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12 Vector Techniques
Bernard Bolzano was born in 1781 in
Prague in what is now the Czech Republic.
During Bolzano’s early life there were two
major influences. The first was his father,
who was active in caring for others and
the second were the monks who taught
him, who were required to take a vow
which committed them to take special care
of young people. In the year 1799–1800
Bolzano undertook mathematical research
with Frantisek Josef Gerstner and
contemplated his future. The result of this
was that in the autumn of 1800, he went
to Charles University to study theology.
During this time he also continued to
work on mathematics and prepared a
doctoral thesis on geometry which led to
him publishing a work on the foundations
of elementary geometry, Betrachtungen über
einige Gegenstände der Elementargoemetrie in 1804.
Bernard Bolzano
In this book Bolzano considers points, lines
and planes as undefined elements, and
defines operations on them. These are key ideas in the concept of linear space, which
then led to the concept of vectors.
Following this, Bolzano entered two competitions for chairs at the Charles University
in Prague. One was for the chair of mathematics and the other for the new chair in
the philosophy of religion. Bolzano was placed first in both competitions, but the
university gave him the chair in the philosophy of religion. In many ways this was the
wrong decision, given the way he was brought up with a belief in social justice and
pacifism and the fact he was a free thinker. His appointment was viewed with
suspicion by the Austrian rulers in Vienna. He criticised the discrimination of the
Czech-speaking Bohemians by the German-speaking Bohemians, against their Czech
fellow citizens and the anti-Semitism displayed by both the German and Czech
Bohemians. It came as no surprise that Bolzano was suspended from his position in
December 1819 after pressure from the Austrian government. He was also suspended
from his professorship, put under house arrest, had his mail censored, and was not
allowed to publish. He was then tried by the Church, and was required to recant his
supposed heresies. He refused to do so and resigned his chair at the university. From
1823 he continued to study, until in the winter of 1848 he contracted a cold which,
given the poor condition of his lungs, led to his death.

305

12 Vector Techniques

12.1 Introduction to vectors
Physical quantities can be classified into two different kinds:
(i) scalar quantities, often called scalars, which have magnitude, but no associated
direction
(ii) vector quantities, often called vectors, which have a magnitude and an associated
direction.
So travelling 20 m is a scalar quantity and is called distance whereas travelling 20 m due
north is a vector quantity and is called displacement.

Vector notation
Vectors can be represented in either two or three dimensions, and are described through
components. Hence if we want to move from the point (1, 2) on the Cartesian plane to
the point (3, 5) we do this by stating we move 2 in the positive x-direction and 3 in the
positive y-direction. There are two possible notations for this, column vector notation
and unit vector notation.

Column vector notation
x
x
In two dimensions a vector can be represented as ¢ ≤ and in three dimensions as £y≥.
y
z

The conventions for x and y in terms of positive and negative are the same as in the
standard two-dimensional Cartesian plane. In three dimensions this is also true, but
we need to define what a standard three-dimensional plane looks like. There are
three different versions, which are all rotations of each other. In all cases they obey
what can be called the “right-hand screw rule”. This means that if a screw were
placed at the origin and turned with a screwdriver in the right hand from the positive
x-axis to the positive y-axis, then it would move in the direction of the positive z-axis.
This is shown below.
z

The axes are always drawn
like this in this book.
Different orientations may
be used on IB examination
papers.

y

x

Unit vector notation
2
The column vector £ 3 ≥ can be represented as 2i 3j 2k using unit vectors.
2

306

12 Vector Techniques

Here the unit vectors i, j and k are vectors of magnitude 1 in the directions x, y and z
respectively. These are shown in the diagram below.
z

y

k
j
x

i

So the vector 2i 3j 2k means 2 along the x-axis, 3 along the y-axis and 2 along
the z-axis.
Hence a vector represents a change in position.

Position vectors, free vectors and tied vectors

There is no advantage to
one notation over the
other. Both are used in IB
examinations and it is
probably best to work in
the notation given in the
question.

