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Mathematics Revision Questionsfor the University of Bristol School of Physics
You will not be surprised to find you have to use a lot of maths in your study of physicsat university! You need to be completely familiar with all the maths you learnt at A-level andbe able to apply it quickly and easily. This will involve algebra, calculus (differentiation andintegration), vectors, . . . . University physics lecturers will assume a familiarity with some basicmathematical tools and techniques. All this requires a lot of practice so we have provided herea brief set of revision notes and revision questions which cover the most important techniquesfor the physics lectures at the beginning of the first year. In addition, you will be asked tocomplete some electronic exercises, testing these skills, throughout the first semester. Weadvise that you practice your Maths skills regularly throughout your first year.
If you find that you can’t do a question or if there is a particular area with which you havedifficulties, do not worry. We suggest you first look through your A level notes but do notpanic if you are completely stuck! Simply make a note of the topic or question and ask yourphysics tutor in tutorial sessions. Another possible source for more information may be therecommended textbook ”Physics for Scientists and Engineers” by Tipler and Mosca.
In recent years, first year students have found it helpful to be given these A-level mathsrevision notes and questions before the start of term to give you an opportunity to refresh yourmemory before lectures begin. Please note that there is no need to complete all the questionsor hand them in. They are for your benefit, so that you can understand what will be expectedof you, what you have forgotten and what you may need to learn.
1 Algebra
Revision MaterialQuadratic equations of the form ax2 + bx+ c = 0 where a, b and c are real valued constants(in mathematical notation this written as a, b, c ∈ R) are solved by the following equation,
x = −b±√b2 − 4ac
2a . (1)
Polynomial long division allows for the solving of problems such as,x3 + 3x2 + 3x+ 1
x+ 1 = x2 + 2x+ 1 ,
1
and can be solved in multiple ways. Two common schemes for solving such expressions are,
x2 + 2x + 1x+ 1
)x3 + 3x2 + 3x + 1
− x3 − x2
2x2 + 3x− 2x2 − 2x
x + 1− x− 1
0
or1 3 3 1
− 1 − 1 − 2 − 11 2 1 0
.
For partial fractions, there are a number of rules which are useful to know:
• For every linear factor such as (ax+b) in the denominator, there will be a partial fractionof the form A/(ax+ b).
• For every repeated factor such as (ax+ b)2 in the denominator, there will be two partialfractions: A/(ax+ b) and B/(ax+ b)2. For higher powers there will be correspondinglymore terms.
• For quadratic factors in the denominator e.g.(ax2+bx+c), there will be a partial fractionof the form: (Ax+B)/(ax2 + bx+ c).
The exponents of variables are combined as follows,
aman = am+n
am
an= am−n .
Logarithms follow the below rules,
ax = n ⇔ x = loga nn = aloga n; loga 1 = 0; loga a = 1
loga(mn) = logam+ loga n; loga(m
n
)= logam− loga n;
loga(mp) = p logam
2
Practice Problems1.1. Solve the following quadratic equations by factorisation:
(a) x2 + 8x+ 12 = 0
(b) x2 + x− 20 = 0
(c) 4x2 + 8x− 12 = 0
(d) x2 − x− 6 = 0
(e) 3x2 − 11x− 20 = 0
(f) x+ 3 = 2x2
(g) 4x2 − 15x+ 9 = 0(h) 36x2 − 48x+ 16 = 0(i) 3(x2 + 2x) = 9(j) x2 + 4x− 5 = 0(k) 2x2 − 7x− 4 = 0
(l) x− 63x
= 2
1.2. Complete the square for the following quadratic equations i.e. rewrite them in the form(x+ a)2 + b = 0, and hence solve for x:
(a) x2 − 4x− 5 = 0(b) x2 − 6x+ 9 = 0(c) x2 − 12x+ 35 = 0(d) x2 − 3x− 5 = 0
(e) 2x2 − 6x+ 1 = 0(f) x2 + 8x+ 17 = 0(g) 3x2 − 3x− 2 = 0(h) 3x2 − 6x− 1 = 0
1.3. Solve the following quadratic equations using the quadratic formula given in Equa-tion (1):
(a) x2 = 8− 3x
(b) (x− 2)(2x− 1) = 3
(c) (x+ 2)2 − 3 = 0
(d) 2x+ 4x
= 7
(e) 3x2 + 2x = 6
(f)(x− 1
2
)2= 1
4(g) 2x(x+ 4) = 1(h) 4x(x− 2) = −3
1.4. Solve the following cubic equations:
(a) x3 − 2x2 − 15x = 0.(b) x3 − 4x2 + x+ 6 = 0, given that x = −1 is a solution.(c) x: x3 − 19x+ 30 = 0, given x = 3 is a solution.
