CHAPTER 11 SOLUTIONS TO EXERCISES IN VECTORS
Note: In these solutions vectors are denoted by underlining – this is how they are often represented in written work anyway.
Exercise on 11.1
Classify the following as scalar or vector quantities:
i) area ii) gravitational force iii) electric field at a point
iv) density v) potential energy vi) work
vii) magnetic field viii) time ix) pressure
x) acceleration xi) voltage xii) momentum.
Solution
i), iv), v), vi), viii), xi) are all scalars, requiring, in each case, just one numerical value to specify them. The rest are all vectors, since they have a directional component as well as a magnitude. Thus, ii), iii), vii) are all essentially forces defined at a particular point in space, acting in some particular direction. Momentum, xii), is mass times velocity and the former is scalar, while the latter is a vector – being speed in a particular direction. So momentum is a vector. Acceleration is rate of change of velocity and so is also a vector. Note that Newton’s second law, Force = mass × acceleration, is thus an equality of vectors.
Exercise on 11.2
Consider vectors in a plane, and use a two-dimensional Cartesian co-ordinate system with the usual rectangular x, y axes. Choosing suitable scales on the axes sketch vectors as follows:-
1. Three free vectors a, b, c all of the same length 2 units and making equal angles with the positive x and y axes.
1. A position vector r of length 3 units in the north-west direction (positive y axis north).
Solution
i)
ii)
Exercises on 11.3
1. Consider the pentagon below:
Express
i) e in terms of a, b, c, d
ii) in terms of e, a, b
iii) in terms of c and d
iv) in terms of and
Solution
i) Since the polygon is closed, we have
e + a + b + c + d = 0
and so, solving for e we have
e = – (a + b + c + d)
ii) By joining appropriate vertices we have
e + a + b + = 0
and so
= – (e + a + b)
iii)Again by joining appropriate vertices we have
+ c + d = – + c + d = 0
and so
= c + d
iv) From the closed figure we have + + = 0 and so
= – –
= –
2. On a two dimensional Cartesian coordinate system, a is the position vector from the origin to the point (1, 0), b is the position vector with length 2 at angle 60o (anti-clockwise) to the positive x – axis. Describe the vector b – a.
Solution
From the figure and the 60-30 triangle b – a is of length and is parallel to the y-axis
Exercise on 11.4
Plot the points (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 0, 0) (–1, 0, 0), (1, 0, – 1), (2, – 1, 1) on a perspective drawing of a Cartesian rectangular coordinate system.
Solution
The figure illustrates the positions of the given points
Exercise on 11.5
Calculate the distances of the points i) (1, 0, 2), ii) (– 2, 1, – 3), iii) (– 1, – 1, – 4) from the origin. Also calculate the distance between each pair of points.
Solution
The distance, r, from the origin O to the point P(x1, y1, z1) is given by
r = OP =
In general the distance between two points P(x, y, z), P´(x´, y, z´) is given by
PP´ =
i)The distance of the point (1, 0, 2) from the origin is therefore
r = =
ii) For (– 2, 1, – 3) we have
r = = =
iii) For (– 1, – 1, – 4) we have
r = =
The distance between I) (1, 0, 2) and ii) (– 2, 1, – 3) is
= =
The distance between I) (1, 0, 2) and iii) (– 1, – 1, – 4) is
= =
The distance between ii) (– 2, 1, – 3) and iii) (– 1, – 1, – 4) is
= =
Exercise on 11.6
Calculate the direction cosines of the position vectors defined by the points in the Exercise on 11.4.
Solution
The position vector of the point (x, y, z) has components relative to the same axes which we also write in the form (x, y, z). The direction cosines are the cosines of each of the angles made by this vector with the axes. For example the x-component has a direction cosine
l =
where r = and similarly for the other components
m = , n =
For (0, 0, 0) we have r = 0 and so in this ‘singular case the DCs are not defined.
For (0, 0, 1), r = 1 and so the DCs are (0, 0, 1). Similarly for (0, 1, 0) and (1, 0, 0), the DCs simply repeat the coordinates. Indeed the same also applies for (–1, 0, 0).
For (1, 0, – 1) we have r = and so in this case the DCs are
( , 0, – )
For (2, – 1, 1), r = and so the DCs are
(, – , )
Exercise on 11.7
Using direction cosines find the acute angles between each pair of lines through the origin defined by the points in Exercises 11.5.
Solution
Direction cosines for i) (1, 0, 2) are (, 0, )
Direction cosines for ii) (– 2, 1, – 3) are (– , , – )
Direction cosines for iii) (– 1, – 1, – 4) are (– , – , – )
So the angle between i) and ii) is given by
cos q = + = –
and so
q = cos– 1 = 17.02° to 2 dp
Between i) and iii)
cos q = + = – = –
and
q = cos– 1 = 18.43° to 2 dp
Between ii) and iii)
cos q = – + =
and
q = cos– 1 = cos– 1 = 35.02° to 2 dp
Exercises on 11.8
1. Express the position vectors with the given endpoints in terms of i, , k vectors.
i) (3, – 1, 2) ii) (1, 0, 1), iii) (a, b, c) iv) (– 1, 1, – 1)
Solution
i) The point (3, – 1, 2) has x-coordinate 3, y-coordinate – 1 and z-coordinate 2, so expressed in terms of i, , k it is defined by the position vector 3i – + 2k
ii) (1, 0, 1) is represented by i + k
iii) (a, b, c) is represented by ai + b+ ck
iv) (– 1, 1, – 1) is represented by – i + – k
2. If (a – b) i + (b + 2c)+ (a – c)k = 2i – k determine a, b, c.
Solution
Equating coefficients, we have
(a – b) i + (b + 2c)+ (a – c)k = 2I – k
a – b = 2 I)
b + 2c = 0 ii)
a – c = – 1 iii)
From i) + ii)
a + 2c = 2
a – c = – 1
Subtract
3c = 3, so c = 1
Then
a = 0
and
b = a – 2 = – 2
So, finally
a = 0, b = – 2, c = 1
Exercises on 11.9
1. If a = µi + 2+ (l – µ)k, b = 2li + u+ 2k
and
2a + b = 0
determine µ, l, u.
Solution
2a + b = 2(µi + 2+ (l – µ)k) + 2li + u+ 2k = 0
= (2µ + 2l) i + (4 + u)+ (l – µ + 1)k
Equating each component to zero now gives
2µ + 2l = 0
4 + u= 0
l – µ + 1 = 0
the solution of which you can confirm is
µ = , u= – 4, l = –
2. Referring to Q1, evaluate
i) |a| ii) a iii) 3b iv) a + 2k v) 3a – b
Solution
Substituting the values of µ = , u= – 4, l = – into a we have
a = i + 2j – k and b = – i – 4j + k
and so
i) |a| = =
ii) a= = =
iii)3b = – 3 i – 12j + 3k
iv) a + 2k = I + 2j – k + 2k = i + 2j + k
v) 3a – b = 3 – ( – i – 4j + k) = i + 10j – 4k
Exercises on 11.10
1. Using the component form for the scalar product prove the properties i) – v)
Solution
In the following we use the standard notation a = a1i + a2j +a3k, b = b1i + b2j +b3k, etc.
i) Since a . b = a1 b1 + a2 b2 + a3 b3
is a simple number it is a scalar quantity.
ii) a. b = a1b1 + a2b2 + a3b3 = b1a1 + b2a2 + b3a3 = b. a
i.e. the scalar product is commutative
iii) a. (b + c) = a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3)
= a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3
= a1b1 + a2b2 + a3b3 + a1c1 + a2c2 + a3c3
= a. b + a. c
i.e. the scalar product is distributive over addition.
iv)(k a). b = (k(a1i + a2j + a3k)). (b1i + b2j + b3k)
= (ka1i + ka2j + ka3k). (b1i + b2j +b3k) = ka1b1 + ka2b2 + ka3b3
= k(a1b1 + a2b2 + a3b3) = k (a. b)
= a1kb1 + a2kb2 + a3kb3 = (a1i + a2j + a3k). (kb1i + kb2j + kb3k)
= a. (kb)
for any scalar k.
v) Actually it is best to use the definition a . b = ab cos q for this result. Thus
a . b = ab cos q = 0
implies (since a and b are non-zero) that cos q = 0 from which q = or and in either case the two vectors are perpendicular to each other.
2. Find all possible scalar products between the vectors a = – i + 2, b = + 2k , c = i + 2+ 3k and determine the angles between each pair of vectors.
Solution
With a = – i + 2, b = + 2k , c = i + 2+ 3k the possible scalar products are
a . b = (– i + 2).(+ 2k) = (– 1)(0) + 2(1) + 0(2) = 2
a . c = (– i + 2).( i + 2+ 3k) = (–1)(1) +2(2) + 0(3) = 3
b . c = (+ 2k). (i + 2+ 3k) = 1(2) + 2(3) = 8
For the angle Ð( a , b ) between a and b we have
cos(Ð( a , b )) = = = 0.4
So
Ð( a , b ) = cos– 1(0.4) = 66.42°
Similarly
Ð( a , c ) = cos– 1= cos– 1= 68.99°
and
Ð( b , c ) = cos– 1= cos– 1= 17.02°
Exercises on 11.11
1. Prove the properties i) ® vi)
Solution
i) a ´ b is a vector
It is clear from the definition
a ´ b = (a2 b3 – a3 b2)i + (a3 b1 – a1 b3)+ (a1 b2 – a2 b1)k
that a ´ b is a vector also.
ii) a ´ (b + c) = a ´ b + a ´ c and (a + b) ´ c = a ´ c + b ´ c
With the usual notation we have
a ´ (b + c) = (a2 (b3 + c3) – a3 (b2 + c2))i + (a3 (b1 + c1) – a1 (b3 + c3))+ (a1 (b2 + c2) – a2 (b1 + c1))k
= (a2 b3 – a3 b2)i + (a3 b1 – a1b3 )+ (a1b2 – a2 b1)k + (a2 c3 – a3 c2)i + (a3 c1 – a1 c3)+ (a1c2 – a2 c1)k
= a ´ c + b ´ c
i.e. the vector product is distributive over vector addition
iii) (k a ) ´ b = k(a ´ b) = a ´ (k b) for any scalar k
(k a) ´ b = (ka2 b3 – ka3 b2)i + (ka3 b1 – ka1 b3)+ (ka1 b2 – ka2 b1)k
= k(a2 b3 – a3 b2)i + k (a3 b1 – a1 b3)+ k(a1 b2 – a2 b1)k = k(a ´ b)
= (a2 (kb3) – a3 (kb2))i + (a3 (kb1) – a1(kb3))+ (a1 (kb2) – a2 (kb1))k
= a ´ (k b)
iv) If a ´ b = 0 and a, b are not zero vectors, then a and b are parallel
a ´ b = (a2 b3 – a3 b2)i + (a3 b1 – a1 b3)+ (a1 b2 – a2 b1)k = 0
gives
a2 b3 – a3 b2 = 0
a3 b1 – a1 b3 = 0
a1 b2 – a2 b1 = 0
From these we get, provided appropriate componentss are non-zero:
= =
So for some constant k:
ai = k bi
and hence
a = k b
ie a and b are parallel.
v) b ´ a = – a ´ b
b ´ a = (b2 a3 – b3 a2)i + (b3 a1 – b1 a3)+ (b1 a2 – b2 a1)k
= – (a2 b3 – a3 b2)i – (a3 b1 – a1 b3)– (a1 b2 – a2 b1)k
= – a ´ b
vi) If we put a = b in v) we get a ´ a = 0
vii) The vector product is not associative, i.e.
a ´ (b ´ c) ¹ (a ´ b) ´ c
We only need to show this for an example and you can check that with a = i + j, b = j, c = k
(i + j) ´ (j´ k) = (i + j) ´ i= j´ i= – k
while
(( i + j) ´ j) ´ k = ( i ´ j) ´ k = k ´ k = 0
2. If a = i + k b = 2i – + 3k c = i + 2+ 3k , evaluate
i) a ´ b ii) a ´ ( b + c) iii) a ´ c iv) b ´ b
v) b ´ a vi) a ´ (b ´ c) vii) (a ´ b) ´ c
Solution
a = i + k b = 2i – + 3k c = i + 2+ 3k
i) a ´ b = = i (1) – (+1)+ (– 1)k = i – j – k
ii) a ´ ( b + c) = ( i + k ) ´ (3 i + + 6k ) =
= i (– 1) – (3)+ k = – i – 3j + k
iii) a ´ c = a ´ ( b + c) – a ´ b = – i – 3j + k – ( i – j – k)
= – 2i – 2j + 2k
iv) b ´ b = 0 from property iv)
v) b ´ a = – a ´ b = – (i – j – k) = – i + j + k
vi) b ´ c = = i (– 9) – (3)+ (5)k = – 9 i – 3+ 5 k
So
a ´ (b ´ c) =
= i (3) – (14)+ (– 3)k = 3i – 14j – 3k
vii) (a ´ b) ´ c = using i)
= i (– 1) – (4)+ (3)k = – i – 4j + 3k ¹ a ´ (b ´ c)
Exercise on 11.12
Sketch the path described by the position vector
f(t) = cos t i + sin t + k
as t varies.
Solution
Since the coefficient of k is unity, the path must always be in a plane lying one unit above the xy-plane. In that plane we always have the parametric form (See UEM 219) x = cos t and y = sin t. In this case we can eliminate t (See Review Question 7.1.8) by squaring and adding to give
x2 + y2 = cos2t + sin2t = 1
This is the equation of a circle, radius unity, centre x = y = 0 (ie centre the z-axis) (See UEM 217). So, as t increases the path lies on a circle in the plane z = 1, with radius unity and centred on the z-axis. This is illustrated in the figure below.
Exercises on 11.13
1. Show that if e is a vector of constant magnitude, then
e . = 0
Solution
Since e is a vector of constant magnitude we have e.e = constant. So, using the product rule we have