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01_HVZ_SM_En_txt Page 1 Wednesday, June 3, 2009 11:58 AM
Vectors in two and
three dimensions
Prep Zone
Prepare for this chapter by attempting the following questions. If you have difficulty with a question,
refer to the section in the Heinemann VCE Zone Enhanced textbook indicated in the icon next to the
question. Fully worked solutions to every question in this Prep Zone are contained in the Student
Worked Solutions book.
-----(a) sin 4π
3
(b) tan 300°
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1 (a) Convert each of the following angles given in degrees to radians.
(i) 47° (correct to three decimal places)
(ii) 240° (in terms of π)
(b) Convert each of the following angles given in radians to degrees.
5 πc
(ii) -------- (exactly)
(i) 1.24c (correct to two decimal places)
6
2 Calculate the exact value of each of the following.
(c) sin (−30°) (d) tan ⎛⎝– 5π
−−−⎞⎠
4
(e) cos (−π)
pl
e
y
32
28°
x
θ
c
38°
73°
c
A
B
★
Methods 1&2 CAS 8.1
★
Advanced General
5.3, 5.5
62° 5′
d
m
Sa
A
Methods 3&4 CAS 3.2
85
4 Calculate the unknown values, correct to one decimal place, in each of the following.
(a)
(b)
(c)
C
C
C
a
★
27
x
7.4
Methods 3&4 CAS 3.1
(f) cos π--6-
3 Calculate the values of each of the pronumerals, correct to two decimal places, in each of the
following right-angled triangles.
(a)
(b)
(c)
100
★
12.3
9.8
35°
c
a
6.8
B
A
72°
4.2
B
3π3πQuadrant 2 ( --π2- < θ < π )
Quadrant 3 ( π < θ < ----Quadrant 4 ( ----)
< θ < 2π )
2
2
sin (2π − θ) = -sin θ
sin (π − θ) = sin θ
sin (π + θ) = -sin θ
cos (2π − θ) = cos θ
cos (π − θ) = -cos θ
cos (π + θ) = -cos θ
tan (2π − θ) = -tan θ
tan (π − θ) = -tan θ
tan (π + θ) = tan θ
• For negative angles:
sin (-θ) = -sin θ
cos (-θ) = cos θ
tan (-θ) = -tan θ
• Triangles that are not right-angled can be solved using the sine and cosine rules.
a
b
c
sine rule: ------------- = ------------ = ------------sin A sin B sin C
cosine rule: a2 = b2 + c2 − 2bc cos A
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1.1
Introduction to vectors
Scalar quantities and scalars
Many mathematical and physical quantities are characterised completely by a single number—
its magnitude, which is called a scalar quantity because it measures the object against a given
scale. Examples are angles, lengths, areas, times, masses and temperatures.
There are other quantities, however, for which such a characterisation is not possible; for
example, the shape of a triangle, the location of a point in space, the acceleration or direction
of motion of a particle, and the state of tension in a body. Several numbers are required to
identify each of these quantities. Gradually, mathematical concepts beyond the continuum of
real numbers have been developed to allow us to represent such quantities by a single symbol.
Vector quantities and vectors
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A vector in a plane is an object that can be described by two pieces of information: a length and
a direction. Some examples are the relative position of two points, the velocity and acceleration
of a particle, and the force acting on a particle.
Geometrically, a vector is essentially a directed straight line segment in the plane (or in
space), characterised by its magnitude (length) and direction. A vector is usually indicated by a
directed line, or arrow, of the required length, pointing in the given direction. Unless otherwise
stated, a vector is assumed to be ‘free’. That is, the location of the beginning of the arrow is not
part of the specifications for the vector. Primary examples of vectors in real-life applications
include velocity, displacement, acceleration and force.
Sa
m
pl
e
Scalar quantities and vector quantities
A scalar quantity can be completely described by its magnitude (size) and, if
appropriate, its unit (e.g. time, length, speed, mass, temperature or density).
Mathematically, a scalar quantity is really just a number.
A vector quantity can be completely described only by specifying its magnitude and
its direction. Examples of quantities that must be specified in this way include:
• displacement (change in position of an object)
• velocity (time rate of change of displacement or position)
• acceleration (time rate of change of velocity)
• force (an action that alters the state of motion of an object)
• weight (the gravitational force acting on an object, as distinct from the mass of an
object)
• momentum (the product of an object’s mass and velocity).
Scalar quantities
• time since launch
• distance travelled
• altitude above
ground
• speed relative to
ground
• mass of balloon and
basket
• temperature of
surrounding air
• density of air inside
balloon
Vector quantities
• displacement from
launch site
• velocity relative to
ground
• acceleration
upwards
• force of wind
• weight of basket
• momentum of balloon
• Earth’s magnetic field
strength
Some scalar and vector quantities
for a balloon in flight
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worked example 1
From its starting point O, a hiking group walks for 12 km on a bearing of N50°E to a camp site, A. Next
day the group heads off in the direction S30°E to another camp site, B, which is 8 km from A. What is
the group’s final position relative to O?
Steps
1. Draw a diagram to illustrate the situation.
Solution
N
A
12 km
30° 8 km
N
50°
B
O
OB2 = OA2 + AB2 − 2 × OA × AB × cos ∠OAB
= 122 + 82 − 2 × 12 × 8 × cos 80°
= 174.66
OB = 13.2 km
sin ∠ AOB sin 80°
-------------------------- = -----------------8
13.2
so ∠ AOB = 36.64° (36.59° with unrounded data)
The bearing from O is N(50° + 36.6°)E = N86.6°E.
The group’s displacement from O is 13.2 km on a
bearing of N86.6°E.
Sa
m
pl
e
4. To find the group’s final position relative to O,
its direction relative to O needs to be
determined. Use the sine rule to find ∠AOB
in ΔOAB.
5. State the final direction of the group relative
to O.
6. State the group’s final position relative to O.
∠OAB = 50° + 30° = 80°
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ge
s
2. Use the properties of angles and triangles to
determine ∠OAB. (Note that the unspecified
part of ∠OAB is the same as ∠NOA.)
3. The distance from O to B can now be
calculated using the cosine rule in ΔOAB.
Note: In Worked Example 1, we needed to define the hiking group’s position or displacement
by specifying its distance and its direction from O.
exercise 1.1
Introduction to vectors
Short answer
1 Vera is orienteering in the state forest. She heads from the starting point O for a distance of
8 km on a bearing of N20°E until she reaches checkpoint A. After a rest, Vera moves in the
direction S77°E until she reaches checkpoint B, which is 8 km from A. What is Vera’s
position relative to her starting position O?
2 Three cyclists are riding in a reserve. From the entry gate, Roslyn rides 500 m west and then
400 m east. Francois rides 600 m south and then 800 m west. Gabrielle rides 300 m north
and then 500 m N30°E.
(a) What is the final displacement of each cyclist from the gate?
(b) What distance has each cyclist ridden altogether?
3 Describe the difference between each of the following.
(a) distance and displacement
(b) speed and velocity.
Worked Example 1
1 ● v e c t ors i n t w o and t hre e
DIMENSIONS
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Multiple choice
4 From his house, Frederick rode his bicycle 300 m east and then 400 m north to Hubert’s
house. What is the displacement of Hubert’s house from Frederick’s house?
A 700 m N37°E
B 700 m N53°E
C 500 m N37°E
D 500 m N53°E
E 700 m NE
5 (a) Which one of the following quantities is a vector quantity?
A area
B force
C mass
D temperature E time
(b) Which one of the following quantities is not a vector quantity?
A displacement B velocity
C speed
D acceleration E weight
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Extended answer
6 Members of the Moroka Walking Club are hiking in the mountains. They leave the starting
point O and head north-east for 2 km across a plain until they reach a track at point A. They
follow this track in a direction N10°E for a further 1 km until they reach a road at point B.
They follow the road N60°E for 2.5 km until they reach a hut at point C.
(a) Determine the group’s position relative to the starting point when they reach the
road at B.
(b) Determine the group’s final position at the hut relative to the starting point O.
(c) In what direction would the group need to head in order to walk in a straight line
back to the starting point?
1.2 Vector algebra
Vectors and vector notation
Vectors are denoted by a directed line segment. The length of the segment
represents the magnitude, and its orientation represents the direction.
pl
e
B
m
The vector AB is defined where A is the initial point (the ‘tail’ of the
vector) and B is the terminal point (the ‘head’ of the vector). Or use a single
letter, with a ‘squiggle’ (tilde) underneath, a.
˜
Magnitude of a vector
a
~
A
Sa
The magnitude of the vector AB or a is written as | AB | or | a|. It is a scalar
˜
˜
quantity.
