12 = 0, or
0z = 12. This equation has no solution so the plane does
not have a z-intercept. Since the plane does not
intersect the z-axis, it must be parallel to it.
c) The plane 3x + 2y − 12 = 0 has x-intercept (4, 0, 0) and
y-intercept (0, 6, 0). It intersects the xy-plane in the line
whose equation in R2 is 3x + 2y − 12 = 0.
15. a) Two vectors in the plane are:
AB = [0 − 1, 2 − 1, 3 − 1]
AC = [−1 − 1, 0 − 1, 1 − 1]
= [−1, 1, 2]
= [−2, −1, 0]
The cross product of these vectors is normal to the plane.
1 2 −1 1
G
n = [−1, 1, 2] × [−2, −1, 0]
−1 0 − 2 −1
= [0 − (−2), −4 − 0, 1 − (−2)]
= [2, −4, 3]
The equation of the plane is of the form:
2x − 4y + 3z + D = 0
Since A(1, 1, 1) lies on the plane:
2(1) −4(1) + 3(1) + D = 0
2−4+3+D=0
D = −1
The equation of the plane is 2x − 4y + 3z − 1 = 0.
b) Two vectors in the plane are:
DE = [1 − 0, 2 − 1, 1 − 2]
DF = [−1 − 0, −1 − 1, 2 − 2]
= [1, 1, −1]
= [−1, −2, 0]
The cross product of these vectors is normal to the plane.
1 −1 1
1
G
n = [1, 1, −1] × [−1, −2, 0]
− 2 0 −1 − 2
= [0 − 2, 1 − 0, −2 − (−1)]
= [−2, 1, −1]
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The vector [2, −1, 1] is also normal to the plane, so the equation of the plane is of the
form:
2x − y + z + D = 0
Since (0, 1, 2) lies on the plane:
2(0) − 1 + 2 + D = 0
−1 + 2 + D = 0
D = −1
The equation of the plane is 2x − y + z − 1 = 0.
c) Two vectors in the plane are:
RS = [0 − 3, 5 − 5, −1 − 2]
RT = [1 − 3, 5 − 5, −3 − 2]
= [−3, 0, −3]
= [−2, 0, −5]
A simpler direction vector is [1, 0, 1].
The cross product of these vectors is normal to the plane.
0
1
1 0
G
n = [1, 0, 1] × [−2, 0, −5]
0 −5 −2 0
= [0, −2 − (−5), 0]
= [0, 3, 0]
The vector [0, 1, 0] is also normal to the plane so the equation of the plane is of the form:
y+D=0
Since R(3, 5, 2) lies on the plane:
5+D=0
D = −5
The equation of the plane is y − 5 = 0.
16. Answers may vary.
There are infinitely many planes that pass through three collinear points. If we try to make a
scalar equation from these points, all sets of direction vectors will be parallel. Since parallel
G
vectors have a cross product of 0 , the normal vector will be [0, 0, 0] and a scalar equation
cannot be formed.
For example, from Example 1 page 137, the points A(3, −4, 1), B(5, −5, 4), C(7, −6, 7) lie on
x−3 y+4 z−1
the line 2 =
= 3 and hence are collinear.
−1
AB = [5 − 3, −5 − (−4,) 4 − 1]
= [2, −1, 3]
AC = [7 − 3, −6 − (−4), 7 − 1]
= [4, −2, 6]
−1 3 2 −1
−2 6 4 −2
= [−6 − (−6), 12 − 12, −4 − (−4)]
= [0, 0, 0]
G
n = AB × AC = [2, −1, 3] × [4, −2, 6]
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17. Answers may vary.
AB = [3 − 1, − 2 − 0, 0 − 1]
AC = [2 − 1, 1 − 0, 5 − 1]
= [2, −2, −1]
= [1, 1, 4]
BC = [2 − 3, 1 − (−2), 5 − 0]
= [−1, 3, 5]
a) Two vector equations of the plane containing A, B, and C are:
[x, y, z] = [1, 0, 1] + s[2, −2, −1] + t[1, 1, 4]
[x, y, z] = [3, −2, 0] + s[1, 1, 4] + t[−1, 3, 5]
b) Two parametric equations of the plane containing A, B, and C are:
x = 1 + 2s + t, y = −2s + t, z = 1 − s + 4t
x = 3 + s − t, y = −2 + s + 3t, z = 4s + 5t
c) The cross product of AB and AC is normal to the plane.
− 2 −1 2 − 2
G
n = [2, −2, −1] × [1, 1, 4]
1 4 1
1
= [−8 − (−1), −1 − 8, 2 − (−2)]
= [−7, −9, 4]
The vector [7, 9, −4] is also normal to the plane so the equation of the plane is of the
form:
7x + 9y − 4z + D = 0
Since A(1, 0, 1) lies on the plane:
7(1) + 9(0) − 4(1) + D = 0
7 −4+D=0
D = −3
The equation of the plane is 7x + 9y − 4z − 3 = 0.
18. a) The angle of intersection of two planes is the acute angle between their normal vectors.
G
b) i) The plane 2x + y − 3z + 7 = 0 has normal n1 = [2, 1, −3].
G
The plane 4x − y + 7z + 5 = 0 has normal n2 = [4, −1, 7].
G G
n1 ⋅ n2
cosθ = G G
n1 n2
[2,1, − 3] ⋅ [4, − 1, 7]
=
2 2 + 12 + (−3) 2 4 2 + (−1) 2 + 7 2
2(4) + 1(−1) − 3(7)
=
14 66
−14
=
924
θ Ñ 117.4°
The angle of intersection is 180° − 117.4° = 62.6°.
G
ii) The plane 2x − y − 2z + 5 = 0 has normal n1 = [2, −1, −2].
G
The plane 3x + 4z + 6 = 0 has normal n2 = [3, 0, 4].
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G G
n ⋅n
cosθ = G1 G 2
n1 n2
=
=
[2, − 1, − 2] ⋅ [3, 0, 4]
2 2 + (−1) 2 + (−2) 2 3 2 + 0 2 + 4 2
2(3) − 1(0) − 2(4)
9 25
−2
= 15
θ Ñ 97.7°
The angle of intersection is 180° − 97.7° = 82.3°.
19. Let P(x, y, z) be a point on the plane.
Since P is equidistant from A and B, AP = BP .
(x −1)2 + (y − 1)2 + z2 = (x −5)2 + (y − 3)2 + (z + 2)2
Square both sides and simplify.
x2 − 2x + 1 + y2 − 2y + 1 + z2 = x2 − 10x + 25 + y2 − 6y + 9 + z2 + 4z + 4
8x + 4y − 4z − 36 = 0
2x + y − z − 9 = 0
The equation of the plane is 2x + y − z − 9 = 0.
20. a) Draw a diagram and label it as shown.
Use points A, B, and C. A is the midpoint of edge
U(2, −2, 2) T(2, −2, −2), so its coordinates are A(2, −2, 0).
B is the midpoint of edge U(2, −2, 2) P(−2, −2, 2), so
its coordinates are B(0, −2, 2). C is the midpoint of edge
P(−2, −2, 2) Q(−2, 2, 2), so its coordinates are C(−2, 0, 2).
Two direction vectors for the plane are:
AB = [0 − 2, −2 − (−2), 2 − 0]
AC = [−2 − 2, 0 − (−2), 2 − 0]
= [−2, 0, 2]
= [−4, 2, 2]
The vectors [−1, 0, 1] and [−2, 1, 1] are also direction vectors
for the plane. The cross product of these vectors is normal to the plane.
0 1
−1
0
G
n = [−1, 0, 1] × [−2, 1, 1]
1 1 −2
1
= [0 − 1, −2 − (−1), −1 − 0]
= [−1, −1, −1]
The vector [1, 1, 1] is also normal to the plane so the equation of the plane is of the
form:
x+y+z+D=0
Since A(2, −2, 0) lies on the plane:
2−2+0+D=0
D=0
The equation of the plane is x + y + z = 0.
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b) The required midpoints are the midpoints of edges PW, TW, and WR.
The midpoint J of edge P(−2, −2, 2)W(−2, −2, −2) has coordinates J(−2, −2, 0).
The midpoint K of edge T(2, −2, −2) W(−2, −2, −2) has coordinates K(0, −2, −2).
The midpoint L of edge W(−2, −2, −2) R(−2, 2, −2) has coordinates L(−2, 0, −2).
Two direction vectors for the plane through these 3 points is:
JK = [0 − (−2), −2 − (−2), −2 − 0]
JL = [−2 − (−2), 0 − (−2), −2 − 0]
= [2, 0, −2]
= [0, 2, −2]
The vectors [1, 0, −1] and [0, 1, −1] are also direction vectors for the plane. The cross
product of these vectors is normal to the plane.
0 −1 1 0
G
n = [1, 0, −1] × [0, 1, −1]
1 −1 0 1
= [0 − (−1), 0 − (−1), 1 − 0]
= [1, 1, 1]
The equation of the plane is of the form:
x+y+z+D=0
Since J(−2, −2, 0) lies on the plane:
−2 − 2 + 0 + D = 0
D=4
The equation of the plane is x + y + z + 4 = 0.
c) The required midpoints are the midpoints of edges UV, QV, and VS.
The midpoint G of edge U(2, −2, 2) V(2, 2, 2) has coordinates G(2, 0, 2).
The midpoint H of edge Q(−2, 2, 2) V(2, 2, 2) has coordinates H(0, 2, 2).
The midpoint I of edge V(2, 2, 2) S(2, 2, −2) has coordinates I(2, 2, 0).
Two direction vectors for the plane through these 3 points is:
GH = [0 − 2, 2 − 0, 2 − 2]
GI = [2 − 2, 2 − 0, 0 − 2]
= [−2, 2, 0]
= [0, 2, −2]
The vectors [−1, 1, 0] and [0, 1, −1] are also direction vectors for the plane. The cross
product of these vectors is normal to the plane.
1
0 −1 1
G
n = [−1, 1, 0] × [0, 1, −1]
1 −1
0 1
= [−1 − 0, 0 − 1, −1 − 0]
= [−1, −1, −1]
The vector [1, 1, 1] is also normal to the plane so the equation of the plane is of the form:
x+y+z+D=0
Since G(2, 0, 2) lies on the plane:
2+0+2+D=0
D = −4
The equation of the plane is x + y + z − 4 = 0.
d) The planes are parallel.
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21. To show that the points are coplanar, find the equation of the plane through A, B, and C.
Then show that D also lies on this plane.
Two direction vectors for the plane are:
AB = [−2 − 1, −4 − 6, −1 − 3]
AC = [3 − 1, 9 − 6, 4 − 3]
= [−3, −10, −4]
= [2, 3, 1]
The cross product of these vectors is normal to the plane.
− 10 − 4 − 3 − 10
G
n = [−3, −10, −4] × [2, 3, 1]
3
1
2
3
= [−10 − (−12), −8 − (−3), −9 − (−20)]
= [2, −5, 11]
Therefore, the equation of the plane is of the form:
2x − 5y + 11z + D = 0
Since A(1, 6, 3) lies on the plane:
2(1) − 5(6) + 11(3) + D = 0
2 − 30 + 33 + D = 0
D = −5
The equation of the plane is 2x − 5y + 11z − 5 = 0.
If (−3, 0, 1) also lies in the plane, its coordinates must satisfy the equation of the plane.
LS = 2(−3) − 5(0) + 11(1) − 5
= −6 + 11 − 5
=0
= RS
Thus D lies on the plane determined by A, B, and C. The four points are coplanar.
22. A direction vector for the plane is:
AB = [2 − 1, 3 − 2, −1 − 3]
= [1, 1, −4]
Hence the normal to the plane is perpendicular to AB .
Since the plane is perpendicular to 3x + y + z + 1 = 0, its normal is also perpendicular to
[3, 1, 1].
1 −4
1 1
G
∴ n = [1, 1, −4] × [3, 1, 1]
1
1 3 1
= [1 − (−4), −12 − 1, 1 − 3]
= [5, −13, −2]
The equation of the plane is of the form:
5x − 13y − 2z + D = 0
Since A(1, 2, 3) lies on the plane:
5(1) − 13(2) − 2(3) + D = 0
5 − 26 − 6 + D = 0
D = 27
The equation of the plane is 5x − 13y − 2z + 27 = 0.
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23. The parametric equations of the plane are:
x = −2 + 2s + t
c
y = 5 + 4s + 4t
d
z = 3 + s + 2t
e
Eliminate s
c: x = −2 + 2s + t
−2 × e: −2z = −6 − 2s − 4t
Add: x − 2z = −8 − 3t
−2 × c: −2x = 4 − 4s − 2t
d:
y = 5 + 4s + 4t
−2x + y = 9 + 2t
Solve for t.
x − 2z + 8
t=
f
−3
Solve for t.
−2x + y − 9
t=
g
2
Equate the expressions for t from equations f and g.
x − 2z + 8 −2x + y − 9
=
2
−3
2x − 4z + 16 = 6x − 3y + 27
4x − 3y + 4z + 11 = 0
This is the scalar equation from page 149.
24. Let P(x, y, z) be a point on the locus.
Since P is equidistant from A and B, AP = BP .
(x −1)2 + (y − 2)2 + (z − 3)2 = (x −3)2 + (y − 2)2 + (z − 4)2
Square both sides and simplify.
x2 − 2x + 1 + y2 − 4y + 4 + z2 − 6z + 9 = x2 − 6x + 9 + y2 − 4y + 4 + z2 − 8z + 16
4x + 2z − 15 = 0
25. a) Let P have coordinates (x, y, z).
Since AP ⋅ OA = 0,
[x− 4, y + 1, z − 3] [4, −1, 3] = 0
4x − 16 − y − 1 + 3z − 9 = 0
4x − y + 3z − 26 = 0
b) The locus is a plane though the point A(4, −1, 3) with normal [4, −1, 3].
26. a) In the scalar equation of a plane, Ax + By + Cz + D = 0, the vector [A, B, C] is a normal
vector for the plane. Similarly in the general equation of a line, Ax + By + C = 0, the vector
[A, B] is a normal vector for the line. For example, consider the line 2x − 3y + 14 = 0 from
pages 124 and 125. The vector [2, −3] is normal to the line since it is perpendicular to
[3, 2], the direction vector of the line.
[2, −3]⋅[3, 2] = 6 − 6
=0
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b) The equations are called linear equations because the exponent of each variable is 1.
27.
Use the diagram above. If P is any point on the plane, then by the Triangle Law:
OP = OA + AP
= OA + s AB + t AC
= OA + s (OB − OA) + t (OC − OA)
= (1 − s − t )OA + s OB + t OC
G G
G
G
p = (1 − s − t )a + sb + tc
G G
G
G
G G
G
The coefficients of a , b , and c add to 1. If we let 1 − s − t = r, then we ave p = ra + sb + tc ,
where r + s + t = 1.
28. a) Use the diagram below.
The plane passes through the midpoints of edges PQ, QV, VS, ST, TW, and WP.
The midpoint of edge PQ is (−2, 0, 2).
The midpoint of edge QV is (0, 2, 2).
The midpoint of edge VS is (2, 2, 0).
Hence, two direction vectors for the plane are [2, 2, 0] and [2, 0, −2], or more simply,
[1, 1, 0] and [1, 0, −1].
Therefore, a normal vector for the plane is:
1
0 1 1
G
n = [1, 1, 0] × [1, 0, −1]
0 −1 1 0
= [−1 − 0, 0 − (−1), 0 − 1]
= [−1, 1, −1]
The vector [1, −1, 1] is also normal to the plane. Hence, the equation of the plane is of the
form:
x−y+z+D=0
Since (−2, 0, 2) lies on the plane:
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−2 − 0 + 2 + D = 0
D=0
The equation of the plane is x − y + z = 0.
b) There are 4 planes that pass through the midpoints of 6 of the 12 edges of the cube.
Call these planes π1, π2, π3, and π4.
π1: The plane through the midpoints of edges PQ, QR, RS, ST, TU, and UP.
From exercise 20a, the equation of this plane is x + y + z = 0.
π2: The plane through the midpoints of edges PQ, QV, VS, ST, TW, and WP.
From part a, the equation of this plane is x − y + z = 0.
π3: The plane through the midpoints of edges UV, VQ, RQ, RW, WT, and TU.
Using the method of part a, the equation of this plane is x + y − z = 0.
π4: The plane through the midpoints of edges PU, UV, VS, SR, RW, and WP.
Using the method of part a, the equation of this plane is x − y − z = 0.
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3. L1 has parametric equations x = 4 + 5t, y = 3t, z = −2 − 2t.
L2 has parametric equations x = 5 + t, y = −1 − 2t, z = 4 + 3t.
L3 has parametric equations x = 2 + 2t, y = 2 − 4t, z = −1 + 7t.
Substitute the parametric equations of each line into the equation of the plane and solve for t.
L1: 4 + 5t − 3(3t) − 2(−2 − 2t) + 2 = 0
4 + 5t − 9t + 4 + 4t + 2 = 0
0t = −10
This equation has no solution so the line and plane do not intersect.
L2: 5 + t − 3(−1 − 2t) − 2(4 + 3t) + 2 = 0
5 + t + 3 + 6t − 8 − 6t + 2 = 0
t = −2
Substitute this value of t into the parametric equations of the line.
x = 5 + (−2)
y = −1 − 2(−2)
z = 4 + 3(−2)
=3
=3
= −2
The point of intersection is (3, 3, −2).
L3: 2 + 2t − 3(2 − 4t) − 2(−1 + 7t) + 2 = 0
2 + 2t − 6 + 12t + 2 − 14t + 2 = 0
0t = 0
This equation is satisfied by all values of t. Therefore, every point on the line lies
on the plane.
a) L2 intersects the plane in one point. Since the equation t = −2 is satisfied by only value of t,
only the point (3, 3, −2) lies on the plane.
b) L3 lies on the plane. Since the equation 0t = 0 is satisfied by all values of t, all points on
the line lie on the plane.
c) L1 is parallel to the plane. Since the equation 0t = −10 is not satisfied by any value of t,
there is no point on the line that lies on the plane.
4. Since the plane is perpendicular to the line, the direction vector of the line is a normal for the
plane.
G
a) A normal for the plane is n = [1, 3, −1] so the equation of the plane is of the form:
x + 3y − z + D = 0
Since A(2, 1, −1) lies on the plane:
2 + 3(1) −(−1) + D = 0
D = −6
The equation of the plane is x + 3y − z − 6 = 0.
G
b) A normal for the plane is n = [3, −1, 0] so the equation of the plane is of the form:
3x − y + D = 0
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Since A(2, 1, −1) lies on the plane:
3(2) − 1 + D = 0
D = −5
The equation of the plane is 3x − y − 5 = 0.
5. Since the lines are parallel to the plane, the cross product of their direction vectors is normal to
the plane.
2
−3
1 2
G
n = [1, 2, −3] × [2, 1, 0]
1
0 2 1
= [0 − (−3), −6 − 0, 1 − 4]
= [3, −6, −3]
The vector [1, −2, −1] is also normal to the plane so the equation of the plane is of the form:
x − 2y − z + D = 0
Since A(5, −3, 6) lies on the plane:
5 − 2(−3) − 6 + D = 0
D = −5
The equation of the plane is x − 2y − z − 5 = 0.
6. a) The parametric equations of the line are x = −1 − 4t, y = 2 + 3t, z = 1 − 2t. At a point of
intersection, these values of x, y, and z satisfy the equation of the plane.
−1 − 4t + 2(2 + 3t) − 3(1 − 2t) + 10 = 0
−1 − 4t + 4 + 6t − 3 + 6t + 10 = 0
8t = −10
5
t = −4
Substitute this value of t into the parametric equations of the line to obtain the point of
intersection.
 5
 5
 5
x = − 1 − 4 − 
y = 2 + 3 − 
z = 1 − 2 − 
 4
 4
 4
7
7
=4
=−4
=2
7 7