A vector can be written as a position vector, a free vector or a tied vector.
A position vector is one that specifies a particular position in space relative to the
¡
2
origin. For example, in the diagram, the position vector of A is OA ¢ ≤.
1

A (2, 1)

2
1

¡
2
OA ¢ ≤ means A is 2
1

1

units to the right of O and
O (0, 0)

1 unit above it.

2

¡
2
Point A with coordinates (2, 1) has position vector OA ¢ ≤.
1

This is true for the position
vector of any point.

2
Now if we talk about a vector a ¢ ≤ then this can be anywhere in space and is
1
therefore a free vector.

a

1

2
¡
2
A vector AB ¢ ≤ is a tied vector since it is specified as the vector that goes from A to B.
1

307

12 Vector Techniques
B

1

A

2

¡
¡
2
2
It is obviously possible that OA a ¢ ≤ and that AB a ¢ ≤, but it must be
1
1
¡

¡

understood that OA and a and AB are slightly different concepts.
Whenever vectors are printed in books or in examination papers, free vectors are always
written in bold, for example a, but in any written work they are written with a bar
underneath, a. Position vectors and tied vectors are always written as the start and end
points of the line representing the vector with an arrow above them, for example
¡

¡

OA , AB .

Forming a tied vector
¡

We now know the vector AB means the vector that takes us from A to B. If we
consider A to be the point 11, 2, 72 and B to be the point (3, 1, 6), then to get from
A to B we need to move 2 along the x-axis, 1 along the y-axis and 13 along the z-axis.
This is shown in the diagram below.

z

B (3, 1, 6)

y

13
O

x

2
A (1, 2, −7)

1
¡

To find BA we
subtract the coordinates of
B from those of A.

2
So AB £ 1≥.
13
¡

¡

More commonly, we think of AB as being the coordinates of A subtracted from the
coordinates of B.

Example
If A has coordinates 12, 3, 12 and B has coordinates 13, 4, 12 find:
¡

a) AB

¡

b) BA

308

12 Vector Techniques

a) To get from 12, 3, 12 to 13, 4, 12 we go 1 in the x-direction, 1 in the
3 2
1
y-direction and 0 in the z-direction. Alternatively, AB £ 4 1 32 ≥ £ 1≥.
1 1
0
¡

¡

b) Similarly with BA , we go 1 in the x-direction, 1 in the y-direction and 0 in
2 3
1
¡
the z-direction. Alternatively, BA £ 3 1 42 ≥ £ 1 ≥.
1 1
0
Notice that in the example
¡

¡

AB BA

This is always true.

The magnitude of a vector
The magnitude (sometimes called the modulus) of a vector is the length of the line
representing the vector. To calculate this we use Pythagoras’ theorem.

Example
5
Find the magnitude of the vector a ¢ ≤.
12
5
Consider the vector a ¢ ≤ in the diagram below.
12

5
12

12

5

The magnitude is given by the length of the hypotenuse. 冟a冟 252 122 13

冟a冟 means the magnitude
of a.

In three dimensions this becomes a little more complicated.

309

12 Vector Techniques

Example
3
Find the magnitude of the vector £4≥.
7
3
The vector £4≥ is shown in the diagram below. OA is the magnitude of the
7
vector.
A
7
3
4
7

C
4

O

B

3

We know by Pythagoras’ theorem that OA 2OC2 AC2.
Applying Pythagoras’ theorem again, OC2 OB2 BC2.
Hence
OA 2OB2 BC2 AC2
1 OA 232 42 72
1 OA 274

Multiplying a vector by a scalar
When we multiply a vector by a scalar we just multiply each component by the scalar.
3
6
Hence the vector changes in magnitude, but not in direction. For example 2¢ ≤ ¢ ≤
4
8
3
has the same direction as ¢ ≤ but has twice the magnitude. This is shown in the
4
diagram below.

3
4

3
4

3
3c
In general, c ¢ ≤ ¢ ≤.
4
4c

310

There is no symbol for
multiplication in this case.
This is important as the
symbols ⴢ and
have specific meanings in
vectors.

