7
In this
In this
cha
chapter
pter
7A
Vectors and scalars
7B
Position vectors in two
and three dimensions
7C
Multiplying two
vectors — the dot
product
7D
Resolving vectors —
scalar and vector
resolutes
7E
Time-varying vectors
syllabus
syllabus
rref
efer
erence
ence
Core topic:
Vectors and applications
Introduction
268
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dVectors and scalars
Introduction
In mathematics, one of the important distinctions that we make is between scalar quan-tities and vector quantities. Scalar quantities have magnitude only; vector quantities have direction as well as magnitude. Most of the quantities that we use are scalar, and include such measurements as time (for example 1.2 s; 15 min), mass (3.4 kg; 200 t) and area (3 cm2; 400 ha).
However, consider the measurement of velocity. A velocity of 20 km/h has both magnitude and direction. One of the boats shown below may travel 20 km/h north from Townsville, while the other one may travel 20 km/h east from the same point. Although they both are travelling at the same speed (magnitude) they are travelling in different directions; they do not end up in the same place!
Now consider the force involved in Daniel and Anna fighting over who gets to use the television remote control.
Daniel exerts a force of 40 N and Anna exerts a force of 50 N and they apply these forces as shown.
In what direction will the remote control move and what is the force in that direction? That is, what is the resultant force?
The resultant force depends not only on the size of each force but the direction in which the forces are applied. In the following discussion we will develop techniques to find the resultant force.
A vector is a quantity that has magnitude and direction.
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269
Vector notation
A vector is shown graphically as a line, with a head (end) and tail (start). The length of the line indicates the magnitude and the orientation of the line indicates its direction.
In the figure at right, the head of the vector is at point B (indicated with an arrow), while the tail is at point A.
When writing this vector we can use the points A and B to indicate the start and end
points with a special arrow to indicate that it is a vector: . Some textbooks use a single letter, in bold, such as w, but this is difficult to write using pen and paper, so can also be used. The symbol (~) is called a tilde.
Equality of vectors
Since vectors are defined by both magnitude and direction:
two vectors are equal if both their magnitude and direction are equal. In the figure, the following statements can be made:
(directions are not equal)
(magnitudes are not equal).
Addition of vectors
Consider a vector which measures the travel from A to B and another vector, , which measures the subsequent travel from B to C. The net result is as if the person travelled directly from A to C (vector ). Therefore we can say that .
To add two vectors, take the tail of one vector and join it to the head of another. The result of this addition is the vector from the tail of the first vector to the head of the second vector.
Returning to Daniel and Anna who are fighting over the television remote control (see p. 268), we see that the forces they apply to the remote control unit can be repre-sented as a sum of two vectors.
From this figure we are able to get a rough idea of the magnitude and direction of the resultant force. In the following sections, we will learn techniques for calculating the resultant magnitude and direction accurately.
The negative of a vector
If is the vector from A to B, then is the vector from B to A.
We can subtract vectors by adding the negative of the second vector to the first vector.
A
B
AB
w ˜
w ~
u ~ v ~
z ~
u ˜ = v˜ u ˜ ≠w˜ u ˜ ≠z˜
u ˜
A
B C
w ~
u ~
v ~
v ˜
w
˜ w˜ = u˜ +v˜
30° 50 N
40 N Resultant
force
Daniel’s force
Anna’s force
u
˜ –u˜
A
B
u ~
–u ~
270
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dMultiplying a vector by a scalar
Multiplication of a vector by a number (scalar) affects only the magnitude of the vector, not the direction. For example, if a vector has a direction of north and a magnitude of 10, then the vector has a direction of north and magnitude of 30.
If the scalar is negative, then the direction is reversed. There-fore, has a direction of south and a magnitude of 20. Using the vectors shown at right, draw the result of:
a b c d .
THINK WRITE
a Move so that its tail is at the head of .
a
Join the tail of to the head of to find .
b Reverse the arrow on to obtain . b
c Reverse to get . c
Join the tail of to the head of to get which is the same as or .
d Reverse to get . The vectors are now ‘aligned properly’ with the head of
joining the tail of .
d
Join the tail of to the head of to get .
Note that this is the same as .
u ~ v ~
u
˜ +v˜ –u˜ u˜ –v˜ v˜–u˜
1 v
˜ u ˜
u ~ v
~ ~v
2 u
˜ v˜
u ˜+v˜
u ~
v ~ u + v ~ ~
u
˜ –u˜
u ~ –u~
1 v
˜ –v˜
u ~ –v~
2 v
˜
– u
˜ v
˜+u˜
– u
˜–v˜ u
˜ +( )–v˜
u ~
–v~ –v~ +u~
1 u
˜ –u˜
u ˜
– v
˜
–u~ v ~
2 u
˜
– v
˜ v
˜–u˜ u
˜+v˜ –
( )
–u~ v ~
v – u ~ ~
1
WORKED
Example
u
~ 3~u –2~u
N
S E W
u ˜ 3u
˜
2u ˜ –
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
271
Use the vectors shown at right to draw the result of:
a b .
THINK WRITE
a Increase the magnitude of by a factor of 2 and by a factor of 3.
a
Move the tail of to the head of . Then join the tail of to the head of
to get .
b Increase the magnitude of by a factor of 2 and by a factor of 4.
b
Reverse the arrow on to get .
Join the tail of to the head of .
r ~
s ~
2r
˜+3s˜ 2s˜–4r˜
1 r
˜ s
˜
2~r
3~s
2 3s
˜ 2r 2r˜
˜ 3s
˜ 2r+3s˜
2~r
3~s 3s
~
2~r+3~s
1 s
˜ r
˜
4~r
2s~
2 4r
˜ –4r˜
–4r ~
2s~
3 4r
˜
– 2s
˜
–4~r 2s ~– 4~r
2s~
2
WORKED
Example
The parallelogram ABCD can be defined by the two vectors and .
In terms of these vectors, find:
a the vector from A to D
b the vector from C to D
c the vector from D to B.
THINK WRITE
a The vector from A to D is equal to the vector from B to C since ABCD is a parallelogram.
a
b The vector from C to D is the reverse of D to C which is .
b
c The vector from D to B is obtained by adding the vector from D to A to the vector from A to B.
c
c ~
b ~
C
B A
D
b
˜
c
˜
AD c
˜
=
b ˜
CD b
˜ –
=
DB c
˜ – b
˜
+ =
b
˜–c˜
=
3
WORKED
Example
272
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dA cube PQRSTUVW can be defined by the three vectors , and as shown at right.
Express in terms of , and :
a the vector joining P to V
b the vector joining P to W
c the vector joining U to Q
d the vector joining S to W
e the vector joining Q to T.
THINK WRITE
All of the opposite sides in a cube are equal in length and parallel. Therefore all opposite sides can be expressed as the same vector. a The vector from P to V is obtained by
adding the vector from P to Q to the vector from Q to V.
a
b The vector from P to W is obtained by adding the vectors P to V and V to W.
b
c The vector from U to Q is obtained by adding the vectors U to P and P to Q.
c
d The vector from S to W is obtained by adding the vectors S to R and R to W.
d
e The vector from Q to T is obtained by adding the vectors Q to P, P to S and S to T.
e
Q W
S
P U
V T
R
b ~ a
~ c ~
a
˜ b˜ c˜
a
˜ b˜ c˜
PV a
˜+b˜
=
PW a
˜ + +b˜ c˜
=
UQ –b
˜+a˜
= a
˜–b˜
=
SW a
˜ +b˜
=
QT –a
˜ + +c˜ b˜
= b
˜+c˜ –a˜
=
4
WORKED
Example
A boat travels 30 km north and then 40 km west.
a Make a vector drawing of the path of the boat.
b Draw the vector that represents the net displacement of the boat.
c What is the magnitude of the net displacement?
d Calculate the bearing (from true north) of this net displacement vector.
THINK WRITE
a Set up vectors (tail to head), one pointing north, the other west.
a
Indicate the distances as 30 km and 40 km respectively.
b Join the tail of the vector with the head of the vector.
b
1 N
S E W
N (30 km)
~ W ~ (40 km)
2
N ˜ W
˜ N + W N ~ (30 km)
~ ~
W ~ (40 km)
5
WORKED
Example
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273
Vectors and scalars
1 a Draw the result of:
b Draw the result of:
i ii iii
i ii iii
THINK WRITE
c Let R km = length of + . c
The length (magnitude) of can be calculated using Pythagoras’ theorem.
d Indicate the angle between and
+ as θ.
d
Use trigonometry to find θ.
The true bearing is 360° minus 53.13°. Therefore the true bearing is:
= 360°− 53.13°
= 306.87°
1 N
˜ W˜
N (30 km)
~ R = N + W
~ ~ ~ W ~ (40 km)
2 R
˜ R = 302+402
900+1600
=
50 km
=
1 N
˜ N
˜ W˜ N + W N ~(30 km)
~ ~
W ~ (40 km)
θ
2
sinθ 40 50
---=
0.8
=
θ = 53.13°
3
remember
1. Definition: A vector is a quantity that has magnitude and direction.
2. Equality of vectors: Two vectors are equal if both magnitude and direction are equal.
3. Addition of vectors: To add two vectors, take the tail of one vector and join it
to the head of the other. The result of addition is the vector from the tail of the
first vector to the head of the second.
4. Subtraction of vectors: Subtract vectors by adding the negative of the second
vector to the first vector.
5. Multiplication of a vector by a scalar: Multiply the magnitude of the vector by the scalar; maintain the direction of the original vector.
remember
7A
r ~
s ~
WORKED Example
1 r˜+s˜ r˜–s˜ s˜–r˜
WORKED Example 2
2r
274
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n d2 The pentagon ABCDE at right can be defined by the four vectors, , , and .
Find in terms of these 4 vectors:
a the vector from A to D
b the vector from A to B
c the vector from D to A
d the vector from B to E
e the vector from C to A.
3
A girl travels 4 km north and then 2 km south. What is the net displacement vector?
4 In the rectangle ABCD, the vector joining A to B is denoted by and the vector joining B to C is . Which pairs of points are joined by:
5
Consider the following relationships between vectors , and .
i ii
Which of the following statements is true?
6 A rectangular prism (box) CDEFGHIJ can be defined by three vectors , and as shown at right.
Express in terms of , and :
a the vector joining C to H
b the vector joining C to J
c the vector joining G to D
d the vector joining F to I
e the vector joining H to E
f the vector joining D to J
g the vector joining C to I
h the vector joining J to C.
7 A pilot plans to fly 300 km north then 400 km east.
a Make a vector drawing of her flight plan.
b Show the resulting net displacement vector.
c Calculate the length (magnitude) of this net displacement vector.
d Calculate the bearing (from true north) of this net displacement vector.
A 6 km north B 6 km south C 2 km north
D 2 km south E −2 km north
a ? b ?
c ? d ?
A B C D E
B
C
D
E A
v
~ ~u
t~ s
~
WORKED Example
3 s˜ ˜t u˜ v˜
m
multiple choiceultiple choice
B C
D A
v ~
u ~
u
˜ v˜
u
˜ +v˜ u˜–v˜
v
˜–u˜ 3u˜+2v˜–2u˜ –v˜
m
multiple choiceultiple choice
u
˜ v˜ w˜ u
˜ = 2v˜+w˜ w
˜ = v˜–u˜
u
˜ = w˜ u˜ = v˜ u˜ 2 3 ---v
˜
= u
˜ 3 2 ---v
˜
= u
˜ = 3v˜
C G
J F
I
E
D H
t~ s
~ r
~
WORKED Example
4 r˜ s˜ ˜t
r
˜ s˜ ˜t
E
XCEL
Spreadshe
et
Position vector
WORKED Example 5
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275
8 Another pilot plans to travel 300 km east, then 300 km north-east. Show that the resultant bearing is 67.5 degrees. How far east of its starting point has the plane travelled?
9 An aeroplane travels 400 km west, then 600 km north. How far is the aeroplane from its starting point? What is the bearing of the resultant displacement?
10 On a piece of graph paper draw a vector, , that is 3 units east and 5 units north of the origin. Draw another vector, , that is 5 units east and 3 units north of the origin.
On the same graph paper, draw the following vectors.
11 Find the direction and magnitude of a vector joining point A to point B, where B is 10 m east and 4 m north of A.
12 Consider a parallelogram defined by the vectors and , and its associated diagonals, as shown at right. Show that the vector sum of the diagonal vectors is .
a b c d
e f g h
i j
a ˜ b
˜
a
˜+b˜ a˜ +3b˜ a˜ –b˜ b˜ –a˜
3b
˜–4a˜ 0.5a˜+2.5b˜ a˜ –2.5b˜ 4a˜ 2.5a
˜–1.5b˜ b˜ –2.5a˜
a ˜ b˜
2a ˜
b ~
a ~
276
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n d13 Show, by construction, that for any vectors and :
(This is called the Distributive Law.)
14 Show, by construction, that for any three vectors , and :
(This is called the Associative Law.)
15 Show, by construction, that for any two vectors and :
16 As you will learn shortly, vectors can be represented by two values: the horizontal (or x) component and the vertical (or y) component.
Consider the vector , defined by joining the origin to the point (4, 5), and the
vector defined by joining the origin to (2, 3). Find the horizontal and vertical com-ponents of each vector.
Demonstrate, graphically, that the sum has an x-component of 6 (that is, 4 + 2), and a y-component of 8 (that is, 5 + 3).
17 Using the same vectors, and , as in question 16, demonstrate graphically that the difference vector, , has an x-component of 2 and a y-component of 2.
18 Using the same vectors, and , as in question 16, demonstrate graphically that:
a the vector has an x-component of 16 and a y-component of 20
b the vector has an x-component of −4 and a y-component of −6.
19 Using the results from questions 16, 17 and 18, what can you deduce about an algebraic method (as opposed to a graphical method) of addition, subtraction and multiplication of vectors?
20
In terms of vectors and in the figure above, the vector joining O to D is given by:
21
In terms of vectors and , the vector joining E to O above is:
A B C D E none of these
A B C D E none of these
u ˜ v˜ 3 u
˜+v˜
( ) 3u
˜+3v˜
=
a
˜ b˜ c˜ a
˜ +b˜
( ) c
˜
+ a
˜+(b˜+c˜)
=
r ˜ s˜ 3r
˜–s˜ = –(s˜–3r˜)
w ˜ v
˜
w ˜ +v˜
w ˜ v˜ w
˜ –v˜
w ˜ v˜ 4w
˜ 2v
˜ –
m
multiple choiceultiple choice
O
D
b ~ a ~
a ˜ b˜ 3a
˜ +3b˜ 2a˜ +4b˜ 3b˜ –2a˜ 2a˜–3b˜
m
multiple choiceultiple choice
O E
b ~ a ~
a ˜ b˜ 3a
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277
22 A girl walks the following route: 400 m north — 300 m east — 200 m north — 500 m west — 600 m south — 200 m east
Make a vector drawing of these six paths. What is the net displacement vector?
23 Which of the following are vector quantities?
speed velocity displacement force volume angle
24 Which of the following are scalar quantities?
speed time acceleration velocity length displacement
25 A 2-dimensional vector can be determined by its length and its angle with respect to (say) true north. What quantities could be used to represent a 3-dimensional vector?
1. Clear or turn off any graph plots or stat plots. Press [FORMAT], scroll to GridOn
and press . Press and select 4:ZDecimal to obtain a grid as shown.
2. Press , scroll to Par and press .
3. To show vector + (without the arrowhead!) press then at X1T= enter 3T and
at Y1T= enter 2T. (To enter T, use the key.)
4. Press and set Tmin= to 0, Tmax= to 1, Tstep= to 1 and then press .
5. Press and scroll to verify that the vector has coordinates (3, 2).
6. You can enter other vectors from the Y= menu. Here is a clever way.
This shows all three vectors + , – – and – .
Graphics Calculator
Graphics Calculator
tip!
tip!
Showing vectors on a grid using
parametric plots
2nd
ENTER ZOOM
MODE ENTER
3i
˜ 2˜j X,T,θ,n Y=
WINDOW GRAPH
TRACE
3i
278
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dPosition vectors in two and three
dimensions
Introduction
As a vector has both magnitude and direction, it can be represented in 2-dimensional planes or 3-dimensional regions in space. (It is easier to discuss 2-dimensional vectors as they fit the page nicely!)
Position vectors in two dimensions
In the figure at right, the vector joins the point A to point B. An identical vector can be considered to join the origin with the point C.
It is easy to see that is made up of two components: one along the x-axis and one parallel to the y-axis. Let be a vector along the
x-axis, with magnitude 1. Similarly, let be a vector along the
y-axis, with magnitude 1.
We can say the vector is the position vector of point C relative to the origin.
With vectors, it is equivalent to travel along from the origin directly to C, or to travel first along the x-axis to D and then along the
y-axis to C. In either case we started at the origin and ended up at C. Clearly then, is made up of some multiple of in the x-direction and some multiple of in the y-direction.
For example, if the point C has coordinates (6, 3) then .
Position vectors in three dimensions
In 3 dimensions, a point in space has 3 coordinates, so a third component, along the z-axis, is needed. Let be a vector along the z-axis, with magnitude 1. The orientation is now such that the
x-axis is coming directly out from the page as shown at right. For example, if the point C has coordinates
(6, −2, 4), then its position vector would be denoted by .
The magnitude of a vector
By using Pythagoras’ theorem on a position vector, we can find its length, or magnitude.
Consider the vector at right.
The magnitude of , denoted as or u, is given by: =
= = 3
B
A u ~
x y
u
˜
u
˜ i
˜
j
˜
C
D u ~
i~
j ~ x y
u
˜
u
˜
u
˜ j i˜
˜
O
3 C
D u
~
6i~
3j ~ y
x
u
˜ = 6˜i+3˜j
0 z
x
y
k
˜
(6, –2, 4)
0 C
v ~
4k~ 6i ~ –2j
~ z
x
y
v
˜ = 6i˜–2˜j+4k˜
C (6, 3) u
~ 3j ~ 6i~ x y
u
˜
u
˜ u˜
u
˜ 6
2
32 +
u
˜ 45 5
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279
Consider, now, the general position vector relative to the origin, for the point with coordinates (x, y):
The magnitude of a vector, , is given by .
The direction of a vector
From what we already know about trigonometry, we can work out the angle (θ) that makes with the positive x-axis (that is anticlock-wise from the positive x-axis). This gives us the direction of .
This angle can be calculated as:
θ= tan−1 ( )
= tan−1 0.5
= 0.464 radians
= 26.6°
The result obtained by this method needs to be adjusted if the angle is in the 2nd, 3rd, or 4th quadrants.
The direction of a vector, , is given by θ= tan−1 with appropriate adjustment depending on the quadrant involved.
u
˜ = xi˜+y j˜
u
˜ = xi˜+y j˜ u˜ x
2+y2
=
C (6, 3)
u ~ 3j
~
6~i
θ
x y
u
˜ u
˜
3 6
---u
˜ = xi˜+y j˜
y x
---Using the vector shown at right, find:
a the magnitude of
b the direction of (express the angle with respect to the positive
x-axis)
c the true bearing of .
THINK WRITE
a Use Pythagoras’ theorem or the rule for magnitude of a vector with the x- and y-components 3 and −5 respectively.
a
Simplify the surd.
b The angle is in the 4th quadrant since x = 3 and y = −5.
b
Use trigonometry to find the angle θ, from the x- and y-component values.
θ= tan−1
Use a calculator to simplify. θ = −59°
c The negative sign implies that the direction is 59° clockwise from the x-axis.
c
The true bearing from north is the angle measurement from the positive y-axis to the vector .
true bearing= 90°+ 59°
= 149°.
(3, –5)
u ~
θ
y
x
u
˜
u
˜
u
˜
1 u
˜ 3
2+( )–5 2
=
2 u
˜ = 9+25
34 (= 5.831 to 3 decimal places)
=
1
2 –˙---35
3 1
2
u ˜
6
WORKED
Example
280
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dUnit vectors
As we have seen, any vector is composed of x and y (and z, in 3 dimensions)
components denoted by , (and ). The vectors, , and are called unit
vectors, as they each have a magnitude of 1. This allows us to resolve a vector into its components.
If a 2-dimensional vector makes an angle of θ with the positive x-axis and it has a magnitude of then we can find its x- and y-components using the formulas:
u
˜
xi
˜ y j˜ zk˜ ˜i ˜j k˜
u
˜
u
˜
x u
˜ cosq =
y u
˜ sinq =
Consider the vector, , whose magnitude is 30 and whose bearing (from N) is 310°. Find its x- and y-components and write in terms of and .
THINK WRITE
Change the bearing into an angle with respect to the positive x-axis (θ). The angle between and the positive
y-axis is 360°− 310°.
Calculate θ. θ= 90°+ 50°
= 140° Find the x- and y-components using
trigonometry.
x=
= 30 cos 140° = –22.98
y=
= 30 sin 140° = 19.28 Express as a vector.
u ~
310°
N
S E W
y
x
u ˜
u
˜ ˜i ˜j
1
u ~
θ
y
x
50°
2 u
˜
3
4 u
˜ cosθ
u
˜ sinθ
5 u
˜ u˜ = –22.98˜i+19.28˜j
7
WORKED
E
xample
A bushwalker walks 16 km in a direction of bearing 050°, then walks 12 km in a direction of bearing 210°. Find the resulting position of the hiker giving magnitude and direction from the starting point.
THINK WRITE
Draw a clear diagram to represent the situation. 1
x y
a~16
km
b ~12 km
50°
30° 210° 240°
8
WORKED
E
xample
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281
Clearly, in 3 dimensions, this is much more difficult as you need two angles (for instance, an angle with respect to the x-axis and another with respect to the z-axis).
Unit vectors can also be found in the direction of any vector. This is merely the original vector divided by its magnitude.
The unit vector of any vector , in the direction of denoted by , is:
THINK WRITE
Express position vectors as angles from the direction of the x-axis.
= 16 cos 40 + 16 sin 40
= 12 cos 240 + 12 sin 240
Simplify position vectors. = 12.2567 + 10.2846
= −6 − 10.3923
Use the Triangle Law of addition of vectors.
+ = (12.2567 − 6) + (10.2846 − 10.3923)
= 6.2567 − 0.1077
Find the angle θ. θ= tan–1
= tan–1
= −0.986°
Find the magnitude. | + |=
= = = 6.26
State the resultant vector in terms of magnitude (distance) and direction (bearing).
Final position is 091° and 6.26 km from the starting point.
2 a
˜ ˜i ˜j
b
˜ ˜i ˜j
3 a
˜ ˜i ˜j
b
˜ ˜i ˜j
4
θ
a ~
~ b
a ~+~b
a
˜ b˜ i ˜i ˜j ˜ ˜j
5 y
x
--0.1077 –
6.2567
---6 a
˜ b˜ x
2
y2
+
6.262+(–0.11)2 39.2
7
u
˜ u˜ uˆ˜
uˆ ˜
u ˜ u ˜
---=
Find the unit vector in the direction of .
Continued over page
THINK WRITE
Express the vector in component form.
Compute the magnitude of the vector, .
C (6, 3) u
~ 3j
~ 6i~
u
˜
1 u
˜ = 6˜i+3˜j
2 u
˜ u˜ 6
2+32
=
45
=
3 5
=
9
WORKED
E
xample
282
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dLocating vectors
In the figure at right, is the position vector of point A
and is the position vector of point B relative to the origin. The vector describing the location of A relative to B is easily found using vector addition as or .
Similarly, the vector describing the location of B relative to A is . This result also applies in 3 dimensions and can be formalised as follows.
If A and B are points defined by position vectors and respectively, then
THINK WRITE
Divide each component of the original vector by the magnitude to get .
= +
= +
Comfirm that has a magnitude of 1. =
=
= = 1
3
uˆ ˜
uˆ ˜
6 3 5 ---i
˜ 3 3 5 ---j
˜ 2 5
5 ---i
˜ 5 5 ---j
˜
4 uˆ
˜
uˆ ˜ x
2
y2
+ 20 25 --- 5
25 ---+
25 25
---a
˜ (OA)
O A
B
a~ b
~ y
x
b
˜ (OB)
BA
( )
–b
˜+a˜ a˜ –b˜
AB
( ) b
˜–a˜
a
˜ b˜
AB b
˜ –a˜
=
a Find the position vector locating point B (3, −3) from point A (2, 5).
b Find the length of this vector.
THINK WRITE
a Express the point A as a position vector . a Let
Express the point B as a position vector . Let
The location of B relative to A is .
b The length of is . b =
=
= (or 8.06)
1 a
˜ OA = a˜ = 2˜i+5˜j
2 b
˜ OB = b˜ = 3i˜–3˜j
3 (AB) b
˜ –a˜ AB = b˜ –a˜ 3i
˜–3˜j–(2˜i+5˜j)
= i
˜–8˜j
=
AB b
˜ –a˜ AB b˜ –a˜
12+( )–8 2
65
10
WORKED
Example
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
283
Magnitudes in three dimensions
Pythagoras’ theorem also applies in the case of a 3-dimensional line or vector. Let x, y
and z be the components of a vector, , in 3-dimensional space, that is
. The magnitude of is
We are now in a position to resolve the problem of finding, accurately, the resultant force acting on the television remote control when Daniel and Anna are pulling on it.
First redraw the diagram to show the addition of vectors.
Taking the direction of Anna’s force as the ~i direction,
Anna’s force, = 50~i + 0j
~
Daniel’s force, =−40 cos 150°~i+ 40 sin 150°j
~
= −34.6410~i+ 20j
~
Resultant force, + = (50 − 34.6410)~i+ 20j
~
= 15.3590~i+ 20j
~
Magnitude =| + |
Magnitude = 25.2 N
Direction θ= tan–1
Direction θ= 52.48°
u
˜
u
˜ = xi˜+y j˜ +zk˜ u˜
u
˜ x
2+y2+z2
=
Consider the point in 3-dimensional space given by the coordinates (2, 4, 3).
Find the magnitude of the position vector, , joining this point to the origin.
THINK WRITE
Express as a position vector.
Since the vector is in 3-dimensional space, use the 3-D version of Pythagoras’ theorem to find the magnitude.
Substitute the components for each direction and compute the magnitude.
(= 5.39 to 2 decimal places)
(2, 4, 3)
u ~ z
x
y u
˜
1 u
˜ u˜ = 2˜i+4˜j+3k˜
2 u
˜ x
2+y2+z2 =
3 u
˜ 2
2+42+32 =
29 =
11
WORKED
E
xample
30° 150° 50 N
40 N Resultant
force
Daniel’s force
Anna’s force
a
˜b
˜
a
˜ b˜
a
˜ b˜
20 15.3590
284
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n d1. Press , scroll to Degree and press ; scroll to Par and press .
2. To show vector – press and at X1T= enter 3T; at Y1T= enter –2T.
3. Press and set Tmin= to 0, Tmax= to 1, Tstep= to 1 and then press
. (You may need to alter other WINDOW settings to get a good view.)
4. Press and scroll to verify that the vector has coordinates (3, –2).
5. Press [FORMAT], scroll to PolarGC
and press .
6. Press and . Note that the magnitude of the vector is stored in R and the angle in θ so that you can use them from the HOME screen.
7. At the HOME screen, enter R (press
[R]), and press , and then
θ (press [θ]) and press . Vector – has magnitude 3.606 and
direction –33.7° to the positive x-direction. If you wish to obtain the magnitude and direction of a vector directly from the HOME
screen, use the following steps:
1. Make sure the MODE setting is for degrees.
2. To find the magnitude of the vector – , press [ANGLE], select
5: R Pr(, enter 3, –2) and press .
3. To find the direction of the vector
– , press [ANGLE], select
6: R Pθ(, enter 3, –2) and press
.
Vector – has magnitude 3.606 and
direction –33.7° to the positive x-direction.
Note: R Pr symbolises going from rectangular coordinates to polar form (which gives the coordinates of the vector as its magnitude and direction).
Graphics Calculator
Graphics Calculator
tip!
tip!
Finding the magnitude and direction of
a plane vector using parametric plots
MODE ENTER ENTER
3i
˜ 2˜j Y=
WINDOW GRAPH
TRACE
2nd ENTER
GRAPH TRACE
ALPHA ENTER
ALPHA ENTER
3i ˜ 2˜j
3i
˜ 2˜j 2nd
▼ ENTER
3i
˜ 2˜j 2nd
▼
ENTER
3i ˜ 2˜j
▼
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
285
Position vectors in two and
three dimensions
1 State the x, y and z components of the following vectors:
2 For each of the following find:
i the magnitude of the vector
ii the direction of each vector. (Express the direction with respect to the positive x-axis.)
3 Find the true bearing of each vector in question 2.
a b c
a b
c d
remember
1. Magnitude of a vector: If , the magnitude is given by
Speed is the magnitude of velocity which is a vector quantity.
2. Direction of a vector (2-D only): If , the direction is given by
θ= tan−1
3. The x- and y-components of a vector: Given magnitude and direction, the
x-and y-components are given by:
4. Unit vector: The unit vector of a vector , in the direction of , is denoted by
and is:
5. Locating vectors: If A and B are points with position vectors and
respectively then .
u
˜ = xi˜+y j˜ +zk˜
u
˜ x
2+y2+z2
=
u
˜ = xi˜+y j˜ y x
--x u
˜ cosθ
=
y u
˜ sinθ
=
u
˜ u˜
uˆ ˜
uˆ ˜
u
˜
u
˜
---=
a
˜ b˜
AB b
˜ –a˜
=
remember
7B
3i
˜+4˜j–2k˜ 6i˜–3k˜ 3.4i˜ 2˜j 1 2 ---k
˜
+ +
EXCEL Spreadshe
et
Position vector
WORKED Example 6a, b
(6, 6)
v ~ y
x
(–4, 7)
w ~
y
x
(–3.4, –3.5)
a ~
y
x
(320, –10)
b ~ y
x
SkillS HEET
7.1
WORKEDExample 6c
286
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n d4 Consider the vector shown at right. Its magnitude is 100 and its bearing is 210° True. Find the x- and y-components of , and express them as exact values (surds).
State the answer in the form .
5
A vector with a bearing of 60 degrees from N and a magnitude of 10 has:
A x-component = , y-component =
B x-component = , y-component =
C x-component = , y-component = 5
D x-component = 5, y-component =
E none of the above
6 An aeroplane travels on a bearing of 147 degrees for 457 km. Express its position as a vector in terms of and .
7
A ship travels on a bearing of 331 degrees for 125 km. Express its position as a vector in terms of and .
8 A pilot flies 420 km in a direction 45° south of east and then 200 km in a direction 60° south of east. Calculate the resultant displacement from the starting position giving both magnitude and direction.
9 The instructions to ‘Black-eye the Pirate’s hidden treasure’ say: Take 20 steps in a north-easterly direction and then 30 steps in a south-easterly direction. However, a rockfall blocks the first part of the route in the north-easterly direction. How could you head directly to the treasure?
10 Two scouts are in contact with home base. Scout A is 15 km from home base in a direction 30° north of east. Scout B is 12 km from home base in a direction 40° west of north. How far is scout B from scout A?
WORKED Example 7
100 210°
N
S E W
y
x
w~
w ˜
w
˜ w
˜ = xi˜+y j˜
m
multiple choiceultiple choice
3 2
--- 1
2 ---1
2
--- 3
2 ---5 3
5 3
i ˜ ˜j
i ˜ ˜j
WORKED Example 8
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
287
11 Find unit vectors in the direction of the given vector for the following:
12
A unit vector in the direction of is:
13 Not all unit vectors are smaller than the original vectors. Consider the vector . Show that the unit vector in the direction of is twice as long as .
14 Find the unit vector in the direction of .
15 Find a unit vector in the direction of for the vector of question 4.
16 Consider the points A (0, 1) and B (4, 5) in the figure at right. A vector joining A to B can be drawn.
a Show that an equivalent position vector is given by: .
b Similarly, show that an equivalent position vector joining B to A is given by: .
17 For each of the following pairs of points find:
i the position vectors locating the second point from the first point
ii the length of this vector.
18 Find the position vectors from question 17, by going from the second point to the first.
19 Find unit vectors in the direction of the position vectors for each of the vectors of question 17.
20 Let and . Find:
a b
c d
e f f =
A B C D E none of these
a (0, 2), (4, −5) b (2, 3), (5, 4) c (4, −5), (0, 2)
d (5, 4), (2, 3) e (3, 7), (5, 7) f (7, −3), (3, −3)
a b c
d e f
g Confirm or reject the statement that
Mathca
d
Unit vector in 2D
WORKED Example
9 (3, 4)
0
a ~ y
x (3, –4)
0
d ~ y
x
b
˜ = 4i˜+3˜j e˜ = –4i˜+3˜j c
˜ = i˜+ 2˜j
3.5i ˜
– 2.7j
˜
+
m
multiple choiceultiple choice
3i ˜–4˜j 3
5 ---i
˜ 4 5 ---j
˜
+ 3
5 ---i
˜ 4 5 ---j
˜
– i
˜–˜j
3 25 ---i
˜ 4 25 ---j
˜ –
v
˜ = 0.3˜i+0.4˜j v˜ v˜
w
˜ = –0.1i˜–0.02˜j
w ˜
(4, 5)
A B
(0, 1)
y
x
4i ˜+4˜j
–4i ˜–4˜j
WORKED Example 10
u
˜ = 5i˜–2˜j e˜ = –2i˜+3˜j u
˜ e˜ uˆ˜
e ˆ
˜ u˜ +e˜ u˜+e˜
u
288
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n d21 Let and . Find:
22 To find the distance between two vectors, and , simply find . Find the distance between these pairs of vectors:
23 A river flows through the jungle from west to east at a speed of 3 km/h. An explorer wishes to cross the river by boat, and attempts this by travelling at 5 km/h due north. Find:
a the vector representing the velocity of the river
b the vector representing the velocity of the boat
c the resultant (net) vector of the boat’s journey
d the bearing of the boat’s journey
e the magnitude of the net vector.
24 Consider the data from question 23. At what bearing should the boat travel so that it arrives at the opposite bank of the river due north of the starting position?
a b c
d e f
g Confirm or reject the statement that .
a and b and
u
˜ = –3i˜+4˜j e˜ = 5i˜–˜j u
˜ e˜ uˆ˜
eˆ
˜ u˜+e˜ u˜ +e˜
u
˜ + e˜ = u˜ +e˜
a
˜ b˜ a˜–b˜
3i
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
289
25 Find the magnitude of the following 3-dimensional vectors.
26 By calculating the difference between two position vectors, a vector representing the separation of the two vectors can be defined. Find the distance between the following 3-dimensional vectors.
27 If four points C, D, E and F in 3-dimensional space are located as follows:
C = (2, 6, 0), D = (3, −1, –2), E = (−4, 8, 10), F = (−2, −6, 6), show that CD is parallel to EF.
28 A boat travels east at 20 km/h, while another boat travels south at 15 km/h. Find:
a a vector representing each boat and the difference between the boats
b the magnitude of the difference vector
c the bearing of the difference vector.
29 Consider the vector and the vector . Find the angles of each of these vectors with respect to the x-axis. Show that these two vectors are perpendicular to each other. Also show that the products of each vector’s corresponding x- and y-components add up to 0. Can you confirm that this is a pattern for all perpendicular vectors?
30 A river has a current of 4 km/h westward. A boat which is capable of travelling at 12 km/h is attempting to cross the river by travelling due north. Find:
a a vector representing the net velocity of the boat
b the bearing of the actual motion of the boat
c how long it takes to cross the river, if the river is 500 m wide (from north to south).
(Hint: The maximum ‘speed’ of the boat is still 12 km/h.)
a b
c d
e f
a and b and
c and d and
WORKED Example 11
(3, 4, –5) 0
z
y
x
(–3, –4, 5)
0
z
y
x
0.5i
˜–2k˜+3˜j 2i˜–2 2˜j+k˜ –7i
˜+14˜j–21k˜ ˜i+ +˜j k˜
4i
˜+3˜j–2k˜ 5˜i–2˜j+k˜ 2i˜+˜j–k˜ 5i˜+ +˜j k˜ –i
˜+2˜j+3k˜ 3i˜+k˜ ˜i+3˜j–k˜ 8i˜+5˜j+2k˜
u
290
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dMultiplying two vectors — the dot product
Introduction
In a previous section we studied the result of multiplying a vector by a scalar. What happens if a vector is multiplied by another vector? There are two possibilities: either the result is a scalar (called the scalar product or dot product) or the result is a vector (called the cross, or vector product). In this course we will study only the former.
The scalar or dot product of two vectors, and , is denoted by .
Calculating the dot product
There are two ways of calculating the dot product. The first method follows from its definition. (The second method is shown later.) Consider the two vectors and below.
By definition, the dot product, is given by:
cosθ [1]
where θ is the angle between (the positive directions of) and .
Note: The vectors are not aligned as for addition or subtraction, but their two tails are joined.
Properties of the dot product
1. The dot product is a scalar. It is the result of multiplying three scalar quantities: the magnitudes of the two vectors and the cosine of the angle between them.
2. The order of multiplication is unimportant (commutative property), thus
3. The dot product is distributive, thus
4. Since the angle between and itself is 0°
Note: An easier method for finding the dot product will now be shown.
u
˜ v˜ u˜
⋅
v˜u ˜ v˜
u ~
v ~
θ u
˜
◊
v˜u˜ • v˜ = u˜ v˜ u
˜ v˜
u
˜ • v˜ = v˜ • u˜
a
˜ • (u˜+v˜) = a˜ • u˜+a˜ • v˜ u
˜ u
˜ • u˜ u˜ 2
=
Let and . Find .
THINK WRITE
Find the magnitudes of and .
Draw a right-angled triangle showing the angle that makes with the positive x-axis since is along the x-axis.
Find cos θ, knowing that u = 5 and the x-component of is 3.
cos θ=
Find using equation 1. =
Simplify. = 5 × 6 ×
= 18
u
˜ = 3˜i+4˜j ˜v = 6˜i u˜ • v˜
1 u
˜ v˜ u˜ 3
2+42
=
5
= v
˜ 6
2
=
6
=
2
u
˜ v
˜ 5
3
u ~ y
x
θ
3
u ˜
3 5
---4 u
˜ • v˜ u˜ • v˜ u˜ × v˜ ×cosθ
5 35
---12
WORKED
Example
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
291
Unit vectors and the dot product
Consider the dot product of the unit vectors , and . Firstly, consider in detail. By definition, and, since the angle between them is 0°, cos θ= 1, thus
. To summarise these results:
(since θ= 0°) (since θ= 0°) (since θ= 0°) (since θ= 90°) (since θ= 90°) (since θ= 90°)
Using this information, we can develop another way to calculate the dot product of any vector. Let and , where x1, y1, z1, x2, y2, z2 are constants. Then we can write as:
Considering the various unit vector dot products (in brackets), the ‘like’ products ( , and , shown underlined) are 1; the rest are 0. Therefore:
[2]
This is a very important result.
We only need to multiply the corresponding x, y and z components of two vectors to find their dot product.
Finding the angle between two vectors
Now that we have two formulas (equations 1 and 2) for calculating the dot product, we can combine them to find the angle between the vectors:
i
˜ ˜j k˜ i˜ • ˜i
i ˜ = 1 i
˜ • ˜i = 1
i ˜ • ˜i = 1
j ˜
• j ˜
1
=
k ˜ • k˜ = 1 i
˜ • ˜j = 0 i
˜ • k˜ = 0 j ˜
• k ˜ = 0
u
˜ = x1˜i+y1˜j+z1uk˜ v˜ = x2˜i+y2˜j+z2k˜ ˜ • v˜
u
˜ • v˜ = (x1˜i+y1˜j+z1k˜) • (x2˜i+y2˜j+z2k˜) x1x2 i
˜ • ˜i
( ) x1y2 i ˜ • ˜j
( ) x1z2 i ˜ • k˜
( ) y1x2 j ˜
• i ˜
( ) y1y2 j ˜
• j ˜
( )
+ + + +
=
+ y1z2 j ˜
• k ˜
( ) z1x2 k ˜ • i˜
( ) z1y2 k ˜ • ˜j
( ) z1z2 k ˜ • k˜
( )
+ + +
i
˜ • i˜ ˜j • ˜j k˜ • k˜
u
˜ • v˜ = x1x2+y1y2+z1z2
Let and . Find .
THINK WRITE
Write down using equation 2.
Multiply the corresponding components.
Simplify.
u
˜ = 3˜i+4˜j+2k˜ v˜ =6˜i–4˜j+k˜ u˜ • v˜
1 u
˜ • v˜ u˜ • v˜ = (3˜i+4˜j+2k˜) • (6˜i–4˜j+k˜)
2
u
˜ • v˜ = 3×6+4×–4+2×1
3 = 18–16+2
4
=
13
WORKED
Example
u
˜ • v˜ = x1x2+y1y2+z1z2 = u
292
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n dRearranging the final two equations, we obtain the result that:
[3]
Note: The angle will always be between 0° and 180° as 180° is the maximum angle between two vectors.
Special results of the dot product
Perpendicular vectors
If two vectors are perpendicular then the angle between them is 90° and equation 1 (page 290) becomes:
(since cos 90° = 0)
If , then and are perpendicular.
cosq x1x2+ y1y2+z1z2
u ˜ v˜
---=
Let and . Find the angle between them to the nearest degree.
THINK WRITE
Find the dot product using equation 2. =
Simplify. = 4 × 2 + 3 × −3 + 1 × −2
= −3
Find the magnitude of each vector. = = = =
Substitute results into equation 3. cosθ =
Simplify the result for cos θ. =
= −0.142 695
Take cos−1 of both sides to obtain θ and
round the answer to the nearest degree. θ= cos
−1 (−0.142 695)
= 98°
u
˜ = 4˜i+3˜j+k˜ v˜ = 2˜i–3˜j–2k˜
1 u
˜ • v˜ (4˜i+3˜j+k˜)
⋅
(2˜i–3˜j–2k˜)2
3 u
˜ 4
2+32+12
26
v
˜ 2
2
3 –
( )2
2 –
( )2
+ + 17
4 –3
26 17
---5 3
442 ---–
6
14
WORKED
Example
u
˜ • v˜ = uvcos 90° uv×0
=
0
=
u
˜ • v˜ = 0 u˜ v˜
Find the constant a if the vectors and are perpendicular.
THINK WRITE
Find the dot product using equation 2.
Simplify.
Set equal to zero since and are perpendicular.
Solve the equation for a. a= 4
u
˜ = 4i˜+3˜j v˜ = –3˜i+a j˜
1 u
˜ • v˜ = (4˜i+3˜j) • (–3i˜+a j˜)
2 = –12+3a
3 u
˜ • v˜ u˜ v˜ u˜ • v˜ = –12+3a = 0
4
15
WORKED
Example
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
293
Parallel vectors
If vector is parallel to vector then = k where k ∈ R.
Note: When applying the dot product to parallel vectors, θ (the angle between them) may be either 0° or 180° depending on whether the vectors are in the same or opposite directions.
Multiplying two vectors —
the dot product
1 Find the dot product of the vectors and using equation 1.
2 Compare the result from question 1 with that obtained by finding the dot product using equation 2. Which is probably the most accurate?
u
˜ v˜ u˜ v˜
Let . Find a vector parallel to such that the dot product is 87.
THINK WRITE
Let the required vector . Let =
=
Find the dot product of . =
Simplify. = 25k + 4k
= 29k Equate the result to the given dot product 87. 29k= 87
Solve for k. k= 3
Substitute k = 3 into vector . =
u
˜ = 5˜i+2˜j u˜
1 v
˜ = ku˜ v˜ k(5˜i+2˜j) 5k i
˜+2k j˜
2 u
˜ • v˜ u˜ • v˜ (5˜i+2˜j) • (5k i˜+2k j˜)
3
4 5
6 v
˜ v˜ 15˜i+6˜j
16
WORKED
Example
remember
1. Scalar (dot) product: The scalar or dot product of two vectors, and is
denoted by .
2. Calculation of dot product:
cosθ (where θ is the angle between the two vectors).
3. Algebraic calculation of dot product:
Let and .
Then . 4. Special results:
(a) If , then and are perpendicular.
(b) If , , then are parallel.
u
˜ v˜
u
˜ • v˜
u
˜ • v˜ = u˜ v˜
u
˜ =x1˜i+y1˜j+z1k˜ v˜ =x2i˜+y2˜j+z2k˜
u
˜ • v˜ = x1x2+y1y2+z1z2
u
˜ • v˜ = 0 u˜ v˜
u
˜ = kv˜ k∈R u˜ and v˜
remember
7C
WORKEDExample
12 3i˜+3˜j 6i˜+2˜j
294
M a t h s Q u e s t M a t h s C Y e a r 1 1 f o r Q u e e n s l a n d3 Find in each of the following cases.
a ,
b ,
c ,
d ,
e ,
f ,
g ,
h ,
4
The dot product of and is:
5
Consider the two vectors shown at right. Their dot product is:
6 Consider the vectors and at right. Their magnitudes are 7 and 8 respectively. Find .
7 Let . Show that .
8 Let and let . Find their dot product.
9 Let , and . Demonstrate, using these vectors,
the property:
Formally, this means that vectors are distributive over subtraction.
10 Repeat question 9 for the property:
Formally, this means that vectors are distributive over addition.
11
If , which of the following is perpendicular to ?
A 0 B 3 C 12 D 21 E 27
A 30 B 21.2 C −21.2 D 0
E There is insufficient data to determine the dot product.
A B C
D E
WORKED Example 13
u ˜ • v˜ u
˜ = 2i˜+3˜j+5k˜ v˜ = 3i˜+3˜j+6k˜ u
˜ = 4i˜–2˜j+3k˜ v˜ = 5i˜+˜j–2k˜ u
˜ = –i˜+4˜j–5k˜ v˜ = 3i˜–7˜j+k˜ u
˜ = 5i˜+9˜j v˜ = 2˜i–4˜j u
˜ = –3i˜+˜j v˜ = ˜j+4k˜ u
˜ = 10i˜ v˜ = –2i˜ u
˜ = 3˜j+5k˜ v˜ = ˜i u
˜ = 6i˜–2˜j+2k˜ v˜ = –i˜–4˜j–k˜
m
multiple choiceultiple choice
u
˜ = 3i˜–3˜j+3k˜ v˜ = i˜–2˜j+6k˜
m
multiple choiceultiple choice
6 5 45°
u ~ ~v
50°
u ~ v ~
u ˜ v˜
u ˜ • v˜
u
˜ = xi˜+y j˜ u˜ • u˜ x 2+y2
=
u
˜ = 2i˜–5˜j+k˜ v˜ = –˜i–2˜j+4k˜
u
˜ = 3˜i+2˜j v˜ = ˜i–2˜j w˜ = 5˜i–2˜j
w
˜ • (u˜–v˜) = w˜ • u˜–w˜ • v˜
w
˜ • (u˜+v˜) = w˜ • u˜ + w˜ • v˜
m
multiple choiceultiple choice
u
˜ = 5i˜+4˜j+3k˜ u˜
–5i
˜–4˜j–3k˜ 3˜i+4˜j+5k˜ –5˜i –3i
C h a p t e r 7 I n t r o d u c t i o n t o v e c t o r s
295
12
If then:
A is parallel to
B and have equal magnitudes
C is perpendicular to
D is a multiple of
E None are true.
13
If then:
A
B must be equal to the zero vector,
C is perpendicular to
D must be equal to
E None are true.
14 Find the dot product of the following pairs of vectors.
15 Find the angle between each pair of vectors in question 14 to the nearest degree.
16
The angle between the vectors and is closest to:
17
The angle between the vectors and is closest to:
18 Find the constant a, if the vectors and are perpendicular.
19 Find the constant a, such that is perpendicular to .
20 Let . Find a vector parallel to such that their dot product is 40.
21 Let . Find a vector parallel to such that their dot product is 80.
a and b and
c and d and
A 0° B 67° C 90°
D 113° E 180°
A 0° B 69° C 90°
D 111° E 180°
m
multiple choiceultiple choice
u ˜–v˜
( ) • u ˜ +v˜
( ) = 0 u
˜ v˜
u ˜ v˜ u
˜ v˜
u
˜ v˜
m
multiple choiceultiple choice
u ˜–v˜
( ) • u ˜ +v˜
( ) v
˜ 2
=
u ˜ = v˜ u
˜ 0˜
u
˜ v˜
u
˜ 2 v˜
4i
˜–3k˜ 7˜j+4k˜ ˜i+2˜j–3k˜ –9i˜+4˜j–k˜ 8i
˜+3˜j 2i˜–3˜j+4k˜ 5i˜–5˜j+5k˜ 5i˜+5˜j–5k˜
WORKED Example 14
m
multiple choiceultiple choice
2i
˜+3˜j 2˜i–3˜j
m
multiple choiceultiple choice
2i
˜–3˜j –4i˜+6