Equality of vectors
Two vectors are said to be equal if and only if they have the same
magnitude and the same direction. For example, consider a
regular hexagon ABCDEF.
A
F
B
C
O
Vectors OA and DO have the same length and direction, so
they are equal vectors. They are also equal to the vectors EF and
CB. Similarly, other groups of equal vectors are
E
D
FO = OC = AB = ED and EO = OB = FA = DC.
Furthermore, | OA | = | OB | = | OC | = | OD | = | OF | = | AB| = | BC | and so on, although these
vectors are not equal because they have different directions.
Note, however, that the effects of two equal vectors may be different. For example, equal
force vectors applied at different points on a see-saw will have different turning effects.
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Parallel vectors
Two vectors are said to be like, parallel vectors if they have the same direction. On the other
hand, two vectors are unlike, parallel vectors if they have opposite directions. For example, in
the hexagon above, AB and FC are like, parallel vectors, whereas AB and CF are unlike,
parallel vectors.
Negative of a vector
B
The negative of a vector AB, or a, is a vector of equal magnitude
˜
but with opposite direction. That is, − AB = BA = − a.
˜
a
~
−a
~
A
Zero vector
B
A
A zero vector is a vector of zero magnitude and no direction. It is represented as a point, and
can be denoted by 0.
˜
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s
Multiplication of a vector by a scalar
Multiplying a vector by a scalar (a number) k results in a vector
with k times the magnitude and either the same or the opposite
direction. For example, the vector 2a is parallel to vector a but
˜
˜
its magnitude is two times as great, whereas the vector − 1--2- a is a
˜
vector parallel to a but with half the magnitude of a and the
˜
˜
opposite direction.
2~
a
−1a
2~
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e
Addition of vectors
a
~
Sa
m
Triangle rule for the addition of
C
b
~
vectors
B
To add two vectors a and b, the vectors are placed head to tail;
a
˜
˜
c
~
i.e. the tail of b is placed at the head of a. The resultant or sum is
~
˜
˜
the vector joining the tail of a, to the head of b (i.e. to form a
˜
˜
triangle, hence the ‘triangle rule for addition’, sometimes called A
the ‘head-to-tail rule’). That is, AB + BC = AC or a + b = c .
˜ ˜ ˜
This method is not limited to the addition of only two vectors; it can be extended to any
number of vectors in space (sometimes called the polygon of vectors).
Parallelogram law for addition
When two vectors are represented by directed line segments and have the same initial point,
we can add them by drawing a parallelogram instead of moving one vector to the head of the
other. The diagonal of the parallelogram, drawn from the initial point, goes to the final point
and represents the sum of the vectors.
Vectors OA = a and OB = b generate the parallelogram OACB, whose diagonal,
˜
˜
OC denotes the sum of OA and OB:
OA + OB = OC or a + b = c
˜ ˜ ˜
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DIMENSIONS
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It can be seen that OC is also the sum of OA and
b
~
A
C
AC or OB and BC.
Since BC is parallel to and equal in magnitude to
c
~
a
~
a
~
OA, then BC = a; and AC is parallel to and equal in
˜
magnitude to OB, then AC = b.
O
B
b
˜
~
Therefore, OA + AC = OC or a + b = c and
˜ ˜ ˜
OB + BC = OC or b + a = c.
˜ ˜ ˜
Commutative law for addition
The order in which two vectors are added is not important; i.e. a + b = b + a. This can be shown
clearly by the parallelogram law for the addition of vectors. ˜ ˜ ˜ ˜
Subtraction of vectors
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Distributive law
If m ∈ R, then m( a + b) = m a + mb.
˜ ˜
˜
˜
Vector subtraction is defined in a similar way to subtraction in algebra. That is, 8 − 3 can be
thought of as 8 + (−3) and we add the negative of the number (or vector) to be subtracted.
The difference between two vectors AB = a and AC = b is found as follows.
˜
˜
AB − AC = AB + (− AC)
= AB + CA = a + (−b)
˜
˜
= CA + AB
B
B
pl
e
= CB = a − b
˜ ˜
a~ − ~
b
a
~
m
a
~
C
C
−b
~
A
Sa
A
b
~
worked example 2
OABCDEFG is a cuboid with OA = a, OC = c and OD = d. Express each of the following vectors in
˜
˜
˜
terms of a, c and d.
