1. No category

Math 133, Chapter 11 Practice
1. The cranes in the following figure are holding a 10650-pound steel girder in midair. Where
F1 is at a 34◦ angle with the horizontal and F2 is at a 53◦ angle with horizontal (in the
diagram below A = 34◦ and B = 53◦ ). Find the magnitudes of the vectors F1 and F2 ,
round your answers to the nearest pound.
Solution: Let a denote the magnitude of F1 and let b denote the magnitude of F2 ,
and let w be the downward weight vector for the girder. Notice also that F1 forms a
(180 − 34)◦ = 146◦ angle in standard position with the positive x-axis. Then
F1 = ahcos(146◦ ), sin(146◦ )i,
F2 = bhcos(53◦ ), sin(53◦ )i,
w = h0, −10650i
Then F1 + F2 + w = 0 because the girder is held in place the vectors sum to the zero
vector. Considering the i-component of the vectors we obtain
a cos(146◦ ) + b cos(53◦ ) = 0 and so b =
(1)
−a cos(146◦ )
.
cos(53◦ )
Now, j component of sum of the vectors F1 , F2 and w is 0 and so we have
(2)
a sin(146◦ ) + b sin(53◦ ) − 10650 = 0 and so a sin(146◦ ) + b sin(53◦ ) = 10650
Substituting the relation from (1) into (2) we obtain
a sin(146◦ ) − a
and so a =
sin(146◦ )
b=
cos(146◦ )
· sin(53◦ ) = 10650
cos(53◦ )
10650
≈ 6418.1258. Thus
− cos(146◦ ) tan(53◦ )
−a cos(146◦ )
−6418.1258 cos(146◦ )
=
≈ 8841.3669
cos(53◦ )
cos(53◦ )
Thus, to the nearest pound kF1 k ≈ 6418 pounds and kF2 k ≈ 8841 pounds.
2. A plane is flying with a bearing of 302◦ (measured clockwise from north). Its speed with
respect to the air is 900 kilometers per hour. The wind at the plane’s altitude is from
the southwest at 100 kilometers per hour. What is the true direction of the plane, and
what is its speed with respect to the ground?
Solution: See 11.1, #83 from 10th edition of Larson at Calc Chat.
3. Find a vector v with magnitude 5 in the same direction as h−1, 8i.
p
√
(−1)2 + 82 = 65, and thus the desired vector v is
5
5
40
v = √ h−1, 8i = − √ , √
65
65 65
Solution: First, kh−1, 8ik =
4. Find a vector v such that kvk = 2 and v forms a 150◦ angle with the positive x-axis.
√
Solution: v = 2hcos(150◦ ), sin(150◦ )i = h− 3, 1i.
5. A gun with a muzzle velocity of 2350 feet per second is fired at an angle 5◦ above
horizontal. Find the horizontal and vertical components of the velocity.
Solution: The horizontal component of velocity is (2350)(cos 5◦ ) ≈ 2341.1 feet per
second.
The vertical component of velocity is (2350)(sin 5◦ ) ≈ 204.8 feet per second.
6. To carry a 100-lb cylindrical weight, two workers lift on the ends of short ropes tied to
an eyelet on the top center of the cylinder. One rope makes a 20◦ angle away from the
vertical and the other makes a 30◦ angle (see figure below).
(a) Find each rope’s tension.
(b) Find the vertical component of each worker’s force.
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Solution: (a) We can write the vectors as F1 = ahsin 30◦ , cos 30◦ i and F2 =
bh− sin 20◦ , cos 20◦ i, and W = h0, −100i. The sum of the forces is h0, 0i and so
−a sin 20◦ + b sin 30◦ = 0
and
a cos 20◦ + b cos 30◦ = 100
The first equation implies b = 2a sin 20◦ . Substituting this into the second implies
a=
cos 20◦
100
≈ 65.27
+ 2 cos 30◦ sin 20◦
and then
b ≈ 44.6476.
Thus the tension of the 20◦ rope is 65.27 pounds, and the tension on the 30◦ rope is
44.65 pounds.
(b) The vertical component on the 20◦ rope is 65.27 cos 20◦ ≈ 61.33 pounds, and the
vertical component on the 30◦ rope is 44.65 cos 30◦ ≈ 38.67 pounds. (These should
add to 100 pounds.)
7. (a) Find the midpoint of (−7, −7, −5) and (3, −5, 5).
(b) Find the distance from the midpoint in (a) to (−7, −7, −5).
(c) Write the equation of a sphere whose end points of a diameter are (−7, −7, −5) and
(3, −5, 5).
Solution: (a) The midpoint is
−7 + 3 −7 + −5 −5 + 5
,
,
= (−2, −6, 0).
2
2
2
p
(x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 for the
(b) Using the distance formula d =
distance between the two points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) we find the distance is
p
√
√
d = (3 − −7)2 + (−5 − −7)2 + (5 − −5)2 = 25 + 1 + 25 = 51
(c) The center of the √
sphere is the midpoint (−2, −6, 0) of the two points, and the
radius is the distance 51 from (b). Recalling that a sphere with radius r and center
(x0 , y0 , z0 ) has equation
(x − x0 )2 + (y − y0 )2 + (z − z0 )2 = r2
we obtain that the equation of the desired sphere is
(x + 2)2 + (x + 6)2 + z 2 = 51
8. Find the center and radius of the sphere given by
x2 − 8x + y 2 + 10y + z 2 = −40
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Solution: We complete the squares to write the equation in standard form
x2 − 8x + 42 + y 2 + 10y + 52 + z 2 = −40 + 42 + 52
and so the equation of the sphere in standard form is
(x − 4)2 + (y + 5)2 + z 2 = 1
√
This sphere has center (4, −5, 0) and radius 1 = 1.
9. A television camera weighing 120 pounds is supported by a tripod. The tops of the legs
join at the √
point (0, 0, 4) in√ the xyz-plane, and the bases of the legs are at the points
(0, −1, 0), ( 23 , 12 , 0) and (− 23 , 12 , 0) respectively. Represent the force exerted on each leg
of the tripod as a vector.
Solution: See Example 7, Section 11.2 of, 10th edition of Larson at at Calc Chat.
10. (a) Find a unit vector in the same direction as w = h−3, 4, −1i.
(b) Find a vector of length 6 in the same direction as w = h−3, 4, −1i.
(c) Find a vector of length 6 in the opposite direction as w = h−3, 4, −1i.
Solution: (a) Given any nonzero vector w, apunit vector in the same direction
as w
√
1
2
2
2
is given by u = kwk w. In this case, kwk = (−3) + (4) + (−1) = 26. So the
desired unit vector is
1
−3
4
−1
u = √ h−3, 4, −1i = √ , √ , √
26
26 26 26
(b) The desired vector is 6u where u is as in (a):
−18 24 −6
6u = √ , √ , √
26 26 26
(c) The desired vector is −6u where u is as in (a):
18 −24 6
−6u = √ , √ , √
26 26 26
11. Use vectors to find the point R that lies
the point Q(−6, 0, 4).
1
20
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of the way from the point P (−3, −1, 2) to
−→
Solution: We will find R by finding the vector v = P Q and then multiplying it by
1
and then adding that vector to the point P . Then
20
−→
v = P Q = h−3, 1, 2i
and so we find
1
R = (−3, −1, 2)+ h−3, 1, 2i =
20
−3
1
2
−3 +
, −1 + , 2 +
= (−3.15, −0.95, 2.10)
20
20
20
12. With the help of dot products, show that ku + vk2 + ku − vk2 = 2kuk2 + 2kvk2 . Explain
why this is called the parallelogram identity.
Solution: Use the fact for any vector w properties of dot products ensure kwk2 =
w · w. Thus
ku + vk2 + ku − vk2 =
=
=
=
(u + v) · (u + v) + (u − v) · (u − v)
u · u + 2(u · v) + v · v + u · u − 2(u · v) + v · v
2(u · u) + 2(v · v)
2kuk2 + 2kvk2 .
This is the parallelogram identity, because a parallelogram whose sides are parallel to
u and v has diagonals u + v and u − v. Thus the sum of the squares of the lengths
of the four sides is equal to the sum of the squares of the lengths of the diagonals.
13. (a) Do the points (2, 9, 1), (3, 11, 4), (0, 10, 2) and (1, 12, 5) form the vertices of a parallelogram? If so, is it a rectangle?
(b) Find a unit vector in the same direction as v = h−4, 3, 7i. Then find a vector of
magnitute 10 in the direction opposite of v.
Solution: (a) Yes, see solution to 11.2#69 of the 10th edition of Larson at Calc Chat,
the points form the vertices of a parallelogram. No, it is not a rectangle because the
dot product of vectors representing sides sharing a vertex is not 0.
√
√
(b) First, kvk = 16 + 9 + 49 = 74, so a unit vector in the same direction as v
is u = √174 h−4, 3, 7i and a vector of magnitude 10 in the opposite direction of v is
w = √174 h4, −3, −7i.
14. Consider the vectors u = h3, 1, 3i and v = h4, 5, 4i.
(a) Find kuk and kvk.
(b) Find the dot product u · v.
(c) Find the angle between u and v. Round your answer to the nearest 0.01 degree.
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Solution: (a) The magnitudes of the vectors are:
p
p
√
√
kuk = (3)2 + (1)2 + (3)2 = 19 and kvk = (4)2 + (5)2 + (4)2 = 57
(b) The dot product is
u · v = (3)(4) + (1)(5) + (3)(4) = 29
(c) The angle θ between the vectors satisfies
u·v
29
29
cos θ =
=√ √ =√
kukkvk
19 57
1083
and so
θ = arccos
29
√
1083
≈ 28.21◦
15. Find the projection of the vector v = h2, −3, 3i onto the vector w = h4, 2, 3i, find also
the component of v that is orthogonal to w.
Solution: For this, we will need the dot product
v · w = (2)(4) + (−3)(2) + (3)(3) = 11
and kwk2 = (4)2 + (2)2 + (3)2 = 29. Then the projection of v onto w is the vector
44 22 33
11
v·w
, ,
w = h4, 2, 3i =
v1 = Projw v =
kwk2
29
29 29 29
Then the component of v that is orthongonal to w is v2 = v − Projw v and so
44 22 33
14 −109 54
, ,
,
,
v2 = h2, −3, 3i −
=
29 29 29
29 29 29
(To check your answer, notice that v1 is clearly parallel to w since v1 is a multiple
of w. Check also that v1 + v2 = v and that v2 · w = 0 and so v2 is orthogonal to w.)
16. A constant force F = h−3, −6, 0i moves an obect along a straight line from the point
P (2, 7, 1) to the point Q(3, 0, −2). Find the work done if distance is measured in meters
and the magnitude of the force is measured in Newtons.
Solution: The vector with intial point P and terminal point Q is given by
−→
v = P Q = h1, −7, −3i
and then the work done is given by F · v, that is
work = h−3, −6, 0i · h1, −7, −3i = (−3)(1) + (−6)(−7) + (0)(−3) = 39 Nm
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17. Suppose u = h4, −1, 4i and v = h4, 4, 0i. Find u × v and v × u
Solution: First we compute
i
j k
u × v = 4 −1 4
4
4 0
= [(−1)(0) − (4)(4)]i − [(4)(0) − (4)(4)]j + [(4)(4) − (4)(−1)]k
= h−16, 16, 20i
From properties of cross products
v × u = −(u × v) = −h−16, 16, 20i = h16, −16, −20i
18. Find the area of a parallelogram with vertices A = (5, 2, 0), B = (3, 6, 4), C = (5, −2, 3)
and D = (3, 2, 7) and find the area of the triangle with vertices ABC.
−→
−→
Solution: Let u = AB = h−2, 4, 4i and v = AC = h0, −4, 3i. Then
i
j
k
4 4 = h28, 6, 8i
u × v = −2
0 −4 3 p
√
(28)2 + (6)2 + (8)2 . The parallelogram has area 884; the
and so ku × vk√=
884
triangle area is
.
2
19. Find the volume of a parallelepiped with adjacent edges u = h1, 3, 1i, v = h0, 6, 6i and
w = h−4, 0, −4i.
Solution: The volume is the absolute value of the triple scalar product
1 3
1
6 = −72.
u · (v × w) = 0 6
−4 0 −4 So the volume is 72 units cubed.
20. A bicycle pedal is pushed straight downward by a foot with a 30 Newton force. The
shaft of the pedal is 22 cm long. If the shaft is 11 degrees above the horizontal, what is
the magnitude of the torque about the point where the shaft is attached to the bicycle?
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Solution: The torque vector can be found by using a cross-product, but the magnitude follows from trigonometry as follows, where 22 cm is converted to 0.22 m:
magnitude of torque = (30)(0.22) cos(11◦ ) ≈ 6.47874 Nm
21. Find (a) parametric and (b) symmetric equations for the line passing through the points
P (−3, −1, −4) and Q(−2, −2, −7).
−→
Solution: A direction vector for the line is the vector v = P Q. Thus
v = h−2 − (−3), −2 − (−1), −7 − (−4)i = h1, −1, −3i
(a) Recalling that parametric equations for a line through a point P (x0 , y0 , z0 ) in a
direction v = ha, b, ci can be obtained as
x = x0 + at, y = y0 + bt z = z0 + ct
we use P (−3, −1, −4) and v = h1, −1, −3i to obtain parametric equations
x = −3 + 1t, y = −1 − 1t, z = −4 − 3t
(b) Solving the parametric equations for the parameter, we obtain the symmetric
equations
x+3
y+1
z+4
=
=
1
−1
−3
22. Find an equation for the plane that passes through the points A(0, −2, 0), B(−3, 2, 2),
and C(0, −3, 3).
−→
−→
Solution: Let u = AB = h−3, 4, 2i and v = AC = h0, −1, 3i. We will use the cross
product u × v to find a normal vector n for the plane. Then
i
j
k
4 2 = h14, 9, 3i
n = u × v = −3
0 −1 3 Recall that a plane with normal vector n = ha, b, ci passing through a point (x0 , y0 , z0 )
has equation
a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0
In our case, we use the point (0, −2, 0) with n = h14, 9, 3i to obtain the equation
14(x − 0) + 9(y + 2) + 3(z − 0) = 0.
and so 14x + 9y + 3z = −18.
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23. Given a point Q(x0 , y0 , z0 ) and a plane ax + by + cz + d = 0, show that the distance from
|ax0 + by0 + cz0 + d|
√
Q to the plane is
.
a2 + b 2 + c 2
Solution: See Section 11.5 in your text.
24. Find the distance from the point P (−3, 0, 4) to the line
x = −4 + 3t, y = −2 + 3t, z = 5 + 1t
Solution: Let Q be the point (−4, −2, 5) that is obtained by letting t = 0 on the
line. Let
−→
u = P Q = h−1, −2, 1i
and let v = h3, 3, 1i be a direction vector for the line. Then the distance from P to
the line is given by
ku × vk
d=
kvk
So we next find u × v:
i
j k u × v = −1 −2 1 3
3 1 = [(−2)(1) − (3)(1)]i − [(−1)(1) − (3)(1)]j + [(−1)(3) − (3)(−2)]k
= h−5, 4, 3i
and thus the distance is
√
kh−5, 4, 3ik
50
=√
d=
kh3, 3, 1k
19
25. Find parametric and equations for the line passing through the point P (2, −2, −5) and
that is perpendicular to the plane 3x + 6y − 13z = −4. The find the point Q at which
the line intersects the xy-plane.
Solution: The line passes through the point (2, −5, −5) and has direction vector
v = h3, 6, −13i. Thus parametric equations for the line are
x = 2 + 3t, y = −2 + 6t, z = −5 − 13t
The line passes through the xy-plane when z = 0. Thus −5 − 13t = 0, and so
t = −5/13. Thus
(3)(−5)
11
=
13
13
−56
and so Q is the point 11
,
,
0
.
13 13
x=2+
and
Page 9
y = −2 +
(6)(−5)
−56
=
13
13
26. The two planes x − 7y − 1z = 2 and −x + 9y + 5z = 0 are not parallel, so the must
intersect along a line. Find parametric equations for the line of intersection.
Solution: (Solution 1 via Systems of Equations). The equations of the planes form a
system of two linear equations in three variables, so the system has either no solution
(the planes are parallel and don’t intersect) or it has infinitely many solutions (the
planes are not parallel and they intersect in a line). The parametric form of the
solutions will be the equation of the
line of intersection. The augmented matrix for
1 −7 −1 2
the system is
which we reduce as follows.
−1
9
5 0
1 −7 −1 2
−1
9
5 0
−→
−→
1 −7 −1 2
R1 + R2
0
2
4 2
1 −7 −1 2
1
R
0
1
2 1
2 2
From the last row have y + 2z = 1 and so we let z = t, and then y = 1 − 2t, and then
the first row implies
x = 2 + 7y + 1z = 2 + 7(1 − 2t) + 1t = 9 − 13t
Thus parametric equations for the line of intersection are
x = 9 − 13t, y = 1 − 2t, z = t
Solution: (Solution 2 using Cross-Products). The line of intersection will be perpendicular to the normal vector of both planes. So a direction vector for the line will
be the cross product of the normal vectors of the planes which are n1 = h1, −7, −1i
and n2 = h−1, 9, 5i. Then
i
j
k n1 × n1 = 1 −7 −1 −1
9
5 = [(−2)(5) − (9)(−1)]i − [(1)(5) − (−1)(−1)]j + [(1)(5) − (9)(−1)]k
= h−26, −4, 2i
any nonzero multiple of the above vector will work, and so we will use
1
v = h−26, −4, 2i = h−13, −2, 1i.
2
as a direction vector for the line of intersection. Now we need a point on the line, so
we need one solution to the equations
x − 7y − 1z = 2
and
Page 10
− x + 9y + 5z = 0
which we put in an augmented matrix.
1 −7 −1 2
−→
−1
9
5 0
−→
1 −7 −1 2
R1 + R2
0
2
4 2
1 −7 −1 2
1
R
0
1
2 1
2 2
The second row implies y + 2z = 1, so we set z = 0, and then y = 1, then the first
row implies x = 2 + (7)(1) = 9. We now have a point and a direction vector, so the
line of intersection has parametric equations
x = 9 − 13t, y = 1 − 2t, z = t
27. Find the distance from the point Q(−4, −3, −3) to the plane 3x − 5y − 7z = 9.
−→
Solution: First, find a point P on the plane, for example, (3, 0, 0). Then P Q =
h7, −3, −3i and let n = h3, −5, −7i be the natural normal vector for the plane. Then
the distance is
−→
|(3)(7) + (−5)(−3) + (−7)(−3)|
57
|P Q · n|
√
=
=√
d=
knk
32 + 52 + 72
83
28. Find the distance between the two parallel planes 4x − 2y + 5z = 4 and 4x − 2y + 5z = 12.
Solution: First, find a point P on the first plane, for example, (1, 0, 0). Find the
distance from that point to the other plane. For that we observe Q(3, 0, 0) is on the
−→
second plane. Then P Q = h−2, 0, 0i and let n = h4, −2, −5i be the natural normal
vector for the plane(s). Then the distance is
−→
|(4)(−2)|
8
|P Q · n|
=√
=√
d=
2
2
2
knk
4 +2 +5
45
29. Sketch the surfaces given by:
x2 y 2 z 2
(i)
+
+
= 1.
9
16
9
(ii) y 2 = 4x2 + 9z 2 .
(iii) 4x2 − 4y + z 2 = 0.
(iv) 4x2 − y 2 + 4z 2 = 4.
(v) 4x2 − y 2 + 4z = 0.
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Solution: See graphs given for Exercises 1 - 6 in Section 11.6. (i) is (c); (ii) is (b);
y2
(iii) is (d); (iv) is (f); (v) is (a) the saddle function z =
− x2 .
4
Page 12