The point of intersection is  4,− ,  .
4 2

b) The parametric equations of the line are x = −3 + 4t, y = 1 + t, z = 5 − 2t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
−3 + 4t + 2(1 + t) + 3(5 − 2t) − 5 = 0
−3 + 4t + 2 + 2t + 15 − 6t − 5 = 0
0t = −9
This equation has no solution. The line and plane do not intersect.
c) The parametric equations of the line are x = 4 + 2t, y = t, z = −1 − t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
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3(4 + 2t) − 2(t) + 4(−1 − t) − 8 = 0
12 + 6t − 2t − 4 − 4t − 8 = 0
0t = 0
This equation is satisfied by all values of t so the line lies on the plane and all points on
the line are points of intersection with the plane.
d) The parametric equations of the line are x = 3 + 4t, y = 2 + 3t, z = −1 − t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
4(3 + 4t) − (−1 − t) + 5 = 0
12 + 16t + 1 + t + 5 = 0
17t = −18
18
t = −17
Substitute this value of t into the parametric equations of the line to obtain the point of
intersection.
 18 
 18 
 18 
x = 3 + 4 − 
y = 2 + 3 − 
z = −1− − 
 17 
 17 
 17 
21
20
1
= − 17
= − 17
= 17
 21 20 1 
The point of intersection is  − ,− ,  .
 17 17 17 
e) The parametric equations of the line are x = t, y = 3 + 7t, z = 1 + 4t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
t − 3(3 + 7t) + 5(1 + 4t) + 4 = 0
t − 9 − 21t + 5 + 20t + 4 = 0
0t = 0
This equation is satisfied by all values of t so the line lies on the plane and all points on
the line are points of intersection with the plane.
f) The parametric equations of the line are x = 2, y = t, z = 5 + 2t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
3(2) − 4(t)+ 2(5 +2t) + 16 = 0
6 − 4t + 10 + 4t + 16 = 0
0t = −32
This equation has no solution. The line and plane do not intersect.
g) The parametric equations of the line are x = 4 + t, y = 1 + 2t, z = 5 + 3t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
5(4 + t) + 3(1 + 2t) + 4(5 + 3t) − 20 = 0
20 + 5t + 3 + 6t + 20 + 12t − 20 = 0
23t = −23
t = −1
Substitute this value of t into the parametric equations of the line to obtain the point of
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intersection.
x=4−1
y =1 + 2(−1)
=3
=−1
The point of intersection is (3, −1, 2).
z = 5 + 3(−1)
=2
7. To show that the two lines form a plane, show that they intersect.
The parametric equations of L1 are x = −5 + 3t, y = 2 + 2t, z = −7 + 6t.
The parametric equations of L2 are x = s, y = −6 − 5s, z = −3 − s.
At a point of intersection, the x-, y-, and z-coordinates of the two lines are the same.
→
s − 3t = −5 c
−5 + 3t = s
→ 5s + 2t = −8 d
2 + 2t = −6 − 5s
→
s + 6t = 4
e
−7 + 6t = −3 − s
c − e: −9t = −9
t=1
Substitute this value of t in equation c and solve for s.
s − 3(1) = −5
s = −2
Verify these values of t and s by substituting them in equation d.
LS = 5(−2) + 2(1)
= −8
= RS
Since the system of equations is consistent, the lines intersect and hence form a plane.
8. At a point of intersection, the x-, y-, and z-coordinates of the line and plane are the same.
→
s − 2t + k = −5 c
4 + k = −1 − s + 2t
→
s − 4t − 2k = −1 d
2 − 2k = 1 − s + 4t
→
3s + t − 3k = 4
e
6 + 3k = 2 + 3s + t
Eliminate s from equations c and d.
c − d: 2t + 3k = −4 f
Eliminate s from equations c and e.
−3 × c: −3s + 6t − 3k = 15
e: 3s + t − 3k = 4
Add:
7t − 6k = 19 g
Eliminate t from equations f and g.
−7 × f: −14t − 21k = 28
2 × g: 14t − 12k = 38
Add:
−33k = 66
k = −2
Substitute this value of k in equation f and solve for t.
2t + 3(−2) = −4
2t = 2
t=1
Substitute these values of t and k in equation c and solve for s.
s − 2(1) − 2 = −5
s − 4 = −5
s = −1
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To obtain the point of intersection, substitute k = −2 in the equation of the line or s = −1 and
t = 1 in the equation of the plane. Substituting k = −2 in the equation of the line gives:
x=4−2
y = 2 − 2(−2)
z = 6 + 3(−2)
=2
=6
=0
The point of intersection is (2, 6, 0).
9. Three points on the plane are A(6, 0, 0), B(0, −12, 0), and C(0, 0, 4), so three direction vectors
for the plane are:
AB = [0 − 6, −12 − 0, 0 − 0]
AC = [0 − 6, 0 − 0, 4 − 0]
= [−6, −12, 0]
= [−6, 0, 4]
BC = [0 − 0, 0 − (−12), 4 − 0]
= [0, 12, 4]
G
G
G
The vectors m1 = [1, 2, 0], m2 = [3, 0, −2], and m3 = [0, 3, 1] are simpler direction vectors for
the plane.
G G
G
a) Use point A and direction vectors m1 , m2 , and m3 to obtain the following equations.
x = 6 + t, y = 2t, z = 0
x = 6 + 3t, y = 0, z = −2t
x = 6, y = 3t, z = t.
G
b) Use direction vector m1 and points A, B, and C to obtain the following equations:
x = 6 + t, y = 2t, z = 0
x = t, y = −12 + 2t, z = 0
x = t, y = 2t, z = 4
10. The angle of intersection of a line and a plane
is the complement of the acute angle
G
between m , the direction vector of the line, and
G
n , the normal to the plane. For example, if α is
G
G
the angle between m and n , then
θ = 90° − α is the angle between the line and
the plane.
G G
m⋅n
a) cos α = G G
mn
=
[2,1, − 1] ⋅ [1, 2, 3]
2 2 + 12 + (−1) 2 12 + 2 2 + 3 2
2(1) + 1(2) − 1(3)
6 14
1
=
84
α Ñ 83.7°
Therefore, the angle between the line and the plane is 90° − 83.7° = 6.3°.
=
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G
G
G
G
b) By inspection, m = − n , so the angle between m and n is 180°. Hence the line is
perpendicular to the plane and the angle between the line and plane is 90°.
11. To obtain the coordinates of P′, determine the point of intersection of the line PP′ and the
plane.
a) The vector [2, −1, 1] is normal to the plane so it is a direction vector for PP′. Hence
parametric equations of PP′ are x = –3 + 2t, y = 1 − t, z = 1 + t. Substituting these equations
into the equation of the plane gives:
2(−3 + 2t) − (1 − t) + 1 + t − 5 = 0
−6 + 4t − 1 + t + 1 + t − 5 = 0
6t = 11
11
t= 6
11
To obtain the coordinates of P′, substitute t = 6 into the parametric equation of
PP′.
11
11
 11 
x = − 3 + 2 
y = 1−
z = 1+
6
6
6
4
5
17
=6
= −6
= 6
2
=3
 2 5 17 
The projection of P in the plane is P′  , − ,  .
3 6 6 
b) The vector [1, 3, −1] is normal to the plane so it is a direction vector for PP′. Hence
parametric equations of PP′ are x = 2 + t, y = 3t, z = 3 − t. Substituting these equations into
the equation of the plane gives:
2 + t + 3(3t) − (3 − t) + 7 = 0
2 + t + 9t − 3 + t + 7 = 0
11t = −6
6
t = −11
6
To obtain the coordinates of P′, substitute t = −11 into the parametric equation of
PP′.
6
 6
 6
x = 2−
y = 3 − 
z = 3 − − 
11
 11 
 11 
16
18
39
= 11
= −11
= 11
 16 18 39 
The projection of P in the plane is P′  , − ,  .
 11 11 11 
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12. If a line lies on a plane, then every point on the line lies on the plane. Therefore, A and B must
lie on the plane and their coordinates must satisfy the equation of the plane:
2x + y + 3z − 10 = 0.
A(1, 5, 1): LS = 2(1) + 5 + 3(1) − 10
B(0, 4, 2): LS = 2(0) + 4 + 3(2) − 10
=0
=0
= RS
= RS
Since A and B lie on the plane, the line through A and B also lies on the plane.
13. a) There are 3 possibilities for the intersection of a line and a plane.
Case 1. The line intersects the plane, so there is a single point of intersection.
Case 2. The line lies in the plane, so there are infinitely many points of intersection.
Case 3. The line is parallel to the plane, so there are no points of intersection.
These are illustrated on page 156.
b) To determine how the line intersects the plane, substitute the parametric equations of the
line into the equation of the plane and solve for the parameter. If the resulting equation has
only one solution, the line intersects the plane at a single point. If it has infinitely many
solutions, all points on the line lie on the plane and if it has no solution, the line and plane
do not intersect.
14. The line through P(1, 2, 3) and Q(−1, 3, −2) has direction vector:
PQ = [−1 − 1, 3 − 2, −2 − 3]
= [−2, 1, −5]
and parametric equations:
x = 1 − 2t, y = 2 + t, z = 3 − 5t
At a point of intersection, these values of x, y, and z must satisfy the equation of the plane.
2(1 − 2t) + 3(2 + t) + 2(3 − 5t) − 3 = 0
2 − 4t + 6 + 3t + 6 − 10t − 3 = 0
−11t = −11
t=1
Substitute this value of t into the parametric equations of the line to obtain the point of
intersection.
x = 1 − 2(1)
y=2+1
z = 3 − 5(1)
= −1
=3
= −2
The point of intersection is (−1, 3, −2).
15. The parametric equations of L1 are x = 3t, y = 2 + t, z = 1 + t.
The parametric equations of L2 are x = 1 + 2s, y = −3 − s, z = s.
a) Skew lines are non-parallel, non-intersecting lines.
By inspection, the direction vectors [3, 1, 1] and [2, −1, 1] are not parallel so the lines are
not parallel.
At a point of intersection, the x-, y-, and z-coordinates of the two lines are the same.
3t = 1 + 2s →
2s − 3t = −1 c
2 + t = −3 − s →
s + t = −5 d
1+t=s
→
s−t=1 e
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d + e: 2s = −4
s = −2
Substitute s = −2 in equation d and solve for t.
− 2 + t = −5
t = −3
Verify these values of t and s by substituting them in equation c.
LS = 2(−2) − 3(−3)
=5
≠ RS
Since the system of equations is inconsistent, the lines do not intersect. Hence they are
skew lines.
b) Since the planes that contain L1 and L2 are parallel, they have the same normal vector.
The normal is perpendicular to L1 and L2 so it is the cross product of their direction
vectors.
1 1
3
1
G
n = [3, 1, 1] × [2, −1, 1]
−1 1
2 −1
= [1 − (−1), 2 − 3, −3 − 2]
= [2, −1, −5]
The equations of the planes containing L1 and L2 are of the form:
2x − y − 5z + D = 0.
The point (0, 2, 1) lies on L1 so it also lies on the plane containing L1. Therefore:
2(0) − 2 −5(1) + D = 0
D=7
The equation of the plane containing L1 is 2x − y − 5z + 7 = 0.
The point (1, −3, 0) lies on L2 so it also lies on the plane containing L2. Therefore:
2(1) −(−3) −5(0) + D = 0
D = −5
The equation of the plane containing L2 is 2x − y − 5z + 5 = 0.
16. Answers may vary.
If a line lies on a plane, its direction vector is perpendicular to the normal of the plane.
For direction vector [a, b, c] and normal [3, −2, 1]:
[a, b, c] ⋅ [3, −2, 1] = 0
3a − 2b + c = 0
Create 2 non-parallel direction vectors, one for each line.
Let a = 1 and b = 1 to obtain 3 − 2 + c = 0, or c = −1. This gives the direction vector [1, 1, −1].
Let a = 1 and b = 0 to obtain 3 + c = 0, or c = −3. This gives the direction vector [1, 0, −3].
Plane π1 has z-intercept (0, 0, 8). Use this point and direction vector [1, 1, −1] to obtain the
following line on π1: [x, y, z] = [0, 0, 8] + t[1, 1, −1].
Plane π2 has z-intercept (0, 0, −4). Use this point and direction vector [1, 0, −3] to obtain a line
on π2 that is skew to the line on π1: [x, y, z] = [0, 0, −4] + t[1, 0, −3].
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17. Answers may vary.
a) Let L2 be the required line. Since L1 and L2 cannot be parallel, the direction vector of L2
cannot be a scalar multiple of [2, −3, 1]. Choose any other direction vector, say [2, 3, 1].
Points on L1 have coordinates of the form (1 + 2t, −2 − 3t, −1 + t). L1 and L2 cannot
intersect so choose a point not on L1. Let t = 1 for the x- and y-coordinates and t = 0 for
the z –coordinate to obtain a point (1 + 2(1), −2 − 3(1), −1 + 0) or (3, −5, −1) not on L1.
x−3 y+5 z+1
Thus L2: 2 = 3 = 1 is skew to L1.
b) By inspection, [2, −3, 1] and [2, 3, 1] are not collinear so the lines are not parallel.
At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
1 + 2t = 3 + 2s
→
2t − 2s = 2, or t − s = 1 c
−2 − 3t = −5 + 3s
→
3t + 3s = 3, or t + s = 1 d
−1 + t = −1 + s
→
t−s=0 e
Use equations c and d.
c + d: 2t = 2
c − d: −2s = 0
t=1
s=0
Verify the values of t and s by substituting them in equation e.
LS = 1 − 0
=1
≠ RS
Since the system of equations is inconsistent, the lines do not intersect.
Hence L1 and L2 are skew lines.
c) Since the planes that contain L1 and L2 are parallel, they have the same normal vector.
The normal is perpendicular to L1 and L2 so it is the cross product of their direction
vectors.
−3
1
2 −3
G
n = [2, −3, 1] × [2, 3, 1]
3
1
2
3
= [−3 − 3, 2 − 2, 6 − (−6)]
= [−6, 0, 12]
The vector [1, 0, −2] is also normal to the planes so the equations of the planes containing
L1 and L2 are of the form:
x − 2z + D = 0.
The point (1, −2, −1) lies on L1 so it also lies on the plane containing L1. Therefore:
1 −2(−1) + D = 0
D = −3
The equation of the plane containing L1 is x − 2z − 3 = 0.
The point (3, −5, −1) lies on L2 so it also lies on the plane containing L2. Therefore:
3 −2(−1) + D = 0
D = −5
The equation of the plane containing L2 is x − 2z − 5 = 0.
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18. The lines are not parallel since their direction vectors [3, 5, 1] and [1, 7, 2] are not collinear.
Therefore, the lines must intersect if they are coplanar.
Parametric equations of L1 are x = 2 + 3t, y = 3 + 5t, z = 1 + t.
Parametric equations of L2 are x = 4 + s, y = 1 + 7s, z = 2s.
At a point of intersection, the x-, y-, and z-coordinates of the two lines are the same.
2 + 3t = 4 + s
→
s − 3t = −2 c
3 + 5t = 1 + 7s → 7s − 5t = 2
d
1 + t = 2s
→
2s − t = 1
e
Use equations c and e.
−2 × c: −2s + 6t = 4
3:
2s − t = 1
Add:
5t = 5
t=1
Substitute t = 1 in equation c and solve for s.
s − 3(1) = −2
s =1
Verify the values of t and s by substituting them in equation d.
LS = 1 − 3(1)
= −2
= RS
The system of equations is consistent so the lines intersect and hence are coplanar.
19.
A is a point on L1 and B is a point on L2 such that line segment AB is perpendicular to both L1
and L2. A lies on L1 so its coordinates are of the form A(3t, 2 + t, 1 + t). Similarly, B’s
coordinates are of the form B(1 + 2s, −3 − s, s).
Thus AB = [1 + 2s − 3t, −3 − s − (2 + t), s − (1 + t)]
= [1 + 2s − 3t, −5 −s − t, −1 + s − t]
Since line segment AB is perpendicular to L1 and L2, AB is collinear to the cross product of
their direction vectors. From Example 2, the cross product is [2, −1, −5], so:
AB = k[2, −1, −5]
c
[1 + 2s − 3t, −5 −s − t, −1 + s − t] = k[2, −1, −5]
The vectors are equal so their components are equal.
→
2s − 3t − 2k = −1 d
1 + 2s − 3t = 2k
−5 − s − t = −k
→
s + t − k = −5 e
−1 + s − t = −5k →
s − t + 5k = 1
f
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Solve the system of equations for k.
e + f: 2s + 4k = −4
s + 2k = −2
g
d: 2s − 3t − 2k = −1
3 × e: 3s + 3t − 3k = −15
Add: 5s
− 5k = −16 h
−5 × g: −5s − 10k = 10
h:
5s − 5k = −16
Add:
−15k = −6
6
2
k = 15 , or 5
2
Substitute k = 5 in c.
2
AB = 5 [2, −1, −5]
4 2