12 Vector Techniques

Example
If A has coordinates 12, 3, 12 and B has coordinates 17, 6, 12, find these
vectors.
¡

a) AB

¡

b) 2 BA

¡

c) p AB

7 2
5
(a) AB £ 6 3 ≥ £3≥
1 1 12
0
¡

2 7
5
(b) BA £ 3 6 ≥ £ 3≥
1 1 12
0
¡

5
10
2 BA 2£ 3≥ £ 6 ≥
0
0
¡

5
5p
(c) p AB p £3≥ £3p≥
0
0
¡

Equal vectors
Vectors are equal if they have the same direction and magnitude.

Example
a 1
2a 2
Find the values of a, b and c for which the vectors 3 £2b 3≥ and £ b 1 ≥
c
5c 2
are equal.
If they are equal then
31a 12 2a 2
1 3a 3 2a 2 1 a 1
312b 32 b 1
8
1 6b 9 b 1 1 b
5
3c 5c 2 1 c 1

Negative vectors
A negative vector has the same magnitude as the positive vector but the opposite
2
2
direction. Hence if a ¢ ≤ then a ¢ ≤.
3
3

311

12 Vector Techniques

Zero vectors
A zero or null vector is a vector with zero magnitude and no directional property. It is
0
denoted by ¢ ≤ or 0i 0j in two dimensions. Adding a vector and its negative vector
0
gives the zero vector, i.e. a 1 a2 zero vector.

Example
If a 2i 4j 7k and 2a b 0i 0j 0k, find b.
2a 4i 8j 14k
b 2a 14i 8j 14k2 4i 8j 14k

If 2a b 0 then
b 2a

Unit vectors
A unit vector is a vector of magnitude one. To find this we divide by the magnitude of
the vector. If n is the vector then the notation for the unit vector is nˆ .

Example
Find a unit vector parallel to m 3i 5j 2k.
The magnitude of 3i 5j 2k is 232 52 1 22 2 238.
1
ˆ
Hence the required vector is m
13i 5j 2k2.
238

Parallel vectors
Since parallel vectors must have the same direction, the vectors must be scalar multiples
of each other.
3
15
15
3
So in two dimensions ¢ ≤ is parallel to ¢
≤ since ¢
≤ 5 ¢ ≤.
2
10
10
2
In three dimensions 4i 2j 5k is parallel to 12i 6j 15k since
12i 6j 15k 314i 2j 5k2.

Example
4
12
Find the value of k for which the vectors £ 2≥ and £ 6≥ are parallel.
8
k
2
4
2
12
1
£ 2≥ 2£ 1≥ and £ 6≥ 6 § k ¥
8
4
k
6

Hence these vectors are parallel when

312

In the case of an equation
like this the zero vector
0i 0j 0k could just
be written as 0.

k
4 1 k 24.
6

12 Vector Techniques

Perpendicular vectors
In the two-dimensional case we use the property that with perpendicular lines the
product of the gradients is 1.

Example
Find a vector perpendicular to ¢

3
≤.
1

From the diagram below we can see that the line representing this vector has
1
a gradient of .
3
3
−1

Hence the line representing the perpendicular vector will have a gradient of 3.
1
Therefore a perpendicular vector is ¢ ≤.
3

There are an infinite
number of perpendicular
vectors.

In three dimensions this is more complicated and will be dealt with later in the chapter.

Exercise 1
1 Find the values of a, b and c.
2
a
a £3≥ 2£b 1≥
4
c 2

a
2 a
c 3£b 1≥ 4£2b 3≥
4
3

1
3a
b £ b ≥ £ 2b2 ≥
2
c 6

2 If the position vector of P is i j and the position vector of Q is 2i 3j, find:
¡

a PQ

¡

b 冨 PQ 冨

2
1
3 If the position vector of A is ¢ ≤ and the position vector of B is ¢ ≤, find:
3
5
¡

a AB

¡

b 冨 AB 冨

4 Write down a vector that is parallel to the line y 3x 5.
5 Find the magnitude of these vectors.
¡
2
x
3
a m 3i 5j
b OP ¢ ≤
c ¢ ≤ ¢ ≤
7
y
9
d a 2i 4j 3k

x
4
e £y≥ £ 1≥
z
2

¡

f OA 2i 7j 2k

1
6 State which of the following vectors are parallel to £ 12≥.
16
1
a £ 12≥
16

4
1
i j– k
b
3
3

5
c £ 15≥
20

d 0.5p 12i 6j 8k2
˛

313

12 Vector Techniques

7 Find the values of c for which the vectors are parallel.
14
18
a i 2j 3k and 3i cj 9k
b £ 35≥ and £ 45≥
c
9
4t
ct
c £ 8t≥ and £ 12t≥
10t
15t
8 A two-dimensional vector has a modulus of 13. It makes an angle of 60°
with the x-axis and an angle of 30° with the y-axis. Find an exact value for
this vector.
5
9 Find a unit vector in the direction of £ 6≥.
1
10 A, B, C and D have position vectors given by
i 3j 2k, 2i j 4k, 3i 2j 7k and 3i 5j 4k. Determine
which of the following pairs of lines are parallel.
a AB and CD
b BC and CD
c BC and AD
11 A triangle has its vertices at the points P (1, 2), Q (3, 5) and R 1 1, 12. Find
˛

¡

¡

¡

the vectors PQ , QR and PR , and the modulus of each of these vectors.
12 A parallelogram has coordinates P(0, 1, 4), Q 14, 1, 32, R(x, y, z) and
S 1 1, 5, 62 .
a Find the coordinates of R.
˛

˛

¡

¡

¡

¡

b Find the vectors PQ , QR , SR and RP .
c Find the magnitude of each of the vectors in part b.
¡

¡

¡

¡

d Hence write down the unit vectors in the directions of PQ , QR , SR and RP .
¡

¡

¡

¡

¡

13 If PQRST is a pentagon, show that PQ QR RS PT TS .
14 Consider the hexagon shown.
a

P

Q
b

U

R

T

S

¡

Find the vector represented by SU .
a b
2a b
≤ is parallel to the x-axis and the vector ¢
≤ is parallel
b
a b
2
to ¢ ≤. Find the values of a and b.
1
16 The vector 1p 2q r2 i 12p q 6r 2j 13p 2q r 2k is parallel
to i j and the vector 1p r2 i 1–q 2r 2j 1p 2q r 2k is
parallel to the z-axis. Find the values of p, q and r.
15 The vector ¢

¡

¡

17 If OP 13x 2y 2p 1x y 32q, OQ 1x y 22p 12x y 1 2q
¡

¡

and 2 OP 3OQ, where vectors p and q are non-parallel vectors, find the
values of x and y.

314

12 Vector Techniques

12.2 A geometric approach to vectors
To add two vectors we just add the x-components, the y-components and the zcomponents. To subtract two vectors we subtract the x-components, the y-components
and the z-components.

Example
If a 2i 6j 12k and b i 6j 7k find a b.
a b 12 12 i 1 6 62j 112 72k
i 12j 19k

Example
3
2
¡
¡
¡
¡
If OA £ 4 ≥ and OB £ 4≥ find OA OB .
2
7
3
2
3 2
5
¡
¡
OA OB £ 4 ≥ £ 4≥ £4 1 42 ≥ £ 0 ≥
2
7
2 1 72
9

We can also look at adding and subtracting vectors geometrically.

Vector addition
¡

¡

Let the vectors p and q be represented by the lines AB and BC respectively as shown
in the diagram.
C
p+q
q

A

B

p

¡

Then the vector represented by the line AC is defined as the sum of p and q and is
written as p q. This is sometimes called the triangle law of vector addition.
Alternatively it can also be represented by a parallelogram. In this case let p and q be
¡

¡

represented by AB and AD .
D

q

A

p

C

p+q

p

q

B

315

12 Vector Techniques
¡

¡

It should be noted that since DC is the same in magnitude and direction as AB , the
¡

¡

line DC can also represent the vector p. Similarly BC can represent the vector q.
¡

Comparing this with the triangle, it is clear that the diagonal AC can represent the sum
¡

¡

AB BC . This is known as the parallelogram law of vector addition.