3
(d) x: x3 + x2 − 22x− 40 = 0, given x = −2 is a solution.(e) x3 − 9x2 + 15x+ 25 = 0, given x = 5 is a solution.
1.5. Solve the following equations:
(a) x1/3 − 3x−1/3 = 2.(b) (2x2 − x)2 − 9(2x2 − x) + 18 = 0.(c) 32x2−x = (2.67)1/x.(d) 32x − 3x+1 + 2 = 0.(e) 52x = 7x+1.
1.6. Express the following in partial fractions:
(a) 4x2 − 4
(b) 2x+ 1(x− 1)(x+ 2)2
(c) 1x2 + 5x+ 6
(d) x
(x+ 1)(x− 1)2
(e) 1x3 − 1
(f) x2 + 2x+ 1x2 + x− 2
(g) 5x3 + 2x2 + 5xx4 − 1
(h) 7x+ 2(x+ 2)2(x− 2)
(i) 10(x+ 1)(x+ 3)(x2 + 1)
(j) 4x+ 1(x− 3)(x2 + x+ 1)
(k) 2x2 + 7(x+ 2)2(x− 3)
(l) x4 + 5x3 + 9x2 + 6x+ 5(x+ 2)(x2 + 1)
1.7. Simplify the following expressions
(a) (x3)2y4
x6y
(b) a2b
abc4
(c) 5× 43n+1 − 20× 82n
(d)√
36x2y4
(e) 3√
27x6y3
4
1.8. Evaluate the following expressions:
(a) 103/2
10−1/2(b) (161/4)3
1.9. Solve the following expressions for x:
(a) 2x = 4(b) 4x = 2
(c)(1
2
)x=√
2
1.10. Questions on logarithms
(a) If loga q = 5 + loga b, and c = a4, prove that q = abc.(b) Simplify (i) aloga x; (ii) a−2 loga x.(c) Write down the values of log2 16 and log8 2.(d) Given that logb a = c and logc b = a, prove that logc a = ac.(e) If logx 10.24 = 2, find x.(f) Without using a calculator, simplify and evaluate the following expression:
log2
(53
)+ log2
(67
)− log2
( 528
)
(g) Solve the following equation: loga(x2 + 3)− loga x = 2 loga 2.
5
2 Trigonometric Equations
Revision MaterialBasic definitions:
tan θ = sin θcos θ ; cosec θ = 1
sin θ ; sec θ = 1cos θ ; cot θ = 1
tan θ .
Pythagorean identities:
sin2 θ + cos2 θ = 1sec2 θ = 1 + tan2 θ
cosec2θ = 1 + cot2 θ
Addition formulae:
sin(A+B) = sinA cosB + cosA sinBsin(A−B) = sinA cosB − cosA sinBcos(A+B) = cosA cosB − sinA sinBcos(A−B) = cosA cosB + sinA sinB
tan(A+B) = tanA+ tanB1− tanA tanB
tan(A−B) = tanA− tanB1 + tanA tanB
Double angle formulae:
sin 2A = 2 sinA cosAcos 2A = cos2 A− sin2 A
cos 2A = 2 cos2 A− 1cos 2A = 1− 2 sin2 A
tan 2A = 2 tanA1− tan2 A
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Sum and difference formulae:
sinA+ sinB = 2 sin(A+B
2
)cos
(A−B
2
)sinA− sinB = 2 cos
(A+B
2
)sin
(A−B
2
)cosA+ cosB = 2 cos
(A+B
2
)cos
(A−B
2
)cosA− cosB = −2 sin
(A+B
2
)sin
(A−B
2
)Formulae using t = tan x
2 :
sin x = 2t1 + t2
; cos x = 1− t21 + t2
; tan x = 2t1− t2 .
Transformation:a cosx+ b sin x = R cos(x− α), where R =
√a2 + b2 and tanα = b/a.
Practice Problems2.1. Find all angles x in the range 0◦ ≤ x ≤ 360◦ which satisfy the following equations:
(a) tan x = −0.4560(b) sin 3x
2 = −0.5678(c) cos 2x
3 = tan 155◦
(d) sin 3x4 = cot 108◦
2.2. Find all values of x (0◦ ≤ x ≤ 360◦) which satisfy the following equations:
(a) 6 sin2 x− 5 cosx− 2 = 0(b) 4 tan x− 2 cotx = 5cosecx(c) tan(45◦ + x) + cot(45◦ + x) = 4
(d) cos 2x+ 7 sin x+ 3 = 0
(e) cos 3x− 3 cosx = cos 2x+ 1
2.3. Using Trigonometric Formulae:
(a) Use addition formulae to show that cot(A−B) = 1 + cotA cotBcotB − cotA .
(b) Show thatsin(A+B+C) = cosA cosB cosC(tanA+tanB+tanC− tanA tanB tanC).
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(c) Show that:i. sin 3A = 3 sinA− 4 sin3 A;
ii. cos 3A = 4 cos3 A− 3 cosA.(d) Show that sin x(cos 2x+ cos 4x+ cos 6x) = sin 3x cos 4x.(e) Show that sin2(A+B)− sin2(A−B) = sin 2A sin 2B.(f) Find the values of x between 0◦ and 180◦ that satisfy: cosx = cos 2x+ cos 4x.(g) Without using a calculator, evaluate cos4 15◦ + sin4 15◦.
2.4. Problems with tan x2 :
(a) If t = tan x2 , find the values of t that satisfy:
(a+ 2) sin(x) + (2a− 1) cosx = 2a+ 1
where a is a non-zero content. Hence find two acute angles satisfying the equationwhen a =
√3.
(b) If tan x2 = cosecx− sin x, prove that tan2 x
2 = −2±√
5.(c) If sec θ − tan θ = x, prove that tan θ
2 = (1− x)/(1 + x).
(d) Find√
(1 + sin θ)(3 sin θ + 4 cos θ + 5) in terms of tan θ2 .
2.5. Transformations:
(a) Express cosx+ sin x in the form R cos(x−α) given the values of R and α. Hencefind the maximum and minimum values of cosx+ sin x.
(b) Find the values of x between 0◦ and 360◦ which satisfy the equations:
(i) 3 cosx+ sin x = 1 (ii) 8 cosx+ 9 sin x = 7.25(iii) cosx+ 7 sin x = 5 (iv) 7 cosx− 6 sin x = 2(v) 5 sin x− 6 cosx = 4 (vi) 12 cosx− 5 sin x+ 3 = 0
(c) Express y = W (sinα+µ cosα) in the form R cos(α−β), giving R and tan β. Showthat the maximum value of y is W
√1 + µ2 and that it occurs when tanα = 1/µ
8
2.6. Solve the following:
(a) Find a positive value of x satisfying the equationtan−1(2x) + tan−1(3x) = 1
4π.(b) Solve the equation cos−1(x
√3) + 2 sin−1 x = 1
2π.
3 Differentiation
Revision MaterialFor a general function y = f(x), the derivative of y with respect to (sometimes abbreviatedto w.r.t) x is denoted by:
dydx or y′ or f ′(x)
and is defined by:
dydx = lim
δx→0
{f(x+ δx)− f(x)
δx
}(2)
While this expression may generally be used to find any derivative from first principles, it ismore usual to rely on standard results, such as given in the following table:
9
Function Derivative Comment
y = xndydx = nxn−1 n may be positive,
negative or fractional.y = ex dy
dx = ex
y = ln x dydx = 1
xln x ≡ loge x
y = sin x dydx = cosx
y = cosx dydx = − sin x
y = tan x dydx = sec2 x
y = sin−1 xdydx = 1√
1− x2
y = tan−1 xdydx = 1
1 + x2
y = f(x).g(x) dydx = g(x)f ′(x) + f(x)g′(x) The Product rule
y = f(x)g(x)
dydx = g(x)f ′(x)− f(x)g′(x)
g2(x) The Quotient rule
y = f [g(x)] dydx = df
dg ×dgdx The Chain rule
(A function of a function)
10
Practice Questions3.1. Differentiate w.r.t. x:
(a) x3
(b) x2 + 4x+ 5(c) x4 − x2
(d) 1x2
(e) 1x4
3.2. Differentiate w.r.t. x:
(a) 1(1 + x)
(b) 1(1− x)
(c) 1(3x+ 2)
3.3. Differentiate w.r.t. x:
(a) sin 3x(b) x+ sin x
(c) cos 5x5
(d) secx(e) tan 2x
3.4. Differentiate w.r.t. x:
(a) (1− x)(2− 3x+ x2)
(b) 5x3 − 3xx2
(c) (x+ 1)2
x
3.5. Solve the following problems:
(a) If y + x = cosx, find dydx .
(b) If y = 2x2 − 4x− 2, find the value of x for which dydx = 0.
(c) Differentiate 3x2 + 6 + 5x−2 w.r.t. x and evaluate the derivative when x = 3.
(d) If y(x+ 1) = 4, find dydx .
11
3.6. Differentiate w.r.t. x:
(a) x5 + 5x4 + 10x3 + 8
(b) 3x5 − cosx+ 2
(c) 3x2 − 1x2
(d) 2x 12 + 2x− 1
2 − 5
3.7. Differentiate w.r.t. x:
(a) (x+ 2)(x2 + 4)(b) (1 + x2)(1− 3x2)(c) (x2 − 2)2
(d) (x2 − 1)2/x
(e) (1− x)3
(f) (3x+ 4)3
(g) (3 + x)(4− x)
3.8. Differentiate w.r.t. x:
(a) sin x cosx
(b) x sin x+ cosx
(c) 12(x2 − 2) sin x+ x cosx
(d) (x2 + 1) tan x(e) secx tan x(f) cosecx cotx(g) x secx+ cosx
3.9. Differentiate w.r.t. x:
(a) 2x1− x2
(b) 2 + x2
1− x
(c) 3x2
(x− 1)(x+ 2)
(d) (3− 2x2)−1
(e) x4
x4 + 4
3.10. Differentiate w.r.t. x:
(a) sin x2x
(b) 3 sin x1 + cos x
(c) 5 + 3 sin x3 + 5 sin x
(d) 2 + 7 cosx3 + 5 cosx
(e) sin x1 + tan x
12
3.11. Differentiate w.r.t. x:
(a) (2x− 3)4
(b)√
2x− 3(c) cos 5x(d) sin(x2)(e) tan(2x+ 5)(f) sec(x3)(g) sin 2x cos 2x
(h) cosec√x
(i)(x+ 1− 1
x
)3
(j) 1 + sin2 x
1− sin2 x
(k) sin3 x sin 3x
(l) 4x√
1 + x2
3.12. Differentiate w.r.t. x:
(a) sin−1 2x(b) 3 tan−1√x
(c) x√
1− x2 + sin−1 x
(d) tan−1(sin2 x)
3.13. Differentiate w.r.t. x:
(a) x(ln x− 1)
(b) ln(1/x)
(c) ln(secx)
(d) 2xln x
(e) ln(sin x+ cosx
sin x− cosx
)(f) e−2x2 ln 3x(g) e2x ln(secx)(h) x2ex tan−1 x
3.14. Solve the following problems:
(a) If x = aθ2 and y = 2aθ, find dy/dx in terms of θ.
(b) If x = b(θ − sin θ) and y = b(1− cos θ), show that(
dydx
)2
= cosec2(θ
2
)− 1.
(c) If x = 3t+ t3 and y = 3− t5/2 express dy/dx in terms of t and prove that, whend2y/dx2 = 0, x has one of the values 0,±6
√3.
(d) Find d2y
dx2 and d3y
dx3 when (i) y = x5 + 2x3 + 4 and (ii) y = sin 4x.
(e) Find the maximum and minimum values of the function(1 + 2x2)e−x2 .
13
4 Integration
Revision MaterialA brief list of standard integrals:∫
xn dx = xn+1
n+ 1 + C, n 6= 1,∫
(ax+ b)n dx = (ax+ b)n+1
(n+ 1)a + C, n 6= 1
∫cos(ax+ b) dx = 1
asin(ax+ b) + C,
∫sin(ax+ b) dx = −1
acos(ax+ b) + C
∫sec2(ax+ b) dx = 1
atan(ax+ b) + C,
∫cosec2(ax+ b) dx = −1
acot(ax+ b) + C
∫ dxa2 + x2 = 1
atan−1 x
a+ C,
∫ dx√a2 − x2
= sin−1 x
a+ C
∫ f ′(x)f(x) dx = ln f(x) + C,
∫eax dx = eax
a+ C
Change of variable (integration by substitution):∫φ(x) dx =
∫φ(x)dx
dudu
Special cases: ∫φ(ax+ b) dx = 1
a
∫φ(u) du, u = ax+ b∫
xφ(x2) dx = 12
∫φ(u) du, u = x2
Useful substitutions:
• If the integrand contains (a2 − x2), x = a sin u or x = a cosu may be useful.
• If the integrand contains (a2 + x2), x = a tan u may be useful.
• If the integrand contains sinm x cosn x, where m and n are positive integers and at leastone of them is odd:
– If n is odd, sin x = u should be used;– If m is odd, cosx = u should be used.
14
Integration by parts:b∫a
udvdx dx = [uv]ba −
b∫a
vdudx dx
Practice Problems4.1. Evaluate the following integrals:
(a)∫
(4x2 − 2x+ 9) dx
(b)3∫
1
( 2x3 + 3
x2 − 5)
dx
(c)∫
(2 sin x+ 5cosec2x) dx
(d)1∫
0
(1− x2)√x dx
(e)∫ x4 + 1
x2 dx
(f)1/2∫−1/2
(4x+ 3√
1− x2
)dx
(g)∫ 3x4 + 5x2 + 2
x2 dx
(h)3∫
0
x3 + 1x+ 1 dx
(i)∫ x4 + x2 + 1
x2 + x+ 1 dx
4.2. Evaluate the following integrals:
(a)∫
sin(1− x) dx
(b)π/2∫0
cos 5x dx
(c)0∫−1
√1− 9x dx
(d)∫
(2x− 3)−3/2 dx
(e)∫ [
1(x− 1)3 −
1(2− x)3
]dx
(f)1∫−1
dx16x2 + 9 dx
(g)1∫−1
dx√16− 9x2
dx
(h)∫ 3dx√
11 + 4x− 4x2dx
(i)∫ 2dx
3x2 − 4x+ 7 dx
15
4.3. Integrate the following functions w.r.t. x using appropriate substitutions:
(a) x√
1− x2
(b) x(1 + x2)3/2
(c)√x
x+ 1
(d)[x+√
1 + x2]2
4.4. Integrate the following using the suggested substitutions:
(a) x2√a3 + x3
, using a3 + x3 = u
(b) x3
1 + x8 , using x4 = tan u
(c) tan x sec2 x, using tan x = u
(d) x2√
1− x6, using x3 = sin u
(e) x
x4 + 9 , using x2 = 3 tan u
(f) x+ 1√x2 + 2x− 9
,
using x2 + 2x− 9 = u
4.5. Integrate by substitution or otherwise:
(a) cos3 x
(b) sin5 x
(c) sin2 x cos2 x
(d) cos4 x sin3 x
(e) sin3 x cos5 x
(f) sin3 2x
(g) sin3 x
cos2 x
(h) cosx√sin x
4.6. Evaluate the following integrals by parts, or otherwise:
(a)∫
cos 3x cos 2x dx
(b)∫
sin 3x cosx dx
(c)π/2∫0
sin 3x sin 2x dx
(d)∫
cos 4x sin 2x dx
(e)π∫
0
x cosx dx
(f)∫x√
1 + x dx
(g)1∫
0
x2√1 + x dx
(h)∫
sin−1 x dx
16
(i)∫ x sin−1 x√
1− x2dx
(j)π∫
0
x sin 2x dx
(k)∫x2 sin x dx
(l)π∫
0
(π − x) sin 3x dx
(m)∫x ln x dx
(n)∫x ln(x+ 4) dx
(o)∫x3 ln 5x dx
4.7. Evaluate the following integrals:
(a)1∫
0
e−3x dx
(b)1∫
0
(ex − e−x)2 dx
(c)2∫
0
(x− 1)ex dx
(d)∫
ex+2 dx
(e)∫x2e2x dx
(f)1∫
0
x3ex2 dx
(g)∫ [
ex + e−xex − e−x
]dx
(h)∫ xe−1 + ex−1
xe + ex dx
Vectors
Revision materialIn the context of this course, a vector is a line segment from a fixed point, called the originand written O, to another point in two- and three-dimensional space. We will write vectorsas bold italic, e.g. v, r. In handwritten script, this is often denoted with a straight underline,v or a curly underline v:. Some books and lecturers will use other notation for a vector, suchas an overarrow, ~v.
Vectors are used to denote quantities which inherently involve position in space or direction:position, velocity, acceleration, force, momentum, . . . are all vectorial quantities. Quantitieswithout a sense of direction are called scalars; examples include energy, temperature, andspeed. Vectorial quantities can have physical dimensions and units; for instance the positionvector r is a length (measured in metres), whereas velocity v has the same units as speed(m s−1), and acceleration a is measured in m s−2, etc.
17
Multiplying a vector by a number (i.e. a scalar) gives a vector parallel to the original vector,but with its length changed. For instance, the vector 2r lies in the same direction as the linefrom O to r, but is twice as far from the origin. Vectors can also be added together, andby the parallelogram rule, the order of addition not not matter, a + b = b + a. In general,vectors can be added together and multiplied by scalars, which looks a bit like the algebraof scalars. However, vector multiplication is different (as we will see) and in particular youcan’t divide by a vector! Although there is a simple geometric meaning to vector additionand scalar multiplication, there is no obvious geometric construction for vector multiplicationor division. Expressing a vector as a sum of other vectors is sometimes called resolving thevector.
Often, we write a vector v as a product of its magnitude (or modulus) |v| and its directionv. The magnitude |v| is a scalar that carries the sense of the length of the vector; it has thesame units as the vector itself. The direction v is defined as
v = v
|v|,
i.e. the vector divided by its scalar magnitude. v is therefore a dimensionless vector of length1 (said to be a unit vector), pointing in the same direction as v. Different vectors, measuringdifferent physical quantities, may point in the same direction (e.g. force and acceleration inthe vectorial version of Newton’s law F = ma – in this case, they have the same direction,i.e. F = a.
Putting these ideas together, we can define the direction vector i in the direction of thex-coordinate, j in the y-coordinate, and k in the z-coordinate. These vectors are defined tobe unit vectors, so
i = i, j = j, k = k, |i| = |j| = |k| = 1.
Therefore we can express a coordinate vector r with components x, y and z in their respectivedirections as
r = xi + yj + zk.
The velocity vector v, with components vx, vy and vz, can similarly be written
v = vxi + vyj + vzk.
The x, y, z axes are sometimes called Cartesian axes (after the French scientist and philosopherRene Descartes), and the vectors i, j,k pointing along these directions are sometimes calledCartesian basis vectors.
18
If we consider vectors in the xy-plane, we can see how some of this fits together. Let’s sayr = xi + yj. By Pythagoras’ theorem, we can see that the modulus of the direction vector|r| =
√x2 + y2, since the x, y axes are mutually perpendicular. To keep the notation simple,
we will write r = |r|. If the vector r makes an angle φ with respect to the x-axis (i.e. the idirection), so that if φ = 90◦ then it is parallel to the y-axis (the j-direction), then we canuse straightforward trigonometry to show that
r = cosφ i + sinφ j.
Since r = rr, we conclude that
x = r cosφ, y = r sinφ.
r is sometimes called the radial coordinate, since itis the radius of the circle centred on O as the an-gle φ varies. It is conventional to define the angle ashere as the direction angle (also called “azimuthal an-gle”, and often denoted by θ) of the two-dimensionalvector. It is given by the formula φ = arctan(y/x)(note you have to be careful that you are not wrongby 180◦, since y/x = (−y)/(−x)).
O
r
�y
x r
With a fixed choice of axes (and basis vectors along them), it is often convenient to writea vector as an array of Cartesian components, e.g. r = (x, y, z). In this sense, we havei = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1). Vector arrays are also often written as columnvectors, i.e. r =
( xyz
). In this course, we will usually write vectors in terms of i, j,k rather
than using arrays or column vectors. You are encouraged to get comfortable with both waysof writing vectors!
The next important notion is that of the scalar product of two vectors (also called the“dot product”): it is a form of multiplying two vectors which gives a scalar quantity (ratherthan a vector). For two vectors a = axi + ayj + azk, b = bxi + byj + bzk, the scalar producta · b is the scalar
a · b = axbx + ayby + azbz.
If the vectors are only in two dimensions (the xy plane), we can proceed as if az, bz = 0.We can work this out assuming a and b are both in the xy plane, with a making an angle
α with the x-axis, and b making an angle β with the x-axis. From what we saw above, we
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can writea = axi + ayj = |a|(cosα i + sinα j),b = bxi + byj = |b|(cos β i + sin β j),
where we have written a for |a| and b for |b|. The scalar product is therefore
a · b = axbx + ayby = |a||b| cosα cos β + |a||b| sinα sin β.
We can idnetify the common scalar factor of |a||b| out at the front, and apply the trigonometricidentity cosα cos β + sinα sin β = cos(α− β). α− β is of course the angle between the twovectors, which is often called θ. This result holds in three dimensions as well, and the generalresult is that
a · b = |a||b| cos θ.xThis is the general way of finding angles between vectors! Although we call this the “scalarproduct”, it is not a multiplication in the standard sense. However, just like standard multi-plication, the order of the vectors in the scalar product does not matter, i.e. a · b = b · a.This can easily be justified by looking at both formulas for the scalar product (the sum ofcomponent products, or using the moduli and angle).
If we take the scalar product of a vector with itself, we get the modulus squared (since θin this case is zero, and its cos is 1), a ·a = |a|2. Therefore we can easily find the modulus ofany vector by taking the scalar product of the vector with itself, and taking the square root:
|a| =√
a · a.
This is consistent with what we found before for the modulus of the position vector r = xi+yjusing geometry.
Since the Cartesian basis vectors i, j,k are mutually perpendicular (also called orthogonal),and each is a unit vector, they have the scalar products
i · i = 1, i · j = 0, i · k = 0,j · i = 0, j · j = 1, j · k = 0,k · i = 0, k · j = 0, k · k = 1.
We can think of the Cartesian components of any vector a = axi + ayj + azk as the scalarproduct with the appropriate Cartesian basis vector, i.e.
ax = a · i, ay = a · j, az = a · k.
Thinking about the angle relation for scalar products, we can understand why some olderbooks refer to a vector’s Cartesian components as “direction cosines”.
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Practice problems4.1. Scalar product.
(a) Find the scalar product A ·B for:i. A = 3i− 6j; B = −4i + 2j;
ii. A = 5i + 5j; B = 2i− 4j;iii. A = 6i + 4j; B = 4i− 6j;
(b) What is the angle between the vectors A and B if A ·B = −|A||B|?
4.2. Find the scalar product A ·B for:
(a) A = 2i + 3j; B = 3i + 4j;(b) A = 2i + 3j − k; B = 3i + j;(c) A = 2i + 3j − k; B = 3i + j + 4k.
4.3. A wall clock has a minute hand that has a length of 0.5 m and an hour hand with a lengthof 0.25 m. Taking the centre of the clock as the origin, and choosing an appropriatecoordinate system, write the position of the hour and minute hands as vectors when thetime reads
(a) 12 : 00;(b) 3 : 30;(c) 6 : 30;(d) 7 : 15.(e) Call the position of the tip of the minute hand A and the position of the tip of the
hour hand B. Find A−B for the times given in the parts above.
4.4. Find the scalar product A ·B for:
(a) A = 2i + 3j; B = 3i + 4j;(b) A = 2i + 3j − k; B = 3i + j;(c) A = 2i + 3j − k; B = 3i + j + 4k.
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4.5. Vector manipulations – transforming between different descriptions of a vector:
(a) Find the Cartesian components of the following vectors which have a magnitudeA and lie in the xy plane and make an angle θ with the x-axis if:
i. A = 10 m, θ = 30◦;ii. A = 5 m, θ = 45◦;
iii. A = 7 km, θ = 60◦;iv. A = 5 km, θ = 90◦;
v. A = 15 km s−1, θ = 150◦;
vi. A = 10m s−1, θ = 240◦;
vii. A = 8m s−2, θ = 270◦.
(b) Find the magnitude and direction of the following vectors:i. A = 5i + 3j;
ii. B = 10i− 7j;iii. C = −2i− 3j + 4k.
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