=  , − , −2 
5 5

2
2
2
 4   2   10 
AB =   +  −  +  − 
5  5  5 
120
= 5
2 30
= 5
2 30
The distance between the 2 lines is 5 units.
20. a)
Determine the equation of L, the line through A perpendicular to the plane. Then find the
coordinates of the B, the point of intersection of the line and plane. The length of line
segment AB is the distance from the point to the plane.
L is perpendicular to the plane, so the normal vector [2, 1, −2] is a direction vector for L.
Since L also passes through A(2, 3, −1), the parametric equations of L are:
x = 2 + 2t
y=3+t
z = −1 − 2t.
Determine the coordinates of B by substituting these equations in the equation of the plane.
2(2 + 2t) + 3 + t −2(−1 − 2t) + 9 = 0
4 + 4t + 3 + t + 2 + 4t + 9 = 0
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9t = −18
t = −2
Substituting t = −2 in the parametric equations of L gives:
x = 2 + 2(−2)
y=3−2
z = −1 − 2(−2)
= −2
=1
=3
B has coordinates (−2, 1, 3).
Therefore, AB = [−2 − 2, 1 − 3, 3 − (−1)]
= [−4, −2, 4]
AB =
(−4)2 + (−2)2 + 42
= 36
=6
Therefore, A is 6 units from the plane.
b) Use the method outlined in part a.
Since L passes through B(0, −2, 1) and has direction vector [3, −1, 1] (the normal of the
plane), the parametric equations of L are x = 3t, y = −2 − t, z = 1 + t. Substitute these
equations in the equation of the plane to obtain the coordinates of P, the point of
intersection of L and the plane.
3(3t) − (−2 − t) + 1 + t − 2 = 0
9t + 2 + t + 1 + t − 2 = 0
11t = −1
1
t = −11
1
Substituting t = −11 in the parametric equations of L gives:
1
 1
 1
x = 3 − 
y = − 2 − − 
z = 1−
11
 11 
 11 
21
10
3
= −11
= 11
= −11
 3 21 10 
The coordinates of P are  − ,− ,  .
 11 11 11 
10 
 3 21
Therefore, BP = − ,− − (−2), − 1
11 
 11 11
1
 3 1
=  − , ,− 
 11 11 11
2
2
2
 3 1  1
BP =  −  +   +  − 
 11   11   11 
1
=
11
11
The distance from B to the plane is 11 units.
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c) Use the method outlined in part a.
Since L passes through P(x1, y1, z1) and has direction vector [A, B, C] (the normal of the
plane), the parametric equations of L are x = x1 + At, y = y1 + Bt, z = z1 + Ct. Substitute
these equations in the equation of the plane to obtain the coordinates of Q, the point of
intersection of L and the plane.
A(x1 + At) + B(y1 + Bt) + C(z1 + Ct) + D = 0
(A2 + B2 + C2)t = − (Ax1 + By1 + Cz1 + D)
Ax1 + By1 + Cz1 + D
t=−
A2 + B2 + C2
Substituting this value of t in the parametric equations of L gives:
Ax1 + By1 + Cz1 + D 
x = x1 −A

A2 + B2 + C2


Ax
+
By
+
Cz
+
D
1
1
 1

y = y1 −B

A2 + B2 + C2


Ax1 + By1 + Cz1 + D 
z = z1 −C

A2 + B2 + C2


so the coordinates of Q are:
A( Ax1 + By1 + Cz1 + D)
B( Ax1 + By1 + Cz1 + D)
C ( Ax1 + By1 + Cz1 + D) 

Q  x1 −
, y1 −
, z1 −

2
2
2
2
2
2
A + B +C
A + B +C
A2 + B 2 + C 2


Therefore:
 − A( Ax1 + By1 + Cz1 + D) − B( Ax1 + By1 + Cz1 + D) − C ( Ax1 + By1 + Cz1 + D) 
PQ = 
,
,

A2 + B 2 + C 2
A2 + B 2 + C 2
A2 + B 2 + C 2


A 2 ( Ax1 + By1 + Cz1 + D) 2 B 2 ( Ax1 + By1 + Cz1 + D) 2 C 2 ( Ax1 + By1 + Cz1 + D) 2
+
+
( A2 + B 2 + C 2 ) 2
( A2 + B 2 + C 2 ) 2
( A2 + B 2 + C 2 ) 2
PQ =
=
( Ax1 + By1 + Cz1 + D) 2 ( A 2 + B 2 + C 2 )
( A2 + B 2 + C 2 ) 2
( Ax1 + By1 + Cz1 + D) 2
=
A2 + B 2 + C 2
Ax1 + By1 + Cz1 + D
=
A2 + B 2 + C 2
The distance from P to the plane is
Ax1 + By1 + Cz1 + D
A2 + B 2 + C 2
units.
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3.5 Exercises, Sample Solutions
5. Answers may vary.
a) x − 2y + 3z = 6 c
2x + y + z = 7 d
Eliminate z.
c:
x − 2y + 3z = 6
−3 × d: −6x −3y − 3z = −21
Add: −5x −5y
= −15, or x + y = 3
Solve for y: y = 3 − x
Let x = t, where t is any real number.
∴y=3−t
Substitute these expressions for x and y in equation d and solve for z.
2t + 3 − t + z = 7
z=4−t
Parametric equations of the line of intersection are: x = t, y = 3 − t, z = 4 − t.
x y−3 z−4
The corresponding symmetric equations are 1 =
=
.
−1
−1
b)
2x + y + z = 5 c
3x + 2y + 2z = 8 d
Eliminate x.
−3 × c: −6x − 3y − 3z = −15
2 × d: 6x + 4y + 4z = 16
Add:
y + z=1
Solve for z: z = 1 − y
Let y = t, where t is any real number.
∴z=1−t
Substitute these expressions for y and z in equation c and solve for x.
2x + t + 1 − t = 5
2x = 4
x=2
Parametric equations of the line of intersection are: x = 2, y = t, z = 1 − t.
y z−1
.
There are no symmetric equations. Instead we can also write x = 2, 1 =
−1
c) 22x + y + 8z = 20 c
11x + 2y + 5z = 18 d
Eliminate y.
−2 × c: −44x −2y − 16z = −40
d: 11x + 2y + 5z = 18
Add:
−33x −11z = −22
Solve for z.
11z = 22 −33x
z = 2 − 3x
Let x = t, where t is any real number.
∴ z = 2 − 3t
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Substitute these expressions for x and z in equation d and solve for y.
11t + 2y + 5(2 − 3t) = 18
11t + 2y + 10 − 15t = 18
2y = 8 + 4t
y = 4 + 2t
Parametric equations of the line of intersection are: x = t, y = 4 + 2t, z = 2 − 3t.
x y−4 z−2
.
The corresponding symmetric equations are 1 = 2 =
−3
6. Since the line is parallel to the line of intersection of the 2 planes, its direction vector is
perpendicular to the normal vector of each plane. Therefore, the cross product of the normal
vectors is a direction vector for the line.
− 3 −1 4 − 3
[4, −3, −1] × [2, 4, 1] = [−3 − (−4), −2 − 4, 16 − (−6)]
4
1 2
4
= [1, −6, 22]
x−7 y+2 z−4
The line passes through A(7, −2, 4) so its symmetric equations are 1 =
= 22 .
−6
7. The cross product of the normal vectors is a direction vector for the line of intersection of the
2 planes.
−1 0
3
−1
[3, −1, 0] × [4, 0, 3] = [−3 − 0, 0 − 9, 0 − (−4)]
0 3
4
0
= [−3, −9, 4]
Since the line of intersection is perpendicular to the plane through A(3, −1, 2), its direction
vector is a normal vector for the plane. Therefore, the equation of the plane is of the form
−3x − 9y + 4z + D = 0
Since the plane passes through A(3, −1, 2):
−3(3) − 9(−1) + 4(2) + D = 0
−9 + 9 + 8 + D = 0
D = −8
The equation of the plane is −3x − 9y + 4z − 8 = 0 or 3x + 9y − 4z + 8 = 0.
8. Answers may vary.
a) 3x − y + 4z = 2 c
x + 6y + 10z = −8 d
Eliminate x.
c: 3x − y + 4z = 2
−3 × d: −3x −18y − 30z = 24
Add:
−19y −26z = 26
Solve for z: 26z = −19y − 26
19
z = − 1 − 26 y
Let y = 26t, where t is any real number.
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∴ z = −1 − 19t
Substitute these expressions for y and z in equation d and solve for x.
x + 6(26t) + 10(−1 − 19t) = −8
x + 156t − 10 −190t = −8
x = 2 + 34t
A vector equation of the line of intersection is [x, y, z] = [2, 0, −1] + t[34, 26, −19].
9. Answers may vary.
The equations of the planes will be of the form 2x + 3y – z + 4 + k(x + 3z – 5) = 0, where k is
any real number.
Let k = 1 to obtain 3x + 3y + 2z – 1 = 0.
Let k = 2 to obtain 4x + 3y + 5z − 6 = 0.
Let k = 3 to obtain 5x + 3y + 8z − 11 = 0.
10. The plane has equation of the form 2x − 3y − z + 1 + k(3x + 5y − 4z + 2) = 0.
Since the plane passes through the point (3, −1, 2):
2(3) − 3(−1) − 2 + 1 + k[3(3) + 5(−1) − 4(2) + 2] = 0
8 − 2k = 0
k=4
Therefore, the equation of the plane is:
2x − 3y − z + 1 + 4(3x + 5y − 4z + 2) = 0
14x + 17y − 17z + 9 = 0
11. The plane has equation of the form:
3x − 2y + 4z − 3 + k(2x + 3z − 5) = 0
c
(3 + 2k)x −2y + (4 + 3k)z + (−3 − 5k) = 0
Since this plane is parallel to the plane 3x + 2y + 5z − 4 = 0, their normal vectors must be
scalar multiples of each other.
[3 + 2k, −2, 4 + 3k] = s[3, 2, 5]
The corresponding components are in proportion, so:
3
5
=
−1
−1=
3 + 2k
4 + 3k
−4 − 3k = 5
3 = −3 − 2k
3k = −9
2k = −6
k = −3
k = −3
Substitute k = −3 in equation c to obtain:
−3x − 2y – 5z + 12 = 0
3x + 2y + 5z − 12 = 0
12. The plane has equation of the form:
4x − 2y + z − 3 + k(2x − y + 3z + 1) = 0
(4 + 2k)x + (−2 − k)y + (1 + 3k)z + (−3 + k) = 0
Since the plane is perpendicular to the plane 3x + y − z + 7 = 0, their normal vectors are
perpendicular and have a dot product of 0.
[4 + 2k, −2 − k, 1 + 3k] ⋅ [3, 1, −1] = 0
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12 + 6k − 2 − k − 1 − 3k = 0
2k + 9 = 0
9
k = −2
Therefore, the equation of the plane is:
9
4x − 2y + z − 3 − 2 (2x − y + 3z + 1) = 0
8x − 4y + 2z − 6 − 18x + 9y − 27z − 9 = 0
−10x + 5y − 25z − 15 = 0
2x − y + 5z + 3 = 0
13. a) k(3x − y + z − 2) + x + 2y − 4z + 1 = 0
Since the plane passes through (3, 1, 3):
k(9 − 1 + 3 − 2) + 3 + 2 − 12 + 1 = 0
9k = 6
6
k=9
2
=3
b) The value of k in part a is the reciprocal of the value of k in Example 3. This is because
when k ≠ 0, k(3x − y + z − 2) + x + 2y − 4z + 1 = 0 is equivalent to
1
3x − y + z − 2 + k (x + 2y − 4z + 1) = 0.
14. The plane passing through the line of intersection of the two planes has equation
x − y + 2z + 5 + k(2x + 3y − z − 1) = 0, where k is any real number.
a) Since the plane passes through (0, 0, 0):
5 + k(−1) = 0
k=5
Therefore, the equation of the plane is:
x − y + 2z + 5 + 5(2x + 3y − z − 1) = 0
11x + 14y −3z = 0
b) Since the plane passes through (1, −1, 4):
1 – (–1) + 2(4) + 5 + k[2(1) + 3(–1) – 4 – 1] = 0
15 – 6k = 0
5
k=2
Therefore, the required equation is:
5
x − y + 2z + 5 + 2(2x + 3y − z − 1) = 0
2x − 2y + 4z + 10 + 5(2x + 3y − z − 1) = 0
12x + 13y − z + 5 = 0
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c) Write the equation in the form Ax + By + Cz + D = 0.
(1 + 2k)x + (−1 +3k)y + (2 − k)z + (5 − k) = 0
The normal to the plane is [1 + 2k, −1 + 3k, 2 − k]. Since the plane is parallel to the z-axis,
G
the normal is perpendicular to k = [0, 0, 1].
∴[1 + 2k, −1 + 3k, 2 − k]⋅[0, 0, 1] = 0
2−k=0
k=2
Therefore, the equation of the plane is 5x + 5y + 3 = 0.
d) From part c, the normal for the plane is [1 + 2k, −1 + 3k, 2 − k]. The plane x + 2y − 2z = 0
has normal [1, 2, −2]. Since the planes are perpendicular, their normal vectors are
perpendicular.
∴[1 + 2k, −1 + 3k, 2 − k]⋅[1, 2, −2] = 0
1 + 2k − 2 + 6k − 4 + 2k = 0
10k = 5
1
k=2
1
Substitute k = 2 in the equation (1 + 2k)x + (−1 +3k)y + (2 − k)z + (5 − k) = 0 to obtain
4x + y + 3z + 9 = 0.
e) A direction vector for line segment AB is:
AB = [3 − 1, 5 − 1, −3 − (−1)]
= [2, 4, −2]
A simpler direction vector is [1, 2, −1].
Since the plane is parallel to line segment AB, its normal vector is perpendicular to AB .
∴[1 + 2k, −1 + 3k, 2 − k]⋅[1, 2, −1] = 0
1 + 2k − 2 + 6k − 2 + k = 0
9k = 3
1
k=3
1
Substitute k = 3 in the equation (1 + 2k)x + (−1 +3k)y + (2 − k)z + (5 − k) = 0 to obtain
5x + 5z + 14 = 0.
f) If the plane has equal y- and z-intercepts, the coefficients of the terms in y and z are equal.
∴−1 + 3k = 2 − k
4k = 3
3
k=4
3
Substitute k = 4 in the equation (1 + 2k)x + (−1 + 3k)y + (2 − k)z + (5 − k) = 0 to obtain
10x + 5y + 5z + 17 = 0.
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15. a) By inspection, the normal vectors [1, 2, −3] and [2, 4, −6] are parallel. Therefore, the
equations represent parallel planes or the same plane. Since the constant terms, 4 and 5,
are different, the equations can’t represent the same plane. Hence the planes are parallel.
b) The equation x + 2y − 3z + 4 + k(2x + 4y − 6z + 5) = 0 is a linear combination of the
equations of 2 parallel planes. Hence it represents a family of planes parallel to
x + 2y − 3z + 4 = 0 and 2x + 4y − 6z + 5 = 0.
x−1 y−2 z+4
16. The line 2 = 3 = 1 is common to both planes, so both planes have common point
P(1, 2, −4) and common direction vector d = [2, 3, 1]. Plane π1 also contains the point
A(2, 1, 1), so AP is another direction vector for π1 and the cross product of AP and d is a
normal for π1. Similarly, BP is another direction vector for π2 and the cross product of BP
and d is a normal for π2.
Equation of π1:
AP = [1 − 2, 2 − 1, −4 − 1]
= [−1, 1, −5]
n = d × AP
= [2, 3, 1] × [−1, 1, −5]
3
1
1 2
− 5 −1
3
1
= [−15 − 1, −1 − (−10), 2 − (−3)]
= [−16, 9, 5]
Therefore, the equation of the plane is of the form:
−16x + 9y + 5z + D = 0
Since A(2, 1, 1) lies on the plane:
−16(2) + 9(1) + 5(1) + D = 0
−32 + 9 + 5 + D = 0
D = 18
The equation of π1 is −16x + 9y + 5z + 18 = 0 or 16x − 9y − 5z – 18 = 0.
Equation of π2:
BP = [1 − 1, 2 − 2, −4 − (−1)]
= [0, 0, −3]
A simpler direction vector is [0, 0, 1].
n = d × BP
= [2, 3, 1] × [0, 0, 1]
3
0
1
1
2
0
3
0
= [3 − 0, 0 − 2, 0 − 0]
= [3, −2, 0]
Therefore, the equation of the plane is of the form:
3x − 2y + D = 0
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Since B(1, 2, −1) lies on the plane:
3(1) − 2(2) + D = 0
3−4+D=0
D=1
The equation of π2 is 3x − 2y + 1 = 0.
17. The line [x, y, z] = [3, 5, 4] + s[2, 3, 1] is common to both planes, so both planes have
common point P(3, 5, 4) and common direction vector d = [2, 3, 1]. Plane π1 also contains
the point A(0, 0, 0), so AP is another direction vector for π1 and the cross product of AP and
d is a normal for π1. Similarly, BP is another direction vector for π2 and the cross product of
BP and d is a normal for π2.
Equation of π1:
AP = [3 − 0, 5 − 0, 4 − 0]
= [3, 5, 4]
n = d × AP
= [2, 3, 1] × [3, 5, 4]
3 1
5 4
2
3
3
5
= [12 − 5, 3 − 8, 10 − 9]
= [7, −5, 1]
Therefore, the equation of the plane is of the form:
7x − 5y + z + D = 0
Since the A(0, 0, 0) lies on the plane, D = 0.
The equation of π1 is 7x − 5y + z = 0.
Equation of π2:
BP = [3 − 1, 5 − 1, 4 − 1]
= [2, 4, 3]
n = d × BP
= [2, 3, 1] × [2, 4, 3]
3
4
1
3
2
2
3
4
= [9 − 4, 2 − 6, 8 − 6]
= [5, −4, 2]
Therefore, the equation of the plane is of the form:
5x − 4y + 2z + D = 0
Since B(1, 1, 1) lies on the plane:
5(1) − 4(1) + 2(1) + D = 0
5−4+2+D=0
D = −3
The equation of π2 is 5x − 4y + 2z − 3 = 0.
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18. Answers may vary.
The plane passes through face ABC of the tetrahedron.
Therefore, the plane intersects the tetrahedron in the lines
AB, AC, and BC.
Line AB: A direction vector for the line is AB = [−2 − 2, 2 − (−2), −2 −(−2)]
= [−4, 4, 0]
A simpler direction vector is [1, −1, 0]. Use point A to obtain parametric
equations for the line: x = 2 + t, y = −2 − t, z = −2.
Line BC: A direction vector for the line is BC = [−2 −(−2), −2 − 2, 2 −(−2)]
= [0, −4, 4]
A simpler direction vector is [0, −1, 1]. Use point B to obtain parametric
equations for the line: x = −2, y = 2 − t, z = −2 + t.
Line AC: A direction vector for the line is AC = [−2 − 2, −2 −(−2), 2 −(−2)]
= [−4, 0, 4]
A simpler direction vector is [1, 0, −1]. Use point A to obtain parametric
equations for the line: x = 2 + t, y = −2 , z = −2 − t.
19. The equation
A1x + B1y + C1z + D1 + k(A2x + B2y + C2z + D2) = 0 …c
represents a family of planes with a common line of intersection. Since the line of
intersection is determined by both of these planes, the plane A2x + B2y + C2z + D2 = 0 cannot
be represented by the equation c.
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3.6 Exercises, Sample Solutions
7. 2x − y + 3z = 2
x − 3y + 2z = −10
3x + y − z = 4
c
d
e
Eliminate x.
c + e:
5x + 2z = 6 f
d: x − 3y + 2z = −10
3 × e: 9x + 3y − 3z = 12
Add: 10x
−z=2 g
Use equations f and g. Eliminate z.
f: 5x + 2z = 6
2 × g: 20x − 2z = 4
Add: 25x
= 10
x= 2
5
2
Substitute x = in equation g and solve for z.
5
 2
10  − z = 2
5
z = 10
5
=2
Substitute x = 2 and z = 2 in equation e and solve for y.
5
 2
3  + y − 2 = 4
5
y = 24
5
 2 24 
The solution is  ,
, 2 .
5 5

8. a)
2x + y − z = 1
x + 3y + z = 10
x + 2y − 2z = −1
c
d
e
Eliminate x.
c:
2x + y − z = 1
−2 × d: −2x − 6y − 2z = −20
Add:
−5y −3z = −19
f
d: x + 3y + z = 10
−e: −x − 2y + 2z = 1
Add:
y + 3z = 11 g
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Use equations f and g. Eliminate z.
f: −5y −3z = −19
g: y + 3z = 11
Add: −4y
= −8
y=2
Substitute y = 2 in equation g and solve for z.
2 + 3z = 11
3z = 9
z=3
Substitute y = 2 and z = 3 in equation d and solve for x.
x + 3(2) + 3 = 10
x=1
The solution is (1, 2, 3).
b) x + 4y + 3z = 5
x + 3y + 2z = 4
x + y − z = −1
c
d
e
Eliminate x.
c: x + 4y + 3z = 5
−d: −x − 3y − 2z = −4
Add:
y+z=1 f
c: x + 4y + 3z = 5
−e: −x − y + z = 1
Add:
3y + 4z = 6 g
Use equations f and g. Eliminate z.
−4 × f: −4y −4z = −4
g:
3y + 4z = 6
Add:
−y
= 2
y = −2
Substitute y = −2 in equation f and solve for z.
−2 + z = 1
z=3
Substitute y = −2 and z = 3 in equation e and solve for x.
x − 2 − 3 = −1
x=4
The solution is (4, −2, 3).
c)
3x + 2y − z = 6
c
x+y+z=5
d
2x − 3y + 2z = −10 e
Eliminate z.
c: 3x + 2y − z = 6
d:
x+y+z=5
Add:
4x + 3y = 11 f
2 × c: 6x + 4y − 2z = 12
e: 2x − 3y + 2z = −10
Add: 8x + y
=2
g
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Use equations f and g. Eliminate x.
−2 × f: −8x − 6y = −22
g: 8x + y = 2
Add:
−5y = −20
y=4
Substitute y = 4 in equation f and solve for z.
4x + 3(4) = 11
4x = −1
1
x = −4
1
Substitute x = −4 and y = 4 in equation d and solve for x.
1
−4 + 4 + z = 5
5
z=4
5
 1
The solution is  − , 4,  .
4
 4
d) 5x + 2y − z = 13
x−y−z=0
2x + y + 3z = −1
c
d
e
Eliminate z.
c: 5x + 2y − z = 13
−x + y + z = 0
−d:
Add:
4x + 3y = 13 f
3 × d: 3x − 3y − 3z = 0
e: 2x + y + 3z = −1
Add:
5x − 2y = −1
g
Use equations f and g. Eliminate y.
2 × f: 8x + 6y = 26
3 × g: 15x − 6y = −3
Add: 23x
= 23
x=1
Substitute x = 1 in equation f and solve for y.
4 + 3y = 13
3y = 9
y=3
Substitute x = 1 and y = 3 in equation d and solve for z.
1−3–z=0
z = –2
The solution is (1, 3, –2).
e) 3x + y + 2z = 5
2x − y + z = −1
4x + 2y − z = −3
c
d
e
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Eliminate z.
c:
3x + y + 2z = 5
−2 × d: −4x + 2y − 2z = 2
Add:
−x + 3y = 7 f
d: 2x − y + z = −1
e: 4x + 2y − z = −3
Add:
6x + y = −4 g
Use equations f and g. Eliminate y.
f:
−x + 3y = 7
−3 × g: −18x − 3y = 12
Add: −19x
= 19
x = −1
Substitute x = −1 in equation g and solve for y.
−6 + y = −4
y=2
Substitute x = −1 and y = 2 in equation d and solve for z.
−2 − 2 + z = −1
z=3
The solution is (−1, 2, 3).
f)
x + y + 2z = −8
3x − y − z = 0
2x + 2y − z = −2
c
d
e
Eliminate y.
c: x + y + 2z = −8
d: 3x − y − z = 0
Add: 4x
+ z = −8 f
2 × d: 6x − 2y − 2z = 0
e: 2x + 2y − z = −2
Add: 8x
− 3z = −2
g
Use equations f and g. Eliminate z.
3 × f: 12x + 3z = −24
g: 8x − 3z = −2
Add: 20x
= −26
13
x = −10
13
Substitute x = −10 in equation f and solve for z.
52
−10 + z = −8
28
z = −10
14
=−5
13
14
Substitute x = −10 and z = − 5 in equation d and solve for y.
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39
14
−10 − y + 5 = 0
11
y = −10
 13 11 14 
The solution is  − , − , −  .
5
 10 10
9. 3x − 3y − 2z = 14
5x + y − 6z = 10
x − 2y + 4z = 9
c
d
e
a) Determine the line of intersection of plane c and plane d. From page 176, c + 3 × d
gives:
9x − 10z = 22
22
10
x= 9 + 9 z
22
f
10
Substitute 9 + 9 z for x in equation d and express y in terms of z.
110 50
9 + 9 z + y − 6z = 10
20 4
y=−9 +9z g
Let z = 9t, where t is any real number. Substitute 9t for z in equations f and g to obtain
22
20
the parametric equations of the line of intersection: x = 9 + 10t, y = − 9 + 4t, z = 9t.
To determine the intersection of this line with plane e, substitute the expressions for x, y,
and z in the equation of the plane.
22
40
9 + 10t + 9 − 8t + 36t = 9
19
38t = 9
1
t = 18
Substitute this value of t in the equations of the line to obtain the point of intersection.
22
x = 9 + 10t
22
5
= 9 +9
=3
20
y = − 9 + 4t
2
20
=−9 +9
= −2
z = 9t
1
=2
1
The point of intersection is (3, −2, 2 ). This agrees with the result obtained on pages 176
and 177.
b) The planes in Example 1 have a point of intersection. In pages 176 and 177, we
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determined this point by finding the lines of intersection of π1 and π2 and π2 and π3. Then
we found the point of intersection of these lines. In part a, we found the point of
intersection by determining the point at which the line of intersection of π1 and π2
intersects π3. In both methods, it was necessary to find the line of intersection of π1 and
π2.
10. 2x − y + 3z = 2
c
x − 3y + 2z = −10 d
5x − 5y + 8z = −6 e
The 3 planes have a line of intersection.
Eliminate y.
−3 × c: −6x + 3y − 9z = −6
d:
x − 3y + 2z = −10
Add: −5x
−7z = −16
Express x in terms of z
5x = 16 − 7z
16
7
x= 5 −5z f
Substitute this expression for x in equation e and solve for y.
16 − 7z − 5y + 8z = −6
−5y = −22 − z
22
1
g
y= 5 +5z
Let z = 5t, where t is a real number. Substitute 5t for z in equations f and g to obtain the
16
22
parametric equations of the point of intersection: x = 5 − 7t, y = 5 + t, z = 5t.
G G G
11. a) If n1 ⋅ n2 × n3 ≠ 0, the three normal vectors are not coplanar and there is only one point of
intersection of the three planes. Since P lies on all three planes, P is the point of
intersection.
G G G
b) If n1 ⋅ n2 × n3 = 0, the three normal vectors are coplanar and if the planes intersect, there is
a line of intersection. Since P lies on all three planes, the intersection is a line passing
through P.
12. Suppose that the planes represented by the system of equations have normal vectors
G G
G
n1 , n 2 , and n3 .
Case 1. The 3 normal vectors are parallel.
Look at constant terms in the equations of the planes.
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•
If the constant terms are the same scalar multiples of each other as the normal vectors,
then the 3 planes are coincident, and the system is consistent. An example of this type of
x + 2 y + 3z = 4 

system is: 2 x + 4 y + 6 z = 8  .
3 x + 6 y + 9 z = 12
•
If the constant terms are not the same scalar multiples of each other as the normal
vectors, then the planes are parallel and distinct, and the system is inconsistent. An
x + 2 y + 3z = 4 

example of this type of system is 2 x + 4 y + 6 z = 5
3 x + 6 y + 9 z = 6 
•
A third possibility is that two of the planes are coincident, and the third plane is parallel
to them; in this case the system is inconsistent. An example of this type of system is
x + 2 y + 3z = 4 

2 x + 4 y + 6 z = 8
3 x + 6 y + 9 z = 5 
Case 2. Exactly two of the normal vectors are parallel.
G
G
For example, suppose n1 and n 2 are parallel. Look at the equations to determine whether
these two planes are coincident or parallel and distinct.
•
If the planes are coincident, then the third plane intersects these planes in a straight line,
x + 2 y + 3z = 4 

and the system is consistent. An example of this type of system is 2 x + 4 y + 6 z = 8
x + y + 9 z = 6 
•
If the two planes are parallel, then the third plane intersects them in a pair of parallel
lines, so the system is inconsistent. An example of this type of system is
x + 2 y + 3z = 4 

2 x + 4 y + 6 z = 7 .
x + y + 9 z = 6 
Case 3. Three distinct normal vectors
G G G
Calculate If n1 ⋅ n2 × n3 .
G G G
•
If n1 ⋅ n2 × n3 ≠ 0, then the three planes intersect at a single point and the system is
consistent. An example of this type of system is given in Example 1, page 175.
G G G
•
If n1 ⋅ n2 × n3 = 0, there may or may not be points of intersection, and we must solve the
system of equations to determine whether they are consistent (intersect in a line) or
inconsistent. An example of an consistent system is given in Example 2 on page 177,
and an inconsistent system is given on the top of page 173.
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13. a) x + 2y + 3z = 2
x + y + z = −5
2x + 3y + 4z = −3
c
d
e
Observe that c + d = e.
Eliminate x.
c − d: y + 2z = 7
f
y = 7 − 2z
Substitute this expression for y in equation 2 and solve for x.
x + 7 − 2z + z = −5
x = −12 + z g
Let z = t, where t is a real number. Substitute t for z in equations f and g to obtain the
parametric equations of the line of intersection: x = −12 + t, y = 7 − 2t, z = t.
b) 2x − y + z = −4
5x + y − z = 10
9x − y + z = 2
c
d
e
Observe that 2 × c + d = e.
Eliminate x.
c + d: 7x = 6
6
x=7
f
Substitute this expression for x in equation e and solve for y.
54
7 −y +z=2
40
y= 7 +z g
Let z = t, where t is a real number. From equations f and g, the parametric equations of
6
40
the line of intersection: x = 7 , y = 7 + t, z = t.
14. a) x + 3y + 2z = 1 c
2x + y − z = 4 d
Eliminate y.
c: x + 3y + 2z = 1
−3 × d: −6x −3y + 3z = −12
Add: −5x
+ 5z = −11
Solve for x.
5x = 11 + 5z
11
x= 5 +z
3
Substitute this expression for x in equation c and solve for y.
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11
5 + z + 3y + 2z = 1
6
3y = −5 − 3z
2
y = −5 − z 4
Let z = t, where t is a real number. Substitute t for z in equations 3 and 4 to obtain the
11
2
parametric equations of the line of intersection: x = 5 + t, y = −5 − t, z = t.
b) 4x − y + 5z = −2
c
d
2x + y + 7z = −3
Eliminate y.
c + d: 6x + 12z = −5
5
x = −6 − 2z 3
Substitute this expression for x in equation d and solve for y.
5
−3 − 4z + y + 7z = −3
4
y = −3 − 3z
4
Let z = t, where t is a real number. Substitute t for z in equations 3 and 4 to obtain the
5
4
parametric equations of the line of intersection: x = −6 − 2t, y = −3 − 3t, z = t.
15. a) 3x + 2y − z = 4 c
x + 3y + z = 5 d
4x + 5y = 8 e
G
G
G
The normal vectors are n1 = [3, 2, −1], n 2 = [1, 3, 1], and n3 = [4, 5, 0].
G
G
G
Since n3 = n1 + n 2 but equation e ≠ equation c + equation d, the system represents
three distinct planes that form a triangular prism. Thus the system of equations is
inconsistent, and the system has no solution.
c
b) x + 2y − 3z = 11
2x + y = 7
d
3x + 6y − 8z = 22 e
Use equations c and e. Eliminate z.
−8 × c: −8x − 16y + 24z = −88
3 × e: 9x + 18y − 24z = 66
Add:
x + 2y
= −22 f
Use equations d and f. Eliminate x.
d:
2x + y = 7
−2 × g: −2x − 4y = 44
Add:
−3y = 51
y = −17
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Substitute y = −17 in equation d and solve for x.
2x − 17 = 7
2x = 24
x = 12
Substitute x = 12 and y = −17 in equation c and solve for z.
12 − 34 −3z = 11
3z = −33
z = −11
The planes intersect at the point (12, −17, –11).
c)
x + 2y − z = −3
2x + y + 3z = 8
2x + 4y − 2z = 5
c
d
e
G
G
G
G
G
The normal vectors are n1 = [1, 2, −1] , n 2 = [2, 1, 3], and n3 = [2, 4, −2]. Since n3 = 2 n1 ,
G
but equation e ≠ 2 × equation c, planes c and e are parallel. However, n 2 is not
G
G
parallel to n1 or n3 , so plane d is not parallel to either plane c or plane e. Therefore,
the system represents a situation similar to that in Case 2 on page 172. Since the planes
intersect in two parallel lines, the system is inconsistent and has no solution.
d) 5x + 2y − 5z = 4
2x + 3y − 4z = 2
x+y+z=3
c
d
e
Eliminate x.
c: 5x + 2y − 5z = 4
−5 × e: −5x − 5y − 5z = −15
Add:
−3y − 10z = −11
3y + 10z = 11 f
d: 2x + 3y − 4z = 2
−2 × e: −2x − 2y − 2z = −6
y − 6z = −4 g
Use equations f and g. Eliminate y.
f: 3y + 10z = 11
−3 × g: −3y + 18z = 12
Add:
28z = 23
23
z = 28
23
Substitute z = 28 in equation g and solve for y.
138
y − 28 = −4
26
y = 28
13
= 14
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13
23
Substitute y = 14 and z = 28 in equation e and solve for x.
13
23
x + 14 + 28 = 3
35
x = 28
5
=4
 5 13 23 
The planes intersect at the point  , ,  .
 4 14 28 
e) x + 3y + 3z = 8
c
d
x − y + 3z = 4
The two planes are distinct, so there is a line of intersection.
Eliminate x.
c − d: 4y = 4
y=1
Substitute in equation d and solve for x.
x − 1 + 3z = 4
x = 5 − 3z
Let z = t, where t is a real number. Therefore, parametric equations of the line of
intersection are x = 5 − 3t, y = 1, z = t.
16. x + 2y + 3z = −4 c
d
x − y − 3z = 8
x + 5y + 9z = −16 e
a) Determine the line of intersection of plane c and plane d.
From page 178, c − d gives:
y + 2z = −4
y = −4 − 2z f
Substitute −4 − 2z for y in equation c and solve for x.
x − 8 − 4z + 3z = −4
x=4+z
g
Let z = t, where t is any real number. Substitute t for z in equations f and g to obtain
the parametric equations of the line of intersection: x = 4 + t, y = −4 − 2t, z = t.
To determine the intersection of this line with plane e, substitute the expressions for x, y,
and z in the equation of the plane.
4 + t − 20 − 10t + 9t = −16
0t = 0
Therefore, the line is contained in the plane. Hence the three planes intersect in the line
x = 4 + t, y = −4 − 2t, z = t. This agrees with the result obtained on pages 177 and 178.
b) The planes in Example 2 have a line of intersection. In pages 176 and 177, we
determined the line of intersection by forming linear combinations of the equations of the
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three planes. In part a, we determined the line of intersection of the π1 and π2 and then
showed that this line was contained in π3. In both methods, it was necessary to find the
line of intersection of π1 and π2.
G
G
G
17. The normal vectors of the planes are n1 = [1, b, c], n 2 = [a, 1, c], and n3 = [a, b, 1].
G G G
If the planes intersect in a line, their normal vectors are coplanar and n1 × n2 ⋅ n3 = 0.
G G G
n1 × n2 ⋅ n3 = 0
[1, b, c] × [a, 1, c] ⋅ [a, b, 1] = 0
b
1
c
c
1
a
b
1
[bc − c, ac − c, 1 − ab] ⋅ [a, b, 1] = 0
a(bc − c) + b(ac − c) + 1 − ab = 0
abc − ac + abc − bc + 1 − ab = 0
2abc + 1 = ab + bc + ac
18. Since the points lie on the parabola, their coordinates satisfy the equation of the parabola.
y = ax2 + bx + c.
A(1, 0):
0=a+b+c
c
B(4, 3):
3 = 16a + 4b + c d
C(6, −5):
−5 = 36a + 6b + c e
Solve the system of equations.
Eliminate c.
c − d: −15a − 3b = −3, or 5a + b = 1
f
c − e: −35a − 5b = 5, or −7a − b = 1
g
Use equations f and g. Eliminate b.
f + g: −2a = 2
a = −1
Substitute a = −1 in equation f and solve for b.
−5 + b = 1
b=6
Substitute a = −1 and b = 6 in equation c and solve for c.
0 = −1 + 6 + c
c = −5
Therefore, the parabola has equation y = −x2 + 6x − 5.
19. π1: 3x − 4y + z = 35 c
π2: ax + by − 5z = −23 d
The line passes through the point (1, −5, 12) and has direction vector [3, 4, 7]. Since π1 and
π2 intersect in this line:
The point (1, −5, 12) must satisfy the equation of each plane. Therefore, from equation d:
a(1) + b(−5) −5(12) = −23
a − 5b = 37 e
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The vector [3, 4, 7] is perpendicular to the normal vector of each plane. Since π2 has normal
[a, b, −5]:
[3, 4, 7]⋅ [a, b, −5] = 0
f
3a + 4b − 35 = 0
From equation e, a = 37 + 5b. g
Substitute this expression for a in equation f and solve for b.
3(37 + 5b) + 4b − 35 = 0
111 + 15b + 4b − 35 = 0
19b = −76
b = −4
Substitute b = −4 in equation g to obtain:
a = 37 + 5(−4)
= 17
Therefore, the planes intersect in a line when a = 17 and b = −4.
G
G
G
G
20. a) The equation ax + b y + c z = d represents the system of equations.
x[1, 2, 3] + y[1, −1, 2] + z[1, 2, 1] = [6, 6, 10]
[x + y + z, 2x − y + 2z, 3x + 2y + z] = [6, 6, 10]
Equate corresponding components to obtain the equations of the linear system.
x+y+z=6
2x − y + 2z = 6
3x + 2y + z = 10
G G G
d ⋅b ×c
b) The solution x = G G G is obtained as follows:
a ⋅ bG × c
G
G
G
ax + b y + c z = d
G
On each side of the equation, take the dot product with b .
G
G
G G G G
ax + b y + c z ⋅ b = d ⋅ b
G
On each side of the equation, take the cross product with c .
G
G
G G
G G G G
[ ax + b y + c z ⋅ b ] × c = d ⋅ b × c
Simplify, to obtain:
G G G
G G G
G G G
G G G
a ⋅ b × cx + b ⋅ b × cy + c ⋅ b × cz = d ⋅ b × c
c
G G
G
G G G G
G G G
Since b × c is perpendicular to both b and c , b ⋅ b × c = 0 and c ⋅ b × c = 0.
G G G
G G G
G G G
d ⋅b ×c
Hence equation c simplifies to a ⋅ b × c x = d ⋅ b × c , or x = G G G .
a ⋅b ×c
(
)
(
)
G G G
a⋅d ×c
The solution y = G G G is obtained as follows:
a ⋅ bG × c
G
G
G
ax + b y + c z = d
G
On each side of the equation, take the dot product with a . On the resulting equation, then
G
take the cross product with c .
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[ a ⋅ (ax + b y + c z )] × c = a ⋅ (d × c )
G
G G
G
G
G G G
Simplify to obtain:
G G G
G G G
G G G
G G G
a ⋅ a × cx + a ⋅ b × cy + a ⋅ c × cz = a ⋅ d × c
d
G G G G G G
G G
G G G G
Since a × c is perpendicular to a , a ⋅ a × c = 0 . Since c × c = 0 , a ⋅ c × c = 0 .
G G G
G G G
G G G
a⋅d ×c
Hence, equation d simplifies to a ⋅ b × c y = a ⋅ d × c , or y = G G G .
a ⋅b ×c
G G G
a ⋅b × d
The solution z = G G G is obtained as follows.
a ⋅ Gb × c
G
G
G
ax + b y + c z = d
G
One each side of the equation, take the cross product with b . On the resulting equation
G
then take the dot product with a .
G
G
G G G
G
G
G
a ⋅ [ b × ax + b y + c z ] = a ⋅ b × d
Simplify to obtain:
G G G
G G G
G G G
G G G
a ⋅ b × ax + a ⋅ b × b y + a ⋅ b × c z = a ⋅ b × d
e
Since the coefficients of x and y simplify to 0, equation e simplifies to
G G G
G G G
G G G
a ⋅b × d
a ⋅ b × c z = a ⋅ b × d or z = G G G .
a ⋅b ×c
(
21.
)
(
)
x + 2y − z = a c
x − y + 2z = b d
3x + 3y + z = c e
Eliminate x.
c − d:
3y − 3z = a − b f
−3 × c: −3x − 6y + 3z = −3a
e:
3x + 3y + z = c
−3y + 4z = c − 3a g
Add:
Use equations f and g. Eliminate y.
f + g: z = −2a − b + c
Substitute this value of z in equation f and solve for y.
3y − 3(−2a − b + c) = a − b
3y + 6a + 3b − 3c = a − b
3y = −5a − 4b + 3c
−5a − 4b + 3c
y=
3
Substitute the values of x and y in equation d and solve for x.
(−5a − 4b + 3c)
x−
+ 2(−2a − b + c) = b
3
Multiply both sides by 3.
3.6 Exercises, Sample Solutions
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GDM, Chapter 3, Sample Solutions
3x + 5a + 4b − 3c −12a − 6b + 6c = 3b
3x = 7a + 5b − 3c
7a + 5b − 3c
x=
3
3.6 Exercises, Sample Solutions
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GDM, Chapter 3, Selected Solutions
3.7 Exercises, Selected Solutions
3 1 5
1. a) Write the system as a matrix. 

 1 1 3
Copy c.
Replace d with c − 3 × d.
1
5
3
0 − 2 − 4 


Copy c.
Divide d by −2.
 3 1 5
0 1 2


Replace c with −d + c.
Copy d.
 3 0 3
0 1 2 


Divide c by 3.
Copy d.
 1 0 1
0 1 2 


The solution is x = 1, y = 2.
 2 − 1 2
b) Write the system as a matrix. 

 1 3 8
Copy c.
Replace d with c − 2 × d.
2
2 − 1
0 − 7 − 14


Copy c.
Divide d by −7.
 2 − 1 2
0
1 2

3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Replace c with d + c.
Copy d.
 2 0 4
 0 1 2


Divide c by 2.
Copy d.
 1 0 2
0 1 2 


The solution is x = 2, y = 2.
5 2 0
c) Write the system as a matrix. 

3 1 5
Copy c.
Replace d with −3 × c + 5 × d.
 5 2 0
0 − 1 25


Replace c with 2 × d + c.
Copy d.
5 0 50
0 − 1 25


Divide c by 5.
Divide d by −1.
10
1 0
0 1 − 25


The solution is x = 10, y = −25.
5
1 3
d) Write the system as a matrix. 

 4 − 1 − 6
Copy c.
Replace d with −4 × c + d.
3
5
1
0 − 13 − 26


Copy c.
Divide d by −13.
 1 3 5
0 1 2 


3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Replace c with −3 × d + c.
Copy d.
 1 0 − 1
0 1 2 


The solution is x = −1, y = 2.
7
5 − 2
2. a) Write the system as a matrix. 
4 − 2
3
Copy c.
Replace d with −3 × c + 5 × d.
7
5 − 2
0 26 − 31


Copy c.
Divide d by 26.
7
5 − 2
0
31 
1 − 26


Replace c with 2 × d + c.
Copy d.
60
5 0
13 

31 
0 1 − 26 
Divide c by 5 and reduce.
Copy d.
12
1 0
13 
0 1 − 31 
26 

The solution is x = 12 , y = – 31 .
13
26
4 − 5
9
b) Write the system as a matrix. 
8
4 − 3
Copy c.
Replace d with −4 × c + 9 × d.
4 − 5
9
0 − 43 92


3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Divide d by −43.
− 5
9 4
0 1 − 92 
43 

Replace c with −4 × d + c.
Copy d.
9 0 153 
43 

0 1 − 92 
43 

Divide c by 9.
Copy d.
17 
1 0
43 

0 1 − 92 
43 

The solution is x = 17 , y = − 92 .
43
43
2 1 − 7
c) Write the system as a matrix. 
2
3 4
Copy c.
Replace d with −3 × c + 2 × d.
2 1 − 7
0 5 25


Copy c.
Divide d by 5.
2 1 − 7
0 1
5

Replace c with −d + c.
Copy d.
2 0 − 12
0 1
5

Divide c by 3.
Copy d.
 1 0 − 6
0 1
5

The solution is x = −6, y = 5.
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
 6 − 5 9
d) Write the system as a matrix. 

 2 − 3 2
Copy c.
Replace d with c − 3 × d.
6 − 5 9 
0
4 3

Copy c.
Divide d by 4.
6 − 5 9 
0
1 3 

4
Replace c with 5 × d + c.
Copy d.
6 0 51
4

0 1 3 
4

Divide c by 6.
Copy d.
 1 0 17 
8

0 1 3 
4

The solution is x = 17 , y = 3 .
8
4
3. a) The coefficients of x and y in one equation were multiples of the coefficients of x and y in
the other equation.
b) Matrix 1
The second row corresponds to the equation 0y = 0, which has infinitely many solutions.
This is because both equations in the system represent the same line, and any point on the
line satisfies both equations.
Matrix 2
The second row corresponds to the equation 0y = 8, which has no solution. This is because
the equations in the system represent a pair of parallel lines.
4. a) If the first equation is multiplied by 2, the result is the second equation. Therefore, the
reduced matrix will have a row of 0s. This corresponds to Matrix 3.
b) The left side of the second equation is twice the left side of the first equation, but the
constant term in the second equation is not twice the constant term in the first equation.
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Therefore in the reduced matrix, all entries in the second row except the last entry will be
0. This corresponds to Matrix 2.
c) The left side of the second equation is not a multiple of the left side of the first equation.
Therefore, the system of equations has one solution. This corresponds to Matrix 1.
1 3 4 19
5. a) Write the system as a matrix. 1 2 1 12
1 1 1 8
Copy c.
Replace d with −c + d.
Replace e with −c + e.
3
4
19
1
0 − 1 − 3 − 7 


0 − 2 − 3 − 11
Copy c.
Copy d.
Replace e with −2 × d + e.
4 19
1 3
0 − 1 − 3 − 7 


0
0
3
3
Copy c.
Copy d.
Divide e by 3.
4 19
1 3
0 − 1 − 3 − 7 


0
0
1
1
Replace c with −4 × e + c.
Replace d with 3 × e + d.
Copy e.
 1 3 0 15
0 − 1 0 − 4 


0
0 1
1
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Replace c with 3 × d + c.
Multiply d by −1.
Copy e.
 1 0 0 3
0 1 0 4 


0 0 1 1
The solution is x = 3, y = 4, z = 1.
1
1 − 4
1
b) Write the system as a matrix.  1 − 1
2 − 13
2
1 −3
15
Copy c.
Replace d with −c + d.
Replace e with −2 × c + e.
1
1 − 4
1
0 − 2
1 − 9

0 − 1 − 5 23
Copy c.
Copy d.
Replace e with −d + 2 × e.
1
1 − 4
1
0 − 2
1 − 9

0
0 − 11 55
Copy c.
Copy d.
Divide e by 11.
1 1 − 4
1
0 − 2 1 − 9 


0
0 1 − 5
Replace c with −e + c.
Replace d with −e + d.
Copy e.
1 0
1
1
0 − 2 0 − 4 


0
0 1 − 5
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Divide d by −2.
Copy e.
1
1 1 0
0 1 0
2

0 0 1 − 5
Replace c with −d + c.
Copy d.
Copy e.
 1 0 0 − 1
0 1 0
2

0 0 1 − 5
The solution is x = −1, y = 2, z = −5.
7
4 2 − 3
c) Write the system as a matrix.  1 3
1
2
 1 4 − 2 − 9
Copy c.
Replace d with c − 4 × d.
Replace e with c − 4 × e.
2 − 3 7
4
0 − 10 − 7 − 1


0 − 14
5 43
Copy c.
Copy d.
Replace e with −14 × d + 10 × e.
2 −3
7
4
0 − 10 − 7 − 1


0
0 148 444
Copy c.
Copy d.
Divide e by 148.
2 − 3 7
4
0 − 10 − 7 − 1


0
0
1 3
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Replace c with 3 × e + c.
Replace d with 7 × e + d.
Copy e.
2 0 16
4
0 − 10 0 20


0
0 1 3
Divide c by 2.
Divide d by −10.
Copy e.
8
2 1 0
 0 1 0 − 2


0 0 1
3
Replace c with −d + c.
Copy d.
Copy e.
2 0 0 10
 0 1 0 − 2


0 0 1
3
Divide c by 2.
Copy d.
Copy e.
5
1 0 0
0 1 0 − 2 


0 0 1
3
The solution is x = 5, y = −2, z = 3.
1 0
 1 1

d) Write the system as a matrix. 16 4
1 3
 1 1 − 1 0
Copy c.
Replace d with −16 × c + d.
Replace e with −c + e.
1
1 0
1
0 − 12 − 15 3


0
0 − 2 0
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Divide d by 3.
Divide e by −2.
1
1 0
1
0 − 4 − 5 1


0
0
1 0
Replace c with −e + c.
Replace d with 5 × e + d.
Copy e.
1 0 0
1
0 − 4 0 1


0
0 1 0
Copy c.
Divide d by −4.
Copy e.
0
1 1 0
0 1 0 − 1 

4
0 0 1
0
Replace c with −d + c.
Copy d.
Copy e.
1
1 0 0
4
0 1 0 − 1 
4

0 0 1
0
The solution is x = 1 , y = − 1 , z = 0.
4
4
 1 − 1 2 7
6. a) Write the system as a matrix. 2
1 − 1 3
 1
1
1 9
Copy c.
Replace d with −2 × c + d.
Replace e with −c + e.
2
7
1 −1
0
3 − 5 − 11

0
2 −1
2
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Copy d.
Replace e with −2 × d + 3 × e.
2
7
1 −1
0
3 − 5 − 11

0
0
7
28
Copy c.
Copy d.
Divide e by 7.
2
7
1 −1
0
3 − 5 − 11

0
0
1
4
Replace c with −2 × e + c.
Replace d with 5 × e + d.
Copy e.
 1 − 1 0 − 1
0
3 0
9

0
0 1 4
Copy c.
Divide d by 3.
Copy e.
 1 − 1 0 − 1
0
1 0
3

0
0 1 4
Replace c with d + c.
Copy d.
Copy e.
 1 0 0 2
0 1 0 3


0 0 1 4
The solution is x = 2, y = 3, z = 4.
1 3 2
− 1

b) Write the system as a matrix.  1 − 3 5 6
 1 − 2 1 2
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Replace d with c + d.
Replace e with c + e.
1 3 2
− 1
 0 − 2 8 8


 0 − 1 4 4
Copy c.
Divide d by −2.
Copy e
1
3
2
− 1
 0
1 − 4 − 4

 0 − 1
4
4
Copy c.
Copy d.
Replace e with d + e.
3
2
− 1 1
 0 1 − 4 − 4


 0 0
0
0
Replace c with −d + c.
Copy d
Copy e.
7
6
− 1 0
 0 1 − 4 − 4


 0 0
0
0
The corresponding system of equations is:
−x + 7z = 6
y − 4z = −4
0z = 0
Let z = t, where t is a real number to obtain the solution: x = −6 + 7t, y = −4 + 4t, z = t.
4 − 6 − 2 10
c) Write the system as a matrix. 2 − 3
1 0
 1 − 9 − 4 5
3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Replace d with −2 × d + c.
Replace e with −4 × e + c.
10
4 − 6 − 2
0
0
0
10

 1 30 14 − 10
The second row is equivalent to 0x + 0y + 0z = 10. Since this equation is never true, the
system has no solution.
1 − 1 3
1

d) Write the system as a matrix. 2 − 1
1 5
 1 − 2
2 6
Copy c.
Replace d with − 2 × c + d.
Replace e with −c + e.
1 − 1 3
1
0 − 3
3 − 1

0 − 3
3
3
Copy c.
Copy d.
Replace e with −d + e.
1 − 1 3
1
0 − 3
3 − 1

0
0
0
4
The last row is equivalent to 0z = 4. Since this equation is never true, the system has no
solution.
1 4 − 1 − 3
7. a) Write the system as a matrix. 
2
1 5 − 3
Copy c.
Replace d with –c + d.
 1 4 − 1 − 3
0 1 − 2
5

3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Replace c with −4 × d + c.
Copy d.
7 − 23
1 0
0 1 − 2
5

The corresponding system of equations is:
x + 7z = −23
y − 2z = 5
Let z = t, where t is a real number, to obtain the solution x = −23 − 7t, y = 5 + 2t, z = t.
2 1 − 4 3
b) Write the system as a matrix. 
3 2
1 1
Copy c.
Replace d with −2 × 2 + 1.
1 −4
3
2
0 − 1 − 10 − 1


Replace c with d + c.
Copy d.
0 − 14
2
2
0 − 1 − 10 − 1


Divide c by 2.
Multiply d by −1.
 1 0 − 7 1
0 1 10 1


The corresponding system of equations is:
x − 7z = 1
y + 10z = 1
Let z = t, where t is a real number, to obtain the solution x = 1 + 7t, y = 1 − 10t, z = t.
 1 2 − 1 3
c) Write the system as a matrix. 

3 1 2 1
Copy c.
Replace d with −3 × c + d.
2 −1
3
1
0 − 5
5 − 8

3.7 Exercises, Selected Solutions
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GDM, Chapter 3, Selected Solutions
Copy c.
Divide d by −5.
 1 2 − 1 3
0 1 − 1 8 
5 

Replace c with −2 × 2 + 1.
Copy d.
1 − 15 
1 0
0 1 − 1
8
5

The corresponding system of equations is:
x+z=–1
5
y−z= 8
5
Let z = t, where t is a real number, to obtain the solution x = – 1 − t, y = 8 + t, z = t.
5
5
1 − 1
5 3
d) Write the system as a matrix. 
2
2 1 − 2
Copy c.
Replace d with −2 × c + 5 × d.
3
1 − 1
5
0 − 1 − 12 12


Replace c with 3 × d + c.
Copy d.
5 0 − 35 35
0 − 1 − 12 12


Divide c by 5.
Multiply d by −1.
7
1 0 − 7
0 1 12 − 12


The corresponding system of equations is:
x − 7z = 7
y + 12z = −12
Let z = t, where t is a real number, to obtain the solution x = 7 + 7t, y = −12 − 12t, z = t.
8. a) The normal vector of one of the equations was a linear combination of the normal vectors
of the other two equations.
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b) Matrix 1
The third row is equivalent to the equation 0z = 0, which has infinitely many solutions.
This is because the three planes intersect in a line.
Matrix 2
The third row is equivalent to the equation 0z = 10, which has no solution. This is because
because either two of the planes are parallel or the three planes form a triangular prism.
9. a) The normal vectors of the planes are [2, −3, 1], [1, 1, 1], and [3, −1, 2].
−3 1 2 −3
[2, −3, 1] × [1, 1, 1] ⋅[3, −1, 2]
1 1 1
1
= [−4, −1, 5] ⋅[3, −1, 2]
= −12 + 1 + 10
= −1
Since the normal vectors are not coplanar, the planes intersect at a single point. Hence the
system has a unique solution. This corresponds to Matrix 1.
b) The third equation is a linear combination of the first two equations. Hence the three
planes intersect in a line and the system has infinitely many solutions described by one
parameter. This corresponds to Matrix 2.
c) Both the second and third equations are multiples of the first equation, so the three
equations represent the same plane. Hence the system has an infinite number of solutions
described by two parameters. This corresponds to Matrix 3.
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1. In R2, the equation y = 3 represents points whose coordinates are of the form (x, 3).
This is a line parallel to the x-axis and 3 units above it (see diagram, below left).
In R3, the equation y = 3 represents points whose coordinates are of the form (x, 3, z). This is
a plane parallel to the xz-plane and 3 units to the right of it (see diagram, below right).
2. In R2, the equation x = 3 represents points whose coordinates are of the form (3, y).
This is a line parallel to the y-axis and 3 units to the right of it (see diagram, below left).
In R3, the equation x = 3 represents points whose coordinates are of the form (3, y, z).
This is a plane parallel to the yz-plane and 3 units in front of it (see diagram, below right).
3. The parametric equations of L1 are:
x = −3 − t, y = 7 + 4t, z = 2 + t
Since L2 passes through A(0, 2, 1) and B(−4, 4, 1), its direction vector is:
AB = [−4 − 0, 4 − 2, 1 − 1]
= [−4, 2, 0]
Use the point A(0, 2, 1) and the simpler direction vector [−2, 1, 0] to obtain the parametric
equations x = −2s, y = 2 + s, z = 1.
At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
−3 − t = −2s
→
t − 2s = −3 c
7 + 4t = 2 + s
→
4t − s = −5 d
2+t=1
→
t = −1 e
Substitute t = −1 in equation d and solve for s.
4(−1) − s = −5
s=1
Verify the values of t and s by substituting them in equation c.
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LS = −1 −2(1)
= −3
= RS
To obtain the point of intersection, substitute t = −1 in the equations of L1 or s = 1 in the
equations of L2.
L1: x = −3 −(−1)
y = 7 + 4(−1)
z = 2 −1
= −2
=3
=1
The lines intersect at the point (−2, 3, 1).
4. Parametric equations of L1 and L2 are:
L1: x = −2 + 3t, y = 1 − t, z = 1 + t
L2: x = 7 − s, y = −4 + s, z = 3
At a point of intersection, the x-, y-, and z-coordinates of the two lines are equal.
−2 + 3t = 7 − s →
3t + s = 9
c
1 − t = −4 + s →
t+s=5
d
1+t=3
→
t=2
e
Substitute t = 2 in equation d and solve for s.
2+s=5
s=3
Verify the values of t and s by substituting them in equation c.
LS = 3(2) +3
=9
= RS
To obtain the point of intersection, substitute t = 2 in the equations of L1 or s = 3 in the
equations of L2.
L1: x = −2 + 3(2)
y=1−2
z=1+2
=4
= −1
=3
The lines intersect at the point (4, −1, 3).
5. a) The vector equation of a line through A(1, 4, 2) and direction vector [2, −1, 0] is:
[x, y, z] = [1, 4, 2] + t[2, −1, 0]
b) From part a, [x, y, z] = [1 + 2t, 4 − t, 2]
Let t = 1 to obtain (1 + 2(1), 4 −1, 2), or (3, 3, 2)
Let t = 2 to obtain (1 + 2(2), 4 − 2, 2), or (5, 2, 2)
Let t = 3 to obtain (1 + 2(3), 4 − 3, 2), or (7, 1, 2)
c) Parametric equations of the line are x = 1 + 2t, y = 4 − t, z = 2
x−1 y−4
, z = 2.
d) There are no symmetric equations, but we can write 2 =
−1
6. Parametric equations of the 3 lines are:
a) L1: x = 1 + 3s, y = 2 − s, z = 3 + 2s
b) L2: x = −2 − 3t, y = 3 + t, z = 1 − 2t
c) L3: x = 7 + 3p, y = − p, z = 6 + 2p
G
G
G
The direction numbers are m1 = [3, −1, 2], m2 = [−3, 1, −2], and m3 = [3, −1, 2] respectively.
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G
G
G
G
By inspection, m1 = − m2 and m1 = m3 , so the lines are either parallel or identical. A point on
L1 is (1, 2, 3). Determine whether this point also lies on L2 and L3.
1 = −2 − 3t
2=3+t
3 = 1 − 2t
L2:
3t = −3
t = −1
2t = −2
t = −1
t = −1
Since t is the same for each coordinate, (1, 2, 3) also lies on L2. Hence L1 and L2
represent the same line.
L3:
1 = 7 + 3p
2 = −p
3 = 6 + 2p
3p = −6
p = −2
2p = −3
p = −2
3
p = −2
Since p is not the same for each coordinate, (1, 2, 3) does not lie on L3. Hence L1 and
L3 are parallel.
Therefore, only L1 and L2, the lines in parts a and b, represent the same line.
7. a) BC is a direction vector for the line.
BC = [3 − (−1), −2 − 3, −1 − 2]
= [4, −5, −3]
Since the line passes through A(1, 2, 3), it has vector equation:
[x, y, z] = [1, 2, 3] + t[4, −5, −3]
b) AC is a direction vector for the line.
AC = [3 − 1, −2 − 2, −1 − 3]
= [2, −4, −4]
A simpler direction vector is [1, −2, −2]
Since the line passes through B(−1, 3, 2), it has parametric equations:
x = −1 + t, y = 3 −2t, z = 2 −2t
c) AB is a direction vector for the line.
AB = [−1 − 1, 3 − 2, 2 − 3]
= [−2, 1, −1]
Since the line passes through C(3, −2, −1), it has symmetric equations:
x−3 y+2 z+1
= 1 =
−1
−2
8. a) The equation of the plane is of the form:
4x − y + 9z + D = 0
Since R(2, −1, −1) lies on the plane:
4(2) −(−1) + 9(−1) + D = 0
D=0
The equation of the plane is 4x − y + 9z = 0.
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b) The cross product of the direction vectors is normal to the plane.
1 5
2
1
G
n = [2, 1, 5] × [−3, 0, 1]
0
1 −3 0
= [1 − 0, −15 − 2, 0 − (−3)]
= [1, −17, 3]
Hence the equation of the plane is of the form:
x − 17y + 3z + D = 0
Since S(4, 0, −1) lies on the plane:
4 −17(0) + 3(−1) + D = 0
D = −1
The equation of the plane is x − 17y + 3z − 1 = 0.
c) Since the plane passes through A, B, and C, AB and AC are direction vectors for the
plane.
AB = [2 − 4, 3 − (−5), 3 − 1]
AC = [0 − 4, 2 − (−5), −4 − 1]
= [−2, 8, 2]
= [−4, 7, −5]
The cross product of these vectors is normal to the plane.
4
1 −1
4
G
n = [−1, 4, 1] × [−4, 7, −5]
7 −5 −4
7
= [−20 − 7, −4 − 5, −7 − (−16)]
= [−27, −9, 9]
The vector [3, 1, −1] is also normal to the plane.
The equation of the plane is of the form:
3x + y − z + D = 0
Since A(4, −5, 1) lies on the plane:
3(4) + 1(−5) − 1 + D = 0
D = −6
The equation of the plane is 3x + y − z − 6 = 0.
9. Planes that are the same have the same scalar equation, so find the scalar equations of π1 and
π2 .
π1 :
The normal to the plane is the cross product of the direction vectors.
2 4 1 2
G
n = [1, 2, 4] × [1, 0, 2]
0 2 1 0
= [4 − 0, 4 − 2, 0 − 2]
= [4, 2, −2]
The vector [2, 1, −1] is also normal to the plane.
The equation of the plane is of the form:
2x + y − z + D = 0
Since (2, 3, 5) lies on the plane:
2(2) + 3 − 5 + D = 0
D = −2
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The equation of the plane is 2x + y − z − 2 = 0 or 2x + y − z = 2. Hence π1 and π3
represent the same plane.
0 2
1
0
G
π2 :
n = [1, 0, 2] × [0, 2, 2]
2 2 0
2
= [0 − 4, 0 − 2, 2 − 0]
= [−4, −2, 2]
The vector [2, 1, −1] is also normal to the plane.
The equation of the plane is of the form:
2x + y − z + D = 0
Since (2, 3, 3) lies on the plane:
2(2) + 3 − 3 + D = 0
D = −4
The equation of the plane is 2x + y − z − 4 = 0 or 2x + y − z = 4. Hence π1 and π3
do not represent the same plane.
Therefore, only π1 and π3 represent the same line.
10. Since the lines are parallel to the plane, the cross product of their direction vectors is normal
to the plane.
2 3
1
2
G
n = [1, 2, 3] × [0, −2, 1]
−2 1
0 −2
= [2 − (−6), 0 − 1, −2 − 0]
= [8, −1, −2]
The equation of the plane is of the form:
8x − y − 2z + D = 0
Since A(−3, 1, 2) lies on the plane:
8(−3) − 1(1) − 2(2) + D = 0
D = 29
The equation of the plane is 8x − y – 2z + 29 = 0.
11. Since the plane passes through points P and Q, PQ is a direction vector for the plane.
PQ = [3 − 2, 2 − 2, 1 − 2]
= [1, 0, −1]
The normal to the plane is perpendicular to PQ and the normal to the plane
4x − y + 2z − 7 = 0 (since perpendicular planes have perpendicular normal vectors).
Therefore:
0 −1 1
0
G
n = [1, 0, −1] × [4, −1, 2]
−1
2 4 −1
= [0 − 1, −4 − 2, −1 − 0]
= [−1, −6, −1]
The vector [1, 6, 1] is also normal to the plane.
The equation of the plane is of the form:
x + 6y + z + D = 0
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Since P(2, 2, 2) lies on the plane:
2 + 6(2) + 2 + D = 0
D = −16
The equation of the plane is x + 6y + z − 16 = 0.
G
G
12. The normal to π1 is n1 = [3, −4, 1] and the normal to π2 is n 2 = [a, b, −5].
Perpendicular planes have perpendicular normal vectors. Hence:
[3, −4, 1] ⋅ [a, b, −5] = 0
3a − 4b − 5 = 0
3a = 4b + 5
4
5
a=3b+3
4
5
If a = 3 b + 3 , π1 will be perpendicular to π2.
13. a) The equation of the plane is of the form:
3x − y + 4z + D = 0
Since A(1, 2, 5) lies on the plane:
3(1) − 2 + 4(5) + D = 0
D = −21
The equation of the plane is 3x − y + 4z − 21 = 0.
b) We are given that the plane passes through A(1, 2, 5). Two other points on the plane are
the x-intercept B(7, 0, 0) and the y-intercept C(0, −21, 0), so AB and AC are direction
vectors for the plane.
AB = [7 − 1, 0 − 2, 0 − 5]
AC = [0 − 1, −21 − 2, 0 − 5]
= [6, −2, −5]
= [−1, −23, −5]
A simpler direction vector is [1, 23, 5]
Use the point A(1, 2, 5) to obtain the parametric equations:
x = 1 + 6s + t
y = 2 − 2s + 23t
z = 5 − 5s + 5t
14. a) The parametric equations of the line are x = −4 + 2t, y = −2 + 3t, z = 3 + 2t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
3(−4 + 2t) −(−2 + 3t) + 2(3 + 2t) − 3 = 0
−12 + 6t + 2 − 3t + 6 + 4t − 3 = 0
7t − 7 = 0
t=1
Substitute this value of t into the parametric equations of the line to obtain the point of
intersection.
x = −4 + 2(1)
y = −2 + 3(1)
z = 3 + 2(1)
= −2
=1
=5
The angle between the line and the plane is the complement of the angle between the
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direction vector of the line [2, 3, 2] and the normal vector of the plane [3, −1, 2].
[2, 3, 2] ⋅ [3, − 1, 2]
cosθ =
2 2 + 3 2 + 2 2 3 2 + (−1) 2 + 2 2
6−3+4
=
17 14
7
=
238
θ Ñ63.0°
Therefore, the angle between the line and the plane is 90° − 63° = 27°.
b) The parametric equations of the line are x = 2 + 3t, y = −1 − 2t, z = −5 − t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
3(2 + 3t) + (−1 − 2t) + 7(−5 − t) + 30 = 0
6 + 9t − 1 − 2t − 35 − 7t + 30 = 0
0t = 0
This equation is satisfied by all values of t, so the line lies on the plane. Thus the
intersection of the line and plane is x = 2 + 3t, y = −1 − 2t, z = −5 − t.
15. a) The parametric equations of the line are x = t, y = 4 − 2t, z = −4 + t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane; that is:
t + 5(4 − 2t) + 9(−4 + t) + 16 = 0
t + 20 − 10t − 36 + 9t + 16 = 0
0t = 0
This equation is satisfied by all values of t. Therefore, the line lies on the plane.
b) The parametric equations of the line are x = 1, y = 1 + 2t, z = 1 + t. At a point of
intersection, these values of x, y, and z must satisfy the equation of the plane.
3(1) − 2(1 + 2t) + 4(1 + t) − 5 = 0
3 − 2 − 4t + 4 + 4t − 5 = 0
0t = 0
This equation is satisfied by all values of t. Therefore, the line lies on the plane.
16. The projection of point P is the point P′ on the plane such that PP′ is perpendicular to the
plane. To determine the coordinates of P′, determine the intersection of the line PP′ and the
plane.
The vector [2, 1, −2] is a normal to the plane and a direction vector to the line PP′. Since P
has coordinates (1, −1, 4), parametric equations of PP′ are x = 1 + 2t, y = −1 + t, z = 4 − 2t.
Substituting these equations into the equation of the plane gives:
2(1 + 2t) −1 + t −2(4 − 2t) − 6 = 0
2 + 4t − 1 + t − 8 + 4t − 6 = 0
9t = 13
13
t= 9
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13
Substitute t = 9 into the parametric equation of PP′ to obtain the coordinates of P′.
13
13
13
x = 1 + 2 9 
y = −1 + 9
z = 4 − 2 9 
 
 
35
4
10
= 9
= 9
= 9
 35 4 10 
The projection of P is the point  , ,  .
 9 9 9
17. a) Skew lines are non-parallel, non-intersecting lines. By inspection, the direction vectors
[−1, −2, 2] and [−2, 2, 1] are not scalar multiples of each other, so the lines are not
parallel. Show that the lines do not intersect.
Parametric equations of the lines are:
L1: x = 4 − t
L2: x = −6 − 2s
y = 2 − 2t
y = −2 + 2s
z = −3 + 2t
z=3+s
At a point of intersection, the x-, y-, and z-coordinates of the lines are equal.
→
t − 2s = 10
c
4 − t = −6 − 2s
→
2t + 2s = 4, or t + s = 2
d
2 − 2t = −2 + 2s
→
2t − s = 6
e
−3 + 2t = 3 + s
Use equations c and d.
c − d: −3s = 8
8
s = −3
8
Substitute s = −3 in equation d and solve for t.
8
t −3=2
14
t= 3
Verify these values of s and t by substituting them in equation e.
8
14
LS = 2 3  − −3
   
36
= 3
= 12
≠ RS
Since the system of equations is inconsistent, the lines do not intersect.
Hence the lines are skew lines.
b) Since the planes that contain L1 and L2 are parallel, they have the same normal vector.
The normal is perpendicular to L1 and L2 and hence is the cross product of their direction
vectors.
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G
n = [−1, −2, 2] × [−2, 2, 1]
−2
2
2
1
−1
−2
−2
2
= [−2 − 4, −4 + 1, −2 − 4]
= [−6, −3, −6]
The vector [2, 1, 2] is also normal to the plane.
The equation of parallel planes containing L1 and L2 is of the form:
2x + y + 2z + D = 0
The point (4, 2, −3) lies on L1 and hence on the plane containing L1. Therefore:
2(4) + 2 + 2(−3) + D = 0
D = −4
The equation of the plane containing L1 is 2x + y + 2z − 4 = 0.
The point (−6, −2, 3) lies on L2 and hence on the plane containing L2. Therefore:
2(−6) − 2 + 2(3) + D = 0
D=8
The equation of the plane containing L2 is 2x + y + 2z + 8 = 0.
18. Answers may vary.
c
a) 3x + 2y − z = 0
d
2x + 2y − 3z = 0
Eliminate z.
−3 × c: −9x − 6y + 3z = 0
d: 2x + 2y − 3z = 0
=0
Add: −7x − 4y
Express y in terms of x: 4y = −7x
7
y = −4 x
Let x = 4t, where t is any real number.
∴ y = −7t
Substitute these expressions for x and y in equation c and solve for z.
3(4t) + 2(− 7t) − z = 0
z = −2t
The vector equation of the line of intersection is [x, y, z] = [0, 0, 0] + s[4, −7, −2]
x
y
z
The corresponding symmetric equations are 4 =
=
.
−7 −2
c
b) 2x − y + 2z = 6
x − 3y + 4z = 1
d
Eliminate y.
−3 × c: −6x + 3y − 6z = −18
d:
x − 3y + 4z = 1
− 2z = −17
Add: −5x
Solve for z: 2z = 17 −5x
17 5
z= 2 −2x
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Let x = 2t, where t is any real number.
17
∴z = 2 − 5t
Substitute these expressions for x and z in equation c and solve for y.
17

2(2t) − y + 2 2 − 5t = 6


y = 4t + 17 − 10t − 6
= 11 − 6t
17 

The vector equation of the line of intersection: [x, y, z] = 0,11,  + t[2, −6, −5]
2

17
z− 2
x y − 11
The corresponding symmetric equations are 2 =
=
.
−6
−5
19. 5x + y + z = 9
x+y−z=1
c
d
Eliminate y and express z in terms of x.
c − d: 4x + 2z = 8
z = 4 − 2x
e
Eliminate z and express y in terms of x.
c + d: 6x + 2y = 10
y = 5 − 3x
f
Let x = t, where t is any real number.
Substitute t for x in equations e and f.
Parametric equations of the line of intersection are x = t, y = 5 − 3t, z = 4 − 2t.
20. The plane has equation of the form 2x + 3y + z − 2 + k(5x − 2y + 2z + 4) = 0.
Since the plane passes through the point (0, 0, 0):
−2 + k(4) = 0
4k = 2
1
k=2
Therefore, the equation of the plane is:
1
2x + 3y + z − 2 + 2 (5x − 2y + 2z + 4) = 0
4x + 6y + 2z − 4 + 5x − 2y + 2z + 4 = 0
9x + 4y + 4z = 0
21. a) By inspection, the normal vectors [3, −2, 7] and [2, 1, −5] are not scalar multiples of each
other so the lines are not parallel. Hence the planes intersect in a line.
b) Answers may vary.
Any linear combination of the equations of the planes contains their line of intersection.
Add the equations to obtain 5x − y + 2z − 7 = 0.
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c) Answers may vary.
A plane parallel to either plane will be parallel to their line of intersection.
One possibility is 3x − 2y + 7z − 10 = 0.
22. a) The plane has equation of the form 3x + 4y − z + 5 + k(2x + y + z + 10) = 0.
Since the plane passes through the point (−2, 5, 1):
3(−2) + 4(5) − 1 + 5 + k[2(−2) + 5 + 1 + 10] = 0
18 + 12k = 0
3
k = −2
Therefore, the equation of the plane is:
3
3x + 4y − z + 5 − 2 (2x + y + z + 10) = 0
6x + 8y − 2z + 10 − 6x − 3y − 3z − 30 = 0
y−z−4=0
b) From part a, the plane has equation of the form:
3x + 4y − z + 5 + k(2x + y + z + 10) = 0
(3 + 2k)x + (4 + k)y + (−1 + k)z + (5 + 10k) = 0
Since the plane is perpendicular to the plane 6x + y + 2z + 10 = 0, their normal vectors are
perpendicular and have a dot product of 0.
[3 + 2k, 4 + k, −1 + k] ⋅ [6, 1, 2] = 0
18 + 12k + 4 + k − 2 + 2k = 0
15k = −20
4
k = −3
Therefore, the equation of the plane is:
4
3x + 4y − z + 5 − 3 (2x + y + z + 10) = 0
9x + 12y − 3z + 15 − 8x − 4y − 4z − 40 = 0
x + 8y − 7z − 25 = 0
23. a) To show that the planes intersect at a single point, show that their normal vectors are not
G
G
G
coplanar. The normal vectors are n1 = [1, 2, 3], n 2 = [1, −1, −3], and n3 = [2, 1, 6].
2
3
1
2
G
G
n1 × n 2 = [1, 2, 3] × [1, −1, −3]
−1 − 3
1
−1
= [−6 − (−3), 3 − (−3), −1 − 2]
= [−3, 6, −3]
G
G G
n1 × n 2 ⋅ n3 = [−3, 6, −3] ⋅ [2, 1, 6]
= −6 + 6 − 18
= −18
G
G G
Since n1 × n 2 ⋅ n3 ≠ 0, the planes intersect at a single point.
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b) Write the system as a matrix.
3 − 4
1 2
1 −1 − 3
8

2
1
6 − 14
Copy c.
Replace d with −c + d.
Replace e with −2 × c + e.
2
3 − 4
1
0 − 3 − 6 12


0 − 3
0 − 6
Copy c.
Divide d by −3.
Divide e by −3.
 1 2 3 − 4
0 1 2 − 4 


0 1 0
2
Replace c with −2 × e + c.
Replace d with −e + d.
Divide e by −3.
 1 0 3 − 8
0 0 2 − 6 


0 1 0
2
Copy c.
Divide d by 2.
Interchange d and e.
Copy d.
 1 0 3 − 8
0 1 0
2

0 0 1 − 3
Replace c with −3 × e + c.
Copy d.
Copy e.
1
1 0 0
0 1 0
2

0 0 1 − 3
The point of intersection is (1, 2, −3).
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G
G
G
24. The normal vectors are n1 = [3, 2, 1], n 2 = [1, 2, 3], and n3 = [1, 1, 1].
We can show that the planes intersect in a triangular prism by showing that one of the normal
vectors is a linear combination of the other two normal vectors, but the equations do not
satisfy the same relationship.
G
G
G
Let n1 = s n 2 + t n3 for some scalars s and t.
[3, 2, 1] = s[1, 2, 3] + t[1, 1, 1]
= [s + t, 2s + t, 3s + t]
Since the vectors are equal, their components are equal.
s+t=3 c
2s + t = 2 d
3s + t = 1 e
Use equations c and d.
c − d: −s = 1, or s = −1
Substitute s = −1 in equation c to determine t.
−1 + t = 3
t=4
Verify s = −1 and t = 4 in equation e.
LS = 3(−1) + 4
=1
= RS
G
G
G
Hence n1 = − n 2 + 4 n3 . However, the equations are not the same linear combination of each
other. Therefore, the planes form a triangular prism.
25. a) Write the system as a matrix.
 2 3 − 4
3 1
1

Copy c.
Replace d with −3 × c + 2 × d.
3 − 4
2
0 − 7 14


Copy c.
Divide d by −7.
 2 3 − 4
 0 1 − 2


Replace c with −3 × d + c.
Copy d.
2
2 0
 0 1 − 2


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Divide c by −7.
Copy d.
1
1 0
0 1 − 2 


The solution is x = 1, y = −2.
b) Write the system as a matrix.
 2 − 1 − 4
3 2
1

Copy c.
Replace d with −3 × c + 2 × d.
 2 − 1 − 4
0 7 14


Copy c.
Divide d by 7.
 2 − 1 − 4
0
1
2

Replace c with d + c.
Copy d.
 2 0 − 2
0 1
2

Divide c by 2.
Copy d.
 1 0 − 1
0 1 2 


The solution is x = −1, y = 2.
c) Write the system as a matrix.
1 7
1 − 4
2 − 6 − 5 − 1


 3
2 − 1 6
Copy c.
Replace d with −2 × c + d.
Replace e with −3 × c + e.
1
7
1 − 4
0
2 − 7 − 15

0 14 − 4 − 15
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Copy c.
Copy d.
Replace e with −7 × d + e.
1
7
1 − 4
0
2 − 7 − 15

0
0 45
90
Copy c.
Copy d.
Divide e by 45.
1
7
1 − 4
0
2 − 7 − 15

0
0
1
2
Replace c with −e + c.
Replace d with 7 × e + d.
Copy e.
5
1 − 4 0
0
2 0 − 1

0
0 1 2
Copy c.
Divide d by 2.
Copy e.
5
1 − 4 0
0
1 0 − 12 

0
0 1
2
Replace c with 4 × d + c.
Copy d.
Copy e.
3
1 0 0
0 1 0 − 1 
2

0 0 1
2
1
The solution is x = 3, y = −2 , z = 2.
d) Write the system as a matrix.
1 − 2 − 14
3
2 − 3
4 − 23

 5
4 − 10 − 13
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GDM, Chapter 3, Sample Solutions
Copy c.
Replace d with −2 × c + 3 × d.
Replace e with −5 × c + 3 × e.
1 − 2 − 14
3
0 − 11
16 − 41

0
7 − 20
31
Copy c.
Copy d.
Replace e with 7 × d + 11 × e.
1
− 2 − 14
3
0 − 11
16 − 41

0
0 − 108
54
Copy c.
Copy d.
Divide e by −108.
3
1 − 2 − 14


0 − 11 16 − 41
0
0
1 − 1
2

Replace c with 2 × e + c.
Replace d with −16 × e + d.
Copy e.
3
1 0 − 15


0 − 11 0 − 33
0
0 1 − 12 

Copy c.
Divide d by −11.
Copy e.
 3 1 0 − 15


3
0 1 0
0 0 1 − 1 
2

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Replace c with −d + c.
Copy d.
Copy e.
 3 0 0 − 18


3
0 1 0
0 0 1 − 1 
2

Divide c by e.
Copy d.
Copy e.
 1 0 0 − 6


3
0 1 0
0 0 1 − 1 
2

1
The solution is x = −6, y = 3, z = −2 .
26. a) Write the system as a matrix.
3
5
1 2
2 − 1 − 4 − 10


 5 7
6
7 
Copy c.
Replace d with −2 × c + d.
Replace e with −5 × c + e.
2
3
5
1
0 − 5 − 10 − 20


0 − 3 − 9 − 18
Copy c.
Divide d by −5.
Divide e with −3.
 1 2 3 5
0 1 2 4 


0 1 3 6
Copy c.
Copy d.
Replace e with –d + e.
 1 2 3 5
0 1 2 4 


0 0 1 2
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Replace c with −3 × e + c.
Replace d with −2 × e + d.
Copy e.
 1 2 0 − 1
0 1 0
0

0 0 1 2
Replace c with –2 × d + c.
Copy d.
Copy e.
 1 0 0 − 1
0 1 0
0

0 0 1 2
The solution is x = −1, y = 0, z = 2.
The 3 planes intersect at a single point.
b) Write the system as a matrix.
5
4 − 3 2
1 − 2
1
3

 3
4 − 1 − 5
Copy c.
Replace d with c − 4 × d.
Replace e with −3 × c + 4 × e.
2
5
4 − 3
0
5 − 2 − 7 

0 25 − 10 − 35
Copy c.
Copy d.
Replace e with −5 × d + e.
2
5
4 − 3
0
5 − 2 − 7 

0
0
0
0
Replace c with 3 × d + 5 × c.
Divide d by 5.
Copy e.
4
4
20 0
 0 1 − 2 − 7
5
5

 0 0
0
0
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Divide c by 20
Copy d.
Copy e.
1
1
1 0
5
5

7
2
0 1 − 5 − 5 
0 0
0
0


The corresponding equations are:
1
1
x+5z=5
7
2
y − 5 z = −5
0z = 0
Therefore, the 3 planes intersect in a line. The parametric equations of the line are:
1 1
7 2
x = 5 − 5 t , y = −5 + 5 t , z = t .
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Chapter 3 Self-Test, Sample Solutions
1. The lines have direction vectors [1, 5] and [2, 1]. By inspection, the direction vectors are not
scalar multiples of each other, so the lines are not parallel. Hence the lines intersect.
The angle of intersection is the acute angle between the direction vectors.
[1, 5] ⋅ [2,1]
cosθ =
12 + 5 2 2 2 + 12
1(2) + 5(1)
=
26 5
7
=
130
θ = 52.1°
The angle of intersection is 52.1°.
2. Since both lines have direction vector [1, −2, 3], they are either parallel or coincident.
If the lines are parallel, points on L1 do not lie on L2.
A point on L1 is (2, 1, −1). Substitute these coordinates into the equation of L2.
x−4 2−4
y+3 1+3
z − 1 −1 − 1
=
1 = 1
3 =
3
−2
−2
2
= −2
= −2
= −3
The ratios are not equal so (2, 1, −1) does not lie on L2.
Hence the lines are parallel.
3. Skew lines are non-parallel, non-intersecting lines.
A direction vector for L1 is:
AB = [1 − 1, −1 − 1, −1 − 1]
= [0, −2, −2]
A simpler direction vector is [0, 1, 1].
A direction vector for L2 is:
CD = [−1 − (−1), 1 − (−1), −1 − 1]
= [0, 2, −2]
A simpler direction vector is [0, 1, −1]. By inspection, [0, 1, 1] and [0, 1, −1] are not scalar
multiples of each other, so L1 and L2 are not parallel.
Verify that the lines do not intersect.
Parametric equations of the lines are:
L1: x = 1
L2: x = −1
y=1+t
y = −1 + s
z=1+t
z=1−s
At a point of intersection, the x-, y- and z-coordinates of the lines are equal. This condition is
not satisfied by the x-coordinates of the lines, so the lines do not intersect.
Hence the lines are skew lines.
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4. AB = [1 – 2, 2 – (–6), –4 – (–1)]
= [–1, 8, –3]
G
n × AB = [3, 1, –2] × [–1, 8, –3]
= [13, 11, 25]
The equation of the plane is of the form:
13x + 11y + 25z + D = 0
Since A(2, −6, −1) lies on the plane:
13(2) + 11(–6) + 25(−1) + D = 0
–65 + D = 0
D = 65
The equation of the plane is 13x + 11y + 25z + 65 = 0.
5. The point B(−1, 1, 2) lies on the line, and hence on the plane.
The direction vector of the line [2, −3, −3] and vector AB are direction vectors for the plane.
AB = [−1 − 3, 1 − (−1), 2 − 1]
= [−4, 2, 1]
The cross product of the direction vectors is normal to the plane.
−3 −3
2 −3
G
n = [2, −3, −3] × [−4, 2, 1]
2
1 −4
2
= [−3 − (−6), 12 − 2, 4 − 12]
= [3, 10, −8]
The equation of the plane is of the form:
3x + 10y − 8z + D = 0
Since A(3, −1, 1) lies on the plane:
3(3) +10(−1) − 8(1) + D = 0
9 − 10 − 8 + D = 0
D=9
The equation of the plane is 3x + 10y − 8z + 9 = 0.
6. a) The parametric equations of the line are x = 4 + 2t, y = −t, z = 11 + t.
At a point of intersection, these values of x, y, and z satisfy the equation of the plane.
4 + 2t + 3(−t) − (11 + t) + 1 = 0
4 + 2t − 3t − 11 − t + 1 = 0
−2t = 6
t = −3
Substitute t = −3 into the equations of the line to obtain the point of intersection.
x = 4 + 2(−3)
y = −(−3)
z = 11 − 3
= −2
=3
=8
The point of intersection is (−2, 3, 8).
b) The parametric equations of the line are x = 1 + 2t, y = −1 + 4t, z = 2 + t.
At a point of intersection, these values of x, y, and z satisfy the equation of the plane.
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4(1 + 2t) −3(−1 + 4t) + 4(2 + t) − 15 = 0
4 + 8t + 3 − 12t + 8 + 4t − 15 = 0
0t = 0
This equation is satisfied by all values of t, so the line lies on the plane. Thus
the intersection of the line and plane is x = 1 + 2t, y = −1 + 4t, z = 2 + t.
7. Answers may vary.
5x+ 4y + 3z = 2
3x + 2y + z = 0
Eliminate y.
: 5x + 4y + 3z = 2
−2 × : −6x − 4y − 2z = 0
z=2
Add: −x +
Solve for z: z = 2 + x
Let x = t, where t is any real number.
∴z=2+t
Substitute these expressions for x and z in equation and solve for y.
3t + 2y + 2 + t = 0
2y = −2 −4t
y = −1 − 2t
The parametric equations of the line of intersection are: x = t, y = −1 − 2t, z = 2 + t.
8. Answers may vary.
To show that three planes intersect at a single point, show that their normal vectors are not
G G G
coplanar. That is, show that n1 × n 2 ⋅ n3 ≠ 0 .
For example, consider the planes:
π1: x + y = 2
π2: y + z = 4
π3: x + z = 6
G
G
G
The normals are n1 = [1, 1, 0], n 2 = [0, 1, 1], and n3 = [1, 0, 1].
1 0 1 1
G G G
n1 × n 2 ⋅ n3 = [1, 1, 0] × [0, 1, 1] ⋅ [1, 0, 1]
1 1 0 1
= [1, −1, 1] ⋅ [1, 0, 1]
=1+1
=2
G G G
Since n1 × n 2 ⋅ n3 ≠ 0 , the planes intersect at a single point.
9. a) Write the system as a matrix.
3 2 − 3
5 4
2

1
2
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GDM, Chapter 3, Sample Solutions
Copy
.
Replace
with −5 ×
Replace
with
+3×
−
.
.
3
0

1

0
.
Copy
 3 2 − 3
0 2 21


Divide 1 by 3.
Divide 2 by 2.
0 − 24
2
21
0 − 8
21
1
2 
21
The solution is x = −8 and y = 2 .
6 − 2 2
1
2 − 5
4 3

7
3 − 1 1
b) Write the system as a matrix.
Copy
Replace
with −2 ×
+
.
Replace
with −7 ×
+
.
Copy
.
Copy
.
Replace
with
with 3 ×
Copy
.
Copy
.
.
+17 ×
.
with −8 ×
Replace
.
.
0 0 0
1
0 − 17 8 − 1


0
0 7 14
0 0
0
1
0 − 17 8 − 1


0
0 1 2
by 7.
Divide
Copy
−2×
.
Replace
Copy
6 − 2 2
1
0 − 17
8 − 1

0
3 −1
1
by (–13).
Divide
Copy
6 −2
2
1
0 − 17
8 − 1

0 − 39 13 − 13
.
.
0 0
0
1
0 − 17 0 − 17 


0
0 1
2
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Copy
by −17.
Divide
Copy
 1 0 0 0
0 1 0 1


0 0 1 2
.
.
The solution is x = 0, y = 1, and z = 2.
10. Let x be the hundred’s digit, y be the ten’s digit, and z be the one’s digit.
From the question, we obtain the following equations:
x + y + z = 21
→
x + y + z = 21
→
9y − 9z = −18, or y − z = −2
100x + 10z + y = 100x + 10y + z + 18
100y + 10x + z = 100x + 10y + z + 180 → 90x − 90y = −180, or x − y = −2
Solve the system of equations using row reduction.
1
1 21
1

Write the system as a matrix.
1 − 1 − 2
0
 1 − 1 0 − 2
Copy .
Copy .
Replace with − +
1
1
21
1
0
1 − 1 − 2

0 − 2 − 1 − 23
Copy .
Copy .
Replace with 2 ×
1
21
1 1
0 1 − 1 − 2 


0 0 − 3 − 27 
.
+
.
Copy .
Copy .
Divide by −3.
1 21
1 1
0 1 − 1 − 2 


0 0
1
9
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Replace with − + .
Replace with + .
Copy .
 1 1 0 12
0 1 0 7 


0 0 1 9
Replace
Copy .
Copy .
1 0 0
0 1 0

0 0 1
with −
+
.
5
7 
9
The solution is x = 5, y = 7, z = 9 so the number is 579.
Chapter 3 Self-Test, Sample Solutions
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