˜
(a) DE ˜ ˜
(b) AB
(c) BC
(d) GC
(e) OB
(f) AC
(g) DB
(h) AG
G
F
D
E
d
~
C
B
c
~
O
6
a
~
A
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Steps
1. Identify vectors in (a)–(g) that have the same
magnitude as the given vectors a, c and d.
˜ ˜
˜
2. For the remaining vectors, write them as the
sum or difference of two or more vectors,
and use vector addition to write them in
terms of a, c and d.
˜ ˜
˜
Solutions
(a) DE = OA = a
˜
(b) AB = OC = c
˜
(c) BC = − OA = - a
˜
(d) GC = − OD = -d
˜
(e) OB = OA + AB
=a+c
˜ ˜
(f) AC = AO + OC
= − OA + OC
(g)
= −a + c
˜ ˜
=c− a
˜ ˜
DB = DO + OB
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= − OD + a + c (from part (e))
˜ ˜
= −d + a + c
˜ ˜ ˜
= a +c−d
˜ ˜ ˜
(h) AG = AO + OD + DG
= −a + d + c
˜ ˜ ˜
=c+d− a
˜ ˜ ˜
Vector algebra
pl
e
exercise 1.2
Short answer
1 In the following diagram the line segments OA, AB, CD, etc. have the same length.
Worked Example 2
(e) CG
H
G
K
(g) KC
(h) EI
E
L
b
~
a
O ~
F
FO
Sa
I
(f)
m
If OA = a and OL = b, express the following vectors in terms of a and b.
˜
˜
˜
˜
(a) OC
(b) OK
(c) AC
(d) AD
D
A
B
C
H
2 ABCDEFGH is a cube with AB = a, as shown in the diagram.
˜ that are equal to a.
(a) Name any other marked vectors
(b) Name any marked vectors that are equal to − a. ˜
(c) Which other marked vectors have magnitude˜ |a|?
3 (a) If the magnitude of a (shown) is 2 (i.e. | a| = 2), ˜find the
˜
˜
value of:
1
(i) 2| a|
(ii) --4- a
(iii) −3| a|
˜
˜
˜
(b) Draw the following vectors.
(i) 2 a
˜
(ii) 1--4- a
˜
(iii) 1.5 a
˜
(iv) − a
˜
G
E
F
D
A
a
~
C
B
a
˜
(v) −1.5 a
˜
(vi) − 1--4- a
˜
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DIMENSIONS
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4 For each of the following triangles, find c in terms of a and b.
˜
˜
˜
(a)
(b)
a
~
b
~
a
~
c
~
b
~
c
~
(c)
(d)
a
~
a
~
b
~
b
~
c
~
c
~
5 Define the following vectors in terms of a, b, c and d, as shown in the diagram.
˜ ˜ ˜
˜
C
c
~
b
~
B
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A
a
~
d
~
D
O
(a) OB
(b) OC
(c) DA
6 Show that OA + AB + BC = OC.
(d) DO
m
pl
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7 Show that AB + BO + OA = 0.
˜
8 In ΔOAB, OA = a and OB = b. M and N are the midpoints of AB and OB respectively. Find
˜
˜
OM, AN and MN in terms of a and b.
˜
˜
9 ABCDEF is a regular hexagon with AB = a, BC = b and CD = c. Find each of the following
˜
˜
˜
in terms of a, b and c.
˜ ˜
˜
(a) AC
(b) FA
(c) AD
(d) AE
(e) FC
Sa
10 OABC is a parallelogram with OA = a and AB = b. Find OB and AC in terms of a and b.
˜
˜
˜
˜
Multiple choice
11 In trapezium ABCD, AB is parallel to and twice the length of DC. If DC = c and CB = b,
˜
˜
then DA equals:
A b+c
B b−c
C c −b
D b + 3c
E b + 1--2- c
˜
˜ ˜
˜ ˜
˜ ˜
˜
˜
˜
1
12 Given that AB = --2- BC, which one of the following statements cannot be true?
A Points A, B and C are the vertices of a triangle.
B Points A, B and C are collinear.
1
C | AB | = --2- BC .
D Points A, B and C are coplanar.
E AB is parallel to BC.
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13 A is a point on a circle with diameter BOC.
If OA = a and OB = b, then AC is equal to:
˜
˜
A 2b + a
B −b − a
C a+b
˜ ˜
D a ˜− b ˜
E b˜ − a˜
˜ ˜
˜ ˜
Extended answer
A
B
C
O
14 In the parallelogram OABC, M is the midpoint of BC. N is a
point on OM such that ON = --23- OM.
C
M
B
N
O
A
pa
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s
(a) If OA = a and AB = b, find each of the following in terms of a and b.
˜
˜
˜
˜
(i) OM
(ii) ON
(iii) CN
(iv) CA
(b) Show that A, N and C are collinear, and that CN = --13- CA.
15 In the two triangles shown in the diagram, OA = a and
˜
OB = b. X is a point on AB such that XA = 1--2- BX and C is a point
˜
such that XC = --12- OX.
(a) Express each of the following in terms of a and b.
˜
˜
(i) BA
(ii) BX
(iii) OX
(v) AC
C
X
a
~
O
b
~
B
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e
(iv) XC
A
(b) Show that AC is parallel to OB and half its length.
m
exam focus 1
VCAA 2006 Specialist Mathematics Units 3 & 4, Exam 2, Section 1,
Question 15
1.5
min
Sa
In the parallelogram shown, |a| = 2|b|.
˜
˜
c
~
d
~
b
~
a
~
Which one of the following statements is true?
A. a = 2b
B. a˜ + b ˜= c + d
C. b˜ − d˜ = 0˜ ˜
D. a˜ + c˜ = 0˜
E. a˜ − b˜ = ˜c − d
˜ ˜ ˜ ˜
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DIMENSIONS
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1.3 Vectors in two dimensions
Unit vectors
a
A vector of magnitude 1 is called a unit vector. It is denoted as â , where â = ----˜ and | a| is the
a
˜
˜
˜
magnitude of a.
˜
˜
worked example 3
(a) Define and draw a unit vector â in the direction of vector a (shown), given that a has a magnitude
˜
˜
˜
of 2.
a
~
(b) Hence draw the following vectors:
(i) 3â
(ii) −â
(iii) 1--2- â
˜
˜
˜
Solutions
(a) Since â = 1-2- a , then â is
˜
˜
˜
parallel to but half the
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s
Steps
(a) Unit vector â has a magnitude of 1.
˜
Select a direction
for the vector, and
draw a and â.
˜
˜
(b) Draw the required vectors in terms of â.
˜
â
~
length of a.
˜
(b) (i) 3â is parallel
to â but three times the length
˜
˜
of â.
˜
aˆ
~
3a
~ˆ
Sa
m
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e
(ii) −â is the same length as â, but with opposite
˜
˜
direction.
aˆ
~
−a
~ˆ
(iii) 1--2- â is parallel to â but half the length of â.
˜
˜
˜
aˆ
~
1
aˆ
2~
Vectors in two-dimensional space
(the i –j system)
˜ ˜ i is defined as a unit vector parallel to the
In the i –j system,
j
˜ ˜
˜
x-axis, and
j is defined as the corresponding unit vector
˜ y-axis. Any vector on the Cartesian plane can
parallel to the
be expressed in terms of i and j.
˜
If r = x i + y j, then: ˜
˜
˜ ˜
y
|r| = x 2 + y 2 and tan θ = -x
˜
where θ is the angle made by the vector r with the positive
˜
x-axis.
10
~
y
~i
~r
yj
~
θ
0
x~i
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We can specify a vector in the Cartesian plane completely by expressing it in the form
xi + yj , or by stating its magnitude and direction.
˜
˜
worked example 4
For each of the vectors (i) r = 4i + 3j and (ii) r = 5i – 7j , find:
˜
˜
˜
˜
˜
˜
(a) the magnitude and direction of the vector
(b) a unit vector parallel to the given vector.
Steps
Solutions
(a) 1. Find the magnitude |r|. If r = xi + yj ,
˜
˜
˜
˜
then |r| = x 2 + y 2.
˜
(i) r = 4i + 3j, so |r| =
˜
˜
˜
˜
42 + 32 =
(ii) r = 5i − 7j, so |r| =
˜
˜
˜
5 2 + ( -7 ) 2 =
74 ≈ 8.60
(i) tan θ = 3-4 , so θ = tan −1 ( 3-4 ) = 36.87°
(ii) tan θ = − -75 , so θ = tan −1 (− -75 ) = −54.46°
pa
ge
s
2. Find the direction. If r = xi + yj , then
˜
˜
˜
y
tan θ = --, where θ is the angle made by
x
the vector r with the positive x-axis.
˜
3. Specify the magnitude and direction
of r.
˜
25 = 5
(i) The vector r = 4i + 3j has magnitude 5 units
˜
˜
˜
and a direction of 36.87°
to the positive x-axis.
(ii) The vector r = 5i − 7j has magnitude
˜
˜
˜
74 ≈ 8.60 units and a direction of −54.46° to
the positive x-axis.
r
(b) The unit vector r is defined by rˆ = ---˜- .
r
˜
˜
˜
m
pl
e
(i) rˆ = 1--5- (4i + 3j ) = 0.8i + 0.6j
˜
˜
˜
˜
˜
1
(ii) rˆ = ---------(5i − 7j )
74
˜
˜
˜
Note: It is vital that you check the quadrant in which the vector lies. For example, in part (a) (ii), the
vector is in the 4th (or −1st) quadrant, so the angle is negative.
Unit vectors using CAS
Sa
By writing vectors in matrix form we can use our CAS to perform some of the arithmetic
associated with vectors. The vector 3i – 2j can be written as a row matrix, 3 -2 or as a
˜ ˜
3
column matrix,
. In this work we will use row matrices to represent our vectors.
-2
Find a unit vector in the direction of b = 3i – 2j .
˜ ˜
˜
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DIMENSIONS
11
01_HVZ_SM_En_txt Page 12 Wednesday, June 3, 2009 11:58 AM
Using the TI-Nspire CAS
1. Write the coefficients of the vector as a row
matrix and store to a variable, in our case b.
pa
ge
s
2. Tap Action > Vector > unitV, enter the variable
name and tap E to get the unit vector
coefficients displayed.
From this result you should also be able to
work out the magnitude of the vector.
pl
e
2. Press b > Matrix & Vector > Vector >
Unit Vector. Then press B ) · to find
the coefficients of the unit vector in the
direction of b.
˜ you should also be able to
From this result
work out the magnitude of the vector.
Using the ClassPad
1. Write the coefficients of the vector as a row
matrix and store to a variable, in our case B.
You find the matrix template in the submenu of the ) keypad.
m
If a vector r of magnitude |r| makes an angle of θ with the positive x-axis, then
˜ θ i + |r| sin θ j. ˜
r = |r| cos
˜ ˜
˜ ˜
˜
j
~
Sa
y
~r
θ
0 |r|cos θ i
~
~
~i
|~r|sin θ j
~
x
In this case, |r| cos θ i is the horizontal component and |r| sin θ j is the vertical component
˜
˜
˜
of the vector r. ˜
˜
The process of specifying a vector of known magnitude and direction in component form is
called resolution, or resolving the vector.
12
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01_HVZ_SM_En_txt Page 13 Wednesday, June 3, 2009 11:58 AM
worked example 5
Resolve the following vectors into components in the form r = xi + yj :
˜ 30°˜ to the
˜ positive x-axis.
(a) r is a vector of magnitude 12 units and has a direction of
˜
(b) r is a vector of magnitude 25.35 units and has a direction of 172° to the positive x-axis.
˜
Solutions
(a) Since |r | = 12 and θ = 30°, then
˜
r = 12 cos 30°i + 12 sin 30°j
˜
˜
˜
1
3
- i + 12 × -- j
= 12 × -----2
2
˜
˜
= 6 3 i + 6j
˜
(b) If a vector r of magnitude |r | makes an
(b) Since |r | ˜= 25.35
and θ = 172°, then
˜
˜
˜
angle of θ with the positive x-axis, then
r = 25.35 cos 172°i + 25.35 sin 172°j
r = |r | cos θi + |r | sin θj .
˜
˜
˜
= -25.10i + 3.53j
˜
˜
˜ ˜
˜
˜
˜
Note: Once again, it is vital that you check the quadrant in which the vector lies. For example, in part
(b) the vector is in the 2nd quadrant, so the i component is negative and the j component is positive.
˜
˜
pa
ge
s
Steps
(a) If a vector r of magnitude |r | makes an
˜ the positive x-axis,
˜
angle θ with
then
r = |r | cos θi + |r | sin θj .
˜
˜
˜ ˜
˜
Using CAS to resolve vectors into Cartesian
(rectangular) form
We can change vectors written in the form magnitude and angle into the form r = xi + yj .
˜
˜
˜
To illustrate this we will use the vector from Worked Example 5(a).
Sa
m
pl
e
Using the TI-Nspire CAS
In the Calculator application enter the vector as
the matrix [12 ∠30°].
You need to use the matrix template which can
be found by pressing / r and find the ∠ in
the symbols template, / k . If your CAS is
set in radians mode you need to enter the ° or
π
you can convert the angle to --- , in this case.
6
To get ¢ Rect you press b > Matrix & Vector >
Vector > Convert to Rectangular. Tap · to
finish.
Using the ClassPad
We use toRect (Action > Vector > toRect) which
requires us to enter the vector in matrix form
and include a ∠ before the angle. In M mode
enter the vector as the matrix [12 ∠(30)].
You need to use the matrix template which can
be found by tapping k ) and you find the ∠ by tapping 9 K .
If you are in Deg mode you enter the angle as
30. If your CAS is set in Rad mode you need to
π
enter --- , in this case. You then tap E when
6
you are done. Both forms of the entry are
shown in the screen below.
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DIMENSIONS
13
01_HVZ_SM_En_txt Page 14 Wednesday, June 3, 2009 11:58 AM
Position vectors
y
j
~
P
The position of a point P relative to the origin O can be defined
by the vector OP. If P is the point P (x, y), then OP = x i + y j,
˜
˜
and OP is the position vector of P relative to O.
~i
yj
~
O
θ
x
x~i
worked example 6
For each of the following points P, find the position vector OP = x i + yj and hence find a unit vector in
˜
˜
the direction of OP.
(a) P(2, 3)
(b) P(-4, -5)
Solutions
(a) 1. Determine the position vector
(a) OP = 2i + 3j
˜
˜
OP = xi + yj.
˜
˜
2. Calculate the magnitude of OP.
ˆ
3. Find the unit vector OP.
(b) 1. Determine the position vector
2
2 +3
2
= 13
ˆ = ---------1
(2i + 3j)
OP
13
˜
˜
(b) OP = −4i − 5j
˜
˜
OP =
2
(−4) + (−5)
2
= 41
ˆ = ---------1
(-4i − 5j )
OP
41
˜
˜
Sa
m
ˆ
3. Find the unit vector OP.
OP =
pl
e
OP = xi + yj.
˜
˜
2. Calculate the magnitude of OP.
pa
ge
s
Steps
Unit vectors in the direction of a given point
To illustrate this process we will redo Worked Example 6(a).
Using the TI-Nspire CAS
Just use unitV which is found by pressing b >
Matrix & Vector > Vector > Unit Vector. You then
enter the position inside the matrix template.
14
Using the ClassPad
Just use unitV which is found by tapping
Action > Vector > unitV. You then enter the
position inside the matrix template.
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01_HVZ_SM_En_txt Page 15 Wednesday, June 3, 2009 11:58 AM
Addition and subtraction of vectors in
component form
Given two vectors a = x1 i + y1 j and b = x2 i + y2 j , then the vectors a + b and a − b are defined
˜
˜ ˜
˜ ˜
˜
˜
˜
˜
˜
as follows.
a + b = (x1 + x2) i + (y1 + y2) j
˜ ˜
˜
˜
a − b = (x1 − x2) i + (y1 − y2) j
˜ ˜
˜
˜
worked example 7
If a = 3i + 8j, b = −5i − 4j and c = i − 2j, find each of the following.
˜
˜
˜
˜
˜
˜ (b) b˜ − c
˜
(a) a + b ˜
(c) 3a + 5c
˜ ˜
˜ ˜
˜
˜
Vectors in two dimensions
m
exercise 1.3
pl
e
(c) 1. To calculate 3 a, multiply the coefficients of
˜
a by 3, and similarly
for 5c.
˜
˜
2. To add vectors in component form, add the
coefficients of the i components, and
˜
likewise add the coefficients
of the j
˜
components.
Solutions
(a) a + b = 3i + 8j + (−5i − 4j)
˜ ˜ = (3˜ + −5)i
˜ + (8˜ + −4)j
˜
˜
˜
= −2i + 4j
˜
(b) b − c = −5i − 4j˜ − (i − 2j)
˜ ˜ = (−5˜ − 1)i˜ + (−4
˜ + ˜2)j
˜
= −6i − 2j˜
˜
˜
(c) 3 a = 3(3i + 8j ) = 9i + 24j
5c˜ = 5(i −˜ 2j ) ˜= 5i −˜ 10j ˜
˜ + (5i˜ − 10j )
3 a˜ + 5c˜ = 9i˜ + 24j
˜
˜ = (9˜ + 5)i ˜+ (24˜− 10)j˜
˜
˜
= 14i + 14j
˜
˜
pa
ge
s
Steps
(a) To add vectors in component form, add the
coefficients of the i components, and likewise
˜ of the j components.
add the coefficients
˜
(b) To subtract vectors in component form,
subtract the coefficients of the i components,
˜
and likewise subtract the coefficients
of the j
˜
components.
Sa
Short answer
1 (a) Find a unit vector in the direction of a (see diagram), given
˜
that a has a magnitude of 3.
a
~
˜
(b) Hence, draw the following vectors.
(i) 2â
(ii) â
(iii) 1--2- â
(iv) −â
˜
˜
˜
˜
2 For each of the following vectors, find:
(a) the magnitude and direction of the vector
(b) a unit vector parallel to the given vector.
(iv) d = −12 i − 9 j
(i) a = 8 i + 6 j
(ii) b = 24 i − 7 j
(iii) c = −5 i + 12 j
˜
˜
˜
˜
˜
˜
˜
˜
˜
˜
˜
˜
3 Resolve the following vectors into components in the form r = x i + y j .
˜
˜
˜ positive x-axis.
(a) r is a vector of magnitude 5 units and has a direction of 60° to the
˜
(b) r is a vector of magnitude 25 units and has a direction of 45° to the positive x-axis.
(c) r˜ is a vector of magnitude 33.21 units and has a direction of 121° to the positive x-axis.
(d) r˜ is a vector of magnitude 765.45 units and has a direction of 98°34′ to the positive
˜x-axis.
4 For each of the points P whose coordinates are given, find the position vector OP = x i + y j
˜
˜
and hence find a unit vector in the direction of OP.
(b) P(12, 5)
(c) P(−2, 2)
(d) P(−3, −7)
(a) P(3, −2)
Worked Example 3
Worked Example 4
Worked Example 5
Worked Example 6
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01_HVZ_SM_En_txt Page 16 Wednesday, June 3, 2009 11:58 AM
5 Given that a = 2 i + j, b = 3 i − 2 j and c = −2 i − 5 j, find:
˜
˜ ˜ ˜
˜
˜
˜
(a) a + b
(b) 3 a +˜ 2c
(c)˜ −3 a − 4b
(d) −2 a − 4(b − 2c)
˜ ˜
˜
˜
˜
˜
˜
˜
˜
6 Given that a = i + 3 j, b = −2 i − j and c = −5 i − 3 j, find:
˜ ˜
˜
˜ ˜
˜
˜
(a) the magnitudes˜ of the three
vectors, i.e. | a|,˜ |b| and | c| respectively
˜ ˜
˜
ˆ and cˆ .
(b) the unit vectors â , b
˜ ˜
˜
Multiple choice
7 The magnitude of a = −4 i + 7 j is:
˜
˜
˜
A − 7--4B 33
C 33
D 65
E
65
8 Which one of the following is a unit vector parallel to 2i − 3 j?
˜
˜2
1
3
- ( 2i – 3j )
- i – ------ j
2i
–
3j
A i− j
B --------C
D ----13 ˜ 13
13
˜ ˜
˜
˜ ˜
˜
˜
2
--- i
5˜
– --35- j
˜
E i− j
˜ ˜
pa
ge
s
9 If a = −3 i + 2 j and b = 2 i − j, then 3 a + 5b equals:
˜
˜
˜ ˜
˜
˜
˜
˜
B 19 i + 11 j
C i+ j
D −9 i + 7 j
A −i + j
˜ ˜
˜
˜ ˜
˜
˜
˜
Extended answer
10 OABC is a parallelogram in which OA = 3 i + 2 j and OC = 2 i − 3 j.
˜
˜
˜
˜
(a) Find AB and CB.
(b) Find the diagonals, OB and CA.
Now let M, N and P be the midpoints of AC, OB and OA respectively.
(c) Find ON and OM.
(d) Find CP and BP.
E
exam focus 2
Worked Example 7
pl
e
VCAA 2007 Specialist Mathematics Units 3 & 4, Exam 2, Section 1,
Question 15
16
Sa
A. 3i + 2j
˜
˜
B. - 1-2 i + 1-3 j
˜
˜
C. 2i − 3j
˜
˜
D. 1-2 i − 1-3 j
˜
˜
E. 2i + 3j
˜
˜
m
In the cartesian plane, a vector perpendicular to the line 3x + 2y + 1 = 0 is
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