This also shows that vector addition is commutative.
¡

¡

¡

¡

¡

q p AD DC AC AB BC p q
Put very simply, vector addition can be thought of as getting
from the start point to the end point by any route. Hence in
the case of the triangle the route along two sides is the
same as the route along the third side because they start
and finish at the same point.

C

A

Hence

¡

¡

B

¡

AB BC AC

This concept can be extended to more than two vectors. To

F
E

get from A to F we can either go directly from A to F or we
can go via B, C, D, and E. Hence
¡

¡

¡

¡

¡

D

¡

AF AB BC CD DE EF .
C
A

B

Example
A quadrilateral has coordinates A 11, 1, 42, B (3, 2, 5), C (1, 2, 0),
D 1 1, 2, 12. Show that:
˛

˛

¡

¡

¡

¡

¡

¡

a) AB BC AC
¡

b) AB BC CD AD
2
2
0
¡
¡
¡
a) AB £3≥, BC £ 0 ≥, AC £ 3 ≥
1
5
4
2
2
0
¡
¡
AB BC £3≥ £ 0 ≥ £ 3 ≥ AC
1
5
4

¡

2
2
¡
¡
b) CD £ 0 ≥, AD £ 3 ≥
1
5
2
2
2
2
¡
¡
¡
AB BC CD £3≥ £ 0 ≥ £ 0 ≥ £ 3 ≥ AD
1
5
1
5

¡

316

12 Vector Techniques

Example
If A has coordinates 10, 1, 22 and B has coordinates 12, 3, 52, find:
a) the position vector of the point C, the midpoint of AB
b) the position vector of the point D which divides the line AB in the ratio of
1 : 2.
a)
2 0
2
AB £ 3 1 12 ≥ £ 2≥
5 2
3
1
2
1
¡
¡
1
1
AC AB £ 2≥ § 3 ¥
2
2
3
2
¡

1
1
0
1 2
The diagram shows that OC OA AC £ 1≥ § 3 ¥ § 7 ¥.
2
2
2
¡

¡

¡

B

C

O

A

b) Since the line AB is divided in the ratio of 1 : 2 the point D is

1
the way
3

along the line.
B
2
D
O

1
A

2
2
3
3
2
¡

1
2
AD AB £ 2≥ ¶ ∑ • 2 ∂
3
3
3
3
3
3
1
3
Therefore the position vector of D is
2
2
3
3
0
¡
¡
¡
5 ∂
OD OA AD £ 1≥ • 2 ∂ •
3
3
2
.
3
1

317

12 Vector Techniques

Vector subtraction
We can now use this principle to look at the subtraction of two vectors. We can first
consider a b to be the same as a 1 b2.
a

D

a ( b)

b

a

A
¡

C

b

B

¡

¡

a 1 b2 is the same as DC CB and hence the diagonal DB can represent
a 1 b2 Thus in terms of the parallelogram one diagonal represents the addition of
two vectors and the other the subtraction of two vectors. This explains geometrically
¡

why to find AB we subtract the coordinates of A from the coordinates of B.
¡

¡

¡

Alternatively if we consider the triangle below we can see that AB AO OB . Hence
¡

¡

¡

AB OA OB a b b a.
B

b a
A
b

a
O

Example
¡

¡

The diagram shows quadrilateral ABCD, where AB 2p, DC p and
¡

AD q.
a) What type of quadrilateral is ABCD?
b) Find these in terms of p and q.

p
D
q

¡

i) BC

A

¡

ii) DB

C

2p

B

¡

iii) AC
a) Since AB and DC are parallel and AD is not parallel to BC, the shape is a
trapezium.
¡
¡
¡
¡
b) (i) BC BA AD DC
2p q p q p
¡

¡

¡

(ii) DB DC CB
p 3 1q p2 4 2p q
¡

¡

¡

(iii) AC AD DC
q p

318

12 Vector Techniques

Example
PQRS is a quadrilateral where X and Y are the midpoints of PQ and RS respectively.
¡

¡

Show that PS
¡

¡

¡

QR

2 XY .

¡

¡

¡

¡

XQ , XQ

Since PX

PX

1 XQ
¡

Similarly YR
¡

Now XY
¡

XP

¡

XY

XQ

SY and
P

¡

¡

¡

¡

¡

XP

¡

PS

PS

¡

SY

¡

XQ

Q

X

RY

¡

1 2 XY

0

¡

QR

Hence 2 XY

SY

¡

PS

¡

S
¡

0 1 RY
¡

Y

0

¡

YS

¡

0

XP

¡

R

¡

QR

RY

QR

Exercise 2
1 If a
a a
e 3a
2 If a
a a

2i
b
3b

j

5k, b i 5j 6k and c 3i 6j
b a b c
c b c
2c

2a

f

i j, b 2i
2b 3c

c the angle that a

3b

7c

g ma

20mb

8k, find:
d 2a b

4c

3mc

3j and c 4i 7j, find:
b a 2b 3c
b

c makes with the x-axis.
1
£1≥, b
1

3 Vectors a, b, c and d are given by a
7
£5≥. If b a is parallel to c
q
ratio of their moduli.
d

¡

5
£ 1≥ and
3

d, find the value of q. Also find the

¡

4 In the triangle shown, OA

2
£4≥, c
0

a and OB

b and C is the midpoint of AB.

B

b

C

O

a

A

Find:
¡

a AB

¡

b AC

¡

c CB

¡

d Hence, using two different methods, find OC .

319

12 Vector Techniques

4
1
≤ and the position vector of B is ¢ ≤,
16
4

5 If the position vector of A is ¢
find:
¡

¡

b 冨 AB 冨

a AB

c the position vector of the midpoint of AB
d the position vector of the point dividing AB in the ratio 2 : 3.
6 If the position vector of P is ¢

7
1
≤ and the position vector of Q is ¢ ≤,
13
3

find:
¡

¡

b 冨 PQ 冨

a PQ

¡

c the position vector of the midpoint of PQ

¡

d the position vector of the point dividing PQ in the ratio 1 : 7.
¡

¡

7 In the triangle shown, OA a and OB b. The point C lies on AB such
that AC : CB 1 : k where k is a constant.
Find:
¡

¡

a AC

¡

b BC

c BA

d

B
k
b

¡

C

OC
O

a

¡

¡

¡

ii) CA

¡

iii) BD

¡

iv) AX

A

¡

8 ABCD is the parallelogram shown, where AB a and BC b.
¡

AX AD .
3
D
a Find:
i) CD

1

C

¡

v) XD

b

¡
2k
4k
≤ and b ¢ ≤, find AC in
3c
c

b If a ¢

X

terms of k and c.
¡

A

¡

9 The trapezium shown has AB a, BC b
¡

a

and AD 3b. E and F are points on BC such that BE : EF : FC m : n : 3.
Find:
¡

a BE

¡

c CF

¡

e ED

b

¡

D

EF

¡

C

d AF

3b

b
B
a

A

320

B

12 Vector Techniques
¡

¡

¡

10 The cuboid ABCDEFGH shown has AB a, AD b and AE c. Find in
terms of a, b and c:
¡

H

¡

a BC

b FH

¡

¡

c AH

E

d AG

¡

G

D

c

e BH

b
F

A

C

a
B

11 If PQR is a triangle and S is the midpoint of PQ, show that
¡

¡

¡

RQ PR 2 PS .
¡

¡

¡

12 ABCDEFGH is a regular octagon in which AB a, BC b and CD c
¡

and DE d. Find in terms of a, b, c and d:
¡

¡

a DG

¡

b AH

c FA

13 T, U and V are the midpoints of the sides PQ, QR and PR of a triangle. Show
¡

¡

¡

¡

¡

¡

that OP OQ OR OT OU OV, where O is the origin.
¡

¡

14 OABC is a rhombus, where O is the origin, OA a and OC c.
¡

¡

¡

¡

a Find AB , BC , AC and OB in terms of a and c.
b What is the relationship between c a and c a?

12.3 Multiplication of vectors
When we multiply two vectors there are two possible answers. One answer is a scalar
and the other is a vector. Hence one is called the scalar product and one is called the
vector product. We use a “dot” to signify the scalar product and a “cross” to signify
the vector product. It is quite common therefore to refer to the “dot product” and
“cross product”.
The reason why there need to be two cases is best seen through physics. Consider the
concept of force multiplied by displacement. In one context this gives the work done,
which is a scalar quantity. In another context it gives the moment of a force (the turning
effect), which is a vector quantity. Hence in the physical world there are two possibilities,
and both need to be accounted for in the mathematical world.

Scalar product
The scalar product or dot product of two vectors a and b inclined at an angle of u is
written as a # b and equals 冟a冟冟b冟 cos u.
A

“Inclined at an angle of u”
means the angle between
the two vectors is u.

b



a

B

321

12 Vector Techniques

The following results are important.

Parallel vectors
If two vectors are parallel, then a # b
So a # b

a b cos 0 or a b cos p

a b

Perpendicular vectors
If two vectors are perpendicular, then a # b

a b cos

So a # b

p
2

0

Commutativity
We know that a # b

a b cos u and b # a
b a cos u then a # b

Since a b cos u

b a cos u.
b # a.

The scalar product is commutative.

Distributivity
The scalar product is distributive across addition.
This means a # 1b

a#b

R2

a # R.

Proof
¡

Consider the diagram below, where OA
¡

¡

a, OB

¡

¡

R. AOˆ B

b and BC
¡

and the angle BC makes with BD, which is parallel to OA, is b.
C
B

D

b

O
¡

Clearly OC

b

Now a # b

a#R

E

a

R
¡ ¡

¡ ¡

OA OB cos u
¡

¡

OA a OE
¡ ¡

OA OF

322

F

OA BC cos b

¡

EF b

A

u, AOˆ C

a,

12 Vector Techniques

and a # 1b R2 冨 OA 冨冨 OC 冟 cos a 冨OA 冨冨 OF 冨
¡ ¡

¡ ¡

So a # 1b R2 a # b b # R, proving the scalar product is distributive over addition.
This result can be extended to any number of vectors.

Scalar product of vectors in component form
Let p a1i b1j c1k and q a2i b2j c2k
˛

˛

˛

˛

˛

˛

Then p # q 1a1i b1j c1k2 # 1a2i b2j c2k2
˛

˛

˛

˛

˛

˛

1a1a2i # i b1b2j # j c1c2k # k2
˛

˛

˛

1a1b2i # j b1c2j # k c1a2k # i b1a2j # i c1b2k # j a1c2i # k2
˛

˛

˛

˛

˛

˛

We need to look at what happens with various combinations of i, j and k:
i # i j # j k # k 112 112 cos0 1
p
i # j j # i i # k k # i j # k k # j 112 112 cos 0
2
Hence
p # q a1a2 b1b2 c1c2
So there are two ways of calculating the scalar product. We normally use this form as we
rarely know the angle between two vectors.

Example
If a 3i j k and b 2i j 6k, find a # b.
a # b 132 122 112 1 12 1 12 162
6 1 6
1

Example
Given that x # p q # x show that x is perpendicular to p q.
x#p q#x
#
1x p q#x 0
x#p x#q 0
x # 1p q2 0

Since the scalar product is zero, x and p q are perpendicular.

Since the scalar product
is commutative
Using the distributive law

323

12 Vector Techniques

Example
Show that the triangle ABC with vertices A(1, 2, 3), B12,
not right-angled.

1, 42 and C13,

3, 22 is

We first write down the vectors representing each side.
1
£ 3≥
1

¡

AB

1
£ 2≥
2

¡

BC

2
£ 5≥
1

¡

AC

1
1
#
£ 3≥ £ 2≥
1
2

¡
Now AB # BC
¡

¡
BC # AC

¡

1
2
#
£ 2≥ £ 5≥
2
1

112 112

112 122

1 32 1 22

1 22 1 52

112 1 22

1 22 1 12

1

2

6

10

2

2

5

14

1
2
#
£ 3≥ £ 5≥ 112 122
1 32 1 52
112 1 12
2 15 1 16
1
1
Since none of the scalar products equals zero, none of the sides are at right angles to
each other, and hence the triangle is not right-angled.
¡
AB # AC

¡

Example
In the triangle ABC shown, prove that c # c

1a

b2 # 1a

b2.

A
c

a

B

Using the scalar product
c2
1 c

2

a#a
a

2

a#b

0

0

b#a
b

b

C

b#b

2

AB
BC
1 AC
This is Pythagoras’ theorem and hence the relationship is proven.
2

324

2

2

Since AB and BC are at
right angles to each other

12 Vector Techniques

Angle between two vectors
If we draw two intersecting vectors, there are two possible angles where one is the
supplement of the other.
Is the angle between the vectors a or b?



Remember that supplement
means “subtract from
180º” or “subtract from p”
depending on whether we
are working in radians or
degrees.



There is a convention for this. The angle between two vectors is the angle between their
directions when those directions both converge or both diverge from a point. Hence in
this case we require a.
Now we know which angle to find, we can find it using the two formulae for scalar
product.

Example
Find the angle u between a 3i 2j 4k and b 2i 4j 7k.
We know that a # b 132 122 1 22 142 142 172 6 8 28 26
So 冟a冟冟b冟 cos u 26
Now 冟a冟 232 1 22 2 42 29 4 16 229
And 冟b冟 222 42 72 24 16 49 269
Hence cos u

26

229269
1 u 54.5°

Example
3
2
If the angle between the vectors a £ 1≥ and b £ 1≥ is 60º, find the
1
x
values of x.
We know that a # b 132 122 1 12 1 12 1 12 1x2 6 1 x 7 x
So 7 x 冟a冟冟b冟 cos 60°
Now 冟a冟 232 1 12 2 1 12 2 29 1 1 211
And 冟b冟 222 1 12 2 x2 24 1 x2 25 x2
Hence 7 x 211 25 x2 cos 60°

21125 x2
2

325

12 Vector Techniques

We can to use a calculator to solve this equation.

The answer is x

2.01 or

10.0.

Exercise 3

1 Given that a
a a#b

e a # 1c

3
2
7
¢ ≤, b ¢ ≤ and c ¢ ≤, find:
1
5
3
b b#c
c a # 1b c 2
d b#i
f 3a # c

b2

2 Given that a
find:
a a#b

2i

3j

g a # 12b

4k, b

b c#b

i

5j

c b#b

c2

2k and c

e 1a b 2 # i
f b # 1a 2b2 g c # 1a 2b2
3 Calculate the angle between each pair of vectors.
a a 2i 4j 5k, b i 3j 8k
b a

2
¢ ≤, b
5

c a

3
£ 3≥, b
4

d a

i

e a

3
£ 1≥, b
4

4
£0≥
0

2t
£ t ≥, b
3t

1
t
2
¶ ∂
t
3
t

f a

2k, b

¢

h b # 12a

3c2

2i

2j

d a # 1c

b2

h b#a

b#c

3

1
1
£ 4≥
3
j

k

4 Find p # q and the cosine of the angle u between p and q if
p
i 3j 2k and q i j 6k.
5 Find which of the following vectors are perpendicular to each other.
a 3i 2j k, b 2i j 4k, c 3i 2j k,
d
36i 27j 54k, e i 2j, f 4i 3j 6k
6 Find the value of l if the following vectors are perpendicular.
a a

326

2
£1≥, b
l

1
£ 1≥
2

k,

12 Vector Techniques

b a 2i 5j 2k, b i 4j lk
c a li 3k, b 2i j 5k
l
l
d a £1≥, b £ l ≥
3
2
7 Show that the triangle ABC is not right-angled, given that A has coordinates
12, 1, 22, B 13, 3, 12 and C 1 2, 1, 42.
¡

¡

8 Find a unit vector that is perpendicular to PