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Vector
5 30°
P
Vectors We drove into the future looking into a rear view mirror. Herbert Marshall McLuhan
●?^ What^ information^ do^ you^ need
to decide how close the aircraft which left these vapour trails passed to each other? A quantity which has both size and direction is called a vector. The velocity of an aircraft through the sky is an example of a vector, having size (e.g. 600 mph) and direction (on a course of 254°). By contrast the mass of the aircraft ( tonnes) is completely described by its size and no direction is associated with it; such a quantity is called a scalar. Vectors are used extensively in mechanics to represent quantities such as force, velocity and momentum, and in geometry to represent displacements. They are an essential tool in three-dimensional co-ordinate geometry and it is this application of vectors which is the subject of this chapter. However, before coming on to this, you need to be familiar with the associated vocabulary and notation, in two and three dimensions.
Vectors in two dimensions
Terminology In two dimensions, it is common to represent a vector by a drawing of a straight line with an arrowhead. The length represents the size, or magnitude, of the vector and the direction is indicated by the line and the arrowhead. Direction is usually given as the angle the vector makes with the positive x axis, with the anticlockwise direction taken to be positive. The vector in figure 8.1 has magnitude 5, direction 30°. This is written (5, 30°) and 254 said to be in magnitude direction form or in polar form. The general form of a vector written in this way is ( r , θ ) where r is its magnitude and θ its direction.
Figure 8.
4
2 )
or 4 i + 2 j
a
θ
Note
In the special case when the vector is representing real travel, as in the case of
the velocity of an aircraft, the direction may be described by a compass bearing
with the angle measured from north, clockwise. However, this is not
done in this chapter, where directions are all taken to be measured
anticlockwise from the positive x direction.
An alternative way of describing a vector is in terms of components in given
directions. The vector in figure 8.2 is 4 units in the x direction, and 2 in the
y direction, and this is denoted by
2
4
Figure 8.
This may also be written as 4 i 2 j , where i is a vector of magnitude 1, a unit
vector , in the x direction and j is a unit vector in the y direction (figure 8.3).
j
i
Figure 8.
In a book, a vector may be printed in bold, for example p or OP , or as a line between two points with an arrow above it to indicate its direction, such as
OP.
When you write a vector by hand, it is usual to underline it, for example, p or OP ,
or to put an arrow above it, as in OP.
To convert a vector from component form to magnitudedirection form, or vice
versa, is just a matter of applying trigonometry to a right-angled triangle.
Write the vector a 4 i 2 j in magnitudedirection form.
SOLUTION
2
4
Figure
8
.
4
EXAMPLE 8.
Vectors
in
two
dimensions
P
4
P
5
60°
290°
10
The magnitude of a is given by the length a in figure 8.4.
a (using Pythagoras’ theorem)
4.47 (to 3 significant figures)
The direction is given by the angle θ.
tan θ
2 0.
θ 26.6° (to 3 significant figures)
The vector a is (4.47, 26.6°).
The magnitude of a vector is also called its modulus and denoted by the
symbols
. In the example a 4 i 2 j , the modulus of a , written a , is 4.47. Another
convention for writing the magnitude of a vector is to use the same letter, but
in italics and not bold type; thus the magnitude of a may be written a.
Write the vector (5, 60°) in component form.
SOLUTION
In the right-angled triangle OPX
OX 5 cos 60° 2.
XP 5 sin 60° 4.33 j
(to 2 decimal places) O
i X
Figure 8.
OP is
or 2.5 i 4.33 j.
This technique can be written as a general rule, for all values of θ.
r^ cos
( r , θ )
r sin
( r cos θ ) i + ( r sin θ ) j
Write the vector (10, 290°) in component form.
SOLUTION
In this case r 10 and θ 290°.
10 cos 290
to 2 decimal places.
10 sin 290
This may also be written 3.42 i 9.40 j.
Figure 8.
256
EXAMPLE 8.
EXAMPLE 8.
Vector s
P
5
4 j length 4 r
j
αθ
In Example 8.3 the signs looked after themselves. The component in the i
direction came out positive, that in the j direction negative, as must be the case for
a direction in the fourth quadrant (270° < θ < 360°). This will always be the case
when the conversion is from magnitudedirection form into component form.
The situation is not quite so straightforward when the conversion is carried out
the other way, from component form to magnitudedirection form. In that
case, it is best to draw a diagram and use it to see the approximate size of the
angle required. This is shown in the next example.
Write 5 i 4 j in magnitudedirection form.
SOLUTION
Figure 8.
In this case, the magnitude r
6.40 (to 2 decimal places).
The direction is given by the angle θ in figure 8.7, but first find the angle α.
tan α
4
α 38.7° (to nearest 0.1°)
so θ 180 α 141.3°
The vector is (6.40, 141.3°) in magnitudedirection form.
257
EXAMPLE 8.
Vectors
in
two
dimensions
P
a
a
a
a
k
i (^) j
Equal vectors
The statement that two vectors a and b are equal means two things.
● The direction of a is the same as the direction of b.
● The magnitude of a is the same as the magnitude of b.
If the vectors are given in component form, each component of a equals the
corresponding component of b.
Position vectors
Saying the vector a is given by 3 i 4 j k tells you the components of the vector, or
equivalently its magnitude and direction. It does not tell you where the vector
is situated; indeed it could be anywhere.
All of the lines in figure 8.9 represent the vector a.
Figure 8.
There is, however, one special case which is an exception to the rule, that of a
vector which starts at the origin. This is called a position vector. Thus the line
joining the origin to the point P(3, 4, 1) is the position vector 4 or 3 i 4 j k.
Another way of expressing this is to say that the point P(3, 4, 1) has the position
^3
vector 4 .
259
Vectors
in
three
dimensions
P
8
P
y 4 3 L 2 N 1
M – 1 Points L, M and N have co-ordinates (4, 3), (2, 1) and (2, 2). (i) Write down, in component form, the position vector of L and the vector MN. (ii) What do your answers to part (i) tell you about the lines OL and MN? SOLUTIO N (i)
^4
.
The position vector of L is
OL
3
^4 The vector MN is also 3
(see figure 8.10). (ii) Since OL MN, lines OL and MN are parallel and equal in length. Figure 8. Note A line joining two points, like MN in figure 8.10, is often called a line segment , meaning that it is just that particular part of the infinite straight line that passes through those two points. The vector MN is an example of a displacement vector. Its length represents the magnitude of the displacement when you move from M to N. The length of a vector In two dimensions, the use of Pythagoras’ theorem leads to the result that a vector a 1 i a 2 j has length a given by a a 2 a 2 . 1 2 260 EXAMPLE 8. Vector s
8
P
a 2 a 4 Write, in component form, the vectors represented by the line segments joining the following points. (i) (2, 3) to (4, 1) (ii) (4, 0) to (6, 0) (iii) (0, 0) to (0, 4) (iv) (0, 4) to (0, 0) (v) (0, 0, 0) to (0, 0, 5) (vi) (0, 0, 0) to (1, 2, 3) (vii) (1, 2 , 3) to (0, 0, 0) (viii) (0, 2, 0) to (4, 0, 4) (ix) (1, 2, 3) to (3, 2, 1) (x) (4, 5, 0) to (4, 5, 1) 5 The points A, B and C have co-ordinates (2, 3), (0, 4) and (2, 1). (i) Write down the position vectors of A and C. (ii) Write down the vectors of the line segments joining AB and CB. (iii) What do your answers to parts (i) and (ii) tell you about (a) AB and OC (b) CB and OA? (iv) Describe the quadrilateral OABC.
Vector calculations
Multiplying a vector by a scalar When a vector is multiplied by a number (a scalar) its length is altered but its direction remains the same. The vector 2 a in figure 8.11 is twice as long as the vector a but in the same direction. Figure 8. When the vector is in component form, each component is multiplied by the number. For example: 2 (3 i 5 j k ) 6 i 10 j 2 k
^3 6
2 –5–10. 1 2
262 The negative of a vector In figure 8.12 the vector a has the same length as the vector a but the opposite direction. Vector s
a – a
5 i + 2 j
2 i 3 i + 5 j 5 j
2 i – 3 j
3 i
P
Figure 8.
When a is given in component form, the components of a are the same as those
for a but with their signs reversed. So
^23 –23
Adding vectors
When vectors are given in component form, they can be added component
by component. This process can be seen geometrically by drawing them on
graph paper, as in the example below.
Add the vectors 2 i 3 j and 3 i + 5 j.
SOLUTION
2 i 3 j + 3 i + 5 j 5 i + 2 j
Figure 8.
The sum of two (or more) vectors is called the resultant and is usually
indicated by being marked with two arrowheads. 263
EXAMPLE 8.
calculations Vector
Q(3, 5)
1
4 )
P(2, 1)
T R
E
C
y
Q
When you find the vector y
represented by the line segment 6 joining two points, you are in
effect subtracting their position
5
vectors. If, for example,
P is the point (2, 1) and Q is the 4
^1
point (3, 5), PQ is (^) 4
, as 3
figure 8.16 shows.
2
You find this by saying
PQ PO OQ p q. (^1)
In this case, this gives
2 3 1 0 1 2 3 4 5
x
PQ –
as expected.
This is an important result:
PQ q p
Figure 8.
where p and q are the position vectors of P and Q.
Geometrical figures
It is often useful to be able to express lines in a geometrical figure in terms of
given vectors.
ACTIVITY 8.1 The diagram shows a cuboid OABCDEFG. P, Q, R, S and T are the mid-points of
the edges they lie on. The origin is at O and the axes lie along OA, OC and OD, as
shown in figure 8.17.
^6
0
OA 0 , OC 5 , OD 0
G S F
D
z B
O x P A
Figure 8.
calculations Vector
P
265
C
CH = BG
FE = BC, EJ = BG, JI = AB
3 i + 5 j (^) This is the unit vector
3 i 5 j
3434
(iv) IJ DE
AB
B
p 8
(v) EF BC
q
D
(vi) BE BC CD DE
q ( q p ) p
2 q 2 p
E
Figure 8.
Notice that BE 2CD.
(vii) AH AB BC CH
p q r
(viii) FI FE EJ JI
q r p
Unit vectors
A unit vector is a vector with a magnitude of 1, like i and j. To find the unit
vector in the same direction as a given vector, divide that vector by its magnitude.
Thus the vector 3 i 5 j (in figure 8.20) has magnitude
the vector
i
j is a unit vector. It has magnitude
34 , and so
The unit vector in the direction of vector a is written as â and read as ‘a hat’.
y
5 j
4 j
3 j
2 j
j
O i
Figure 8.
2 i 3 i 4 i x
267
calculations Vector
P
AB
3
AC
BC
AB
AB + AC
BC.
AC
9 This is themagnitude^ of^ AB.
Relative to an origin O, the position vectors of the points A, B and C are given by
OA 3 , OB 1 and OC 3 .
(i) Find the unit vector in the direction AB.
(ii) Find the perimeter of triangle ABC.
SOLUTION
For convenience call OA a , OB b and OC c.
(i) AB b a 1 3 2
To find the unit vector in the direction AB, you need to divide AB by its
magnitude.
^2
2
3
So the unit vector in the direction AB is
1 2
2
3
1
3
(ii)
1
^2 ^2
^0
AC c a 3 3 0
^ ^
^2
^2
BC c b 3 1 2
^
(^)
Perimeter of ABC ^ ^ +
+ BC
EXAMPLE 8.
Vector s
The perimeter of the triangle is given +
P
Verify that AB BC
a
M
1 Simplify the following.
P
(i) ^2
^4 (ii)
(iii)
^3
(iv) 3
(v) 6(3 i 2 j ) 9(2 i j )
2 The vectors p , q and r are given by
p 3 i 2 j k q 2 i 2 j 2 k r 3 i j 2 k.
Find, in component form, the following vectors.
(i) p + q + r (ii) p − q (iii) p + r
(iv) 3( p − q ) + 2( p + r ) (v) 4 p − 3 q + 2 r
3 In the diagram, PQRS is a parallelogram and PQ a , PS b.
(i) Write, in terms of a and b , Q R the following vectors.
(a) QR (b) PR
(c) QS
(ii) The mid-point of PR is M. Find
(a) PM (b) QM. P (^) b S
(iii) Explain why this shows you that the
diagonals of a parallelogram bisect each other.
4 In the diagram, ABCD is a kite. B AC and BD meet at M.
AB i j and
AD i 2 j j
(i) Use the facts that the diagonals
of a kite meet at right angles
i
and that M is the mid-point of
AC to find, in terms of i and j , D
(a) AM (b) AC
(ii)
(c) BC (d) CD.
(^)
AD CD.
269
EXERCISE 8B
8B Exercise
and
A C
NM
8
P
5 In the diagram, ABC is a triangle. A L, M and N are the mid-points of the sides BC, CA and AB. AB p and AC q (i) Find, in terms of p and q , BC, MN, LM and LN. B L C (ii) Explain how your results from part (i) show you that the sides of triangle LMN are parallel to those of triangle ABC, and half their lengths. 6 Find unit vectors in the same directions as the following vectors. (i)
(ii) 3 i 4 j (iii)
(iv) 5 i 12 j 7 Find unit vectors in the same direction as the following vectors.
^1
(i) 2
(ii) 2 i – 2 j + k (iii) 3 i – 4 k
^4
(iv) 4 (v) 5 i – 3 j + 2 k (vi) 0
8 Relative to an origin O, the position vectors of the points A, B and C are given by
^2
^1
OA 1 , OB 4 and OC 2 .
Find the perimeter of triangle ABC. 9 Relative to an origin O, the position vectors of the points P and Q are given by OP 3 i j 4 k and OQ i x j 2 k. Find the values of x for which the magnitude of PQ is 7. 10 Relative to an origin O, the position vectors of the points A and B are given by
^4
^3
OA 1 and OB 2 .
(i) Given that C i s the point such that AC = 2 AB, find the unit vector in the direction of OC.
^1
The position vector of the point D is given by OD = 4 , where k is a k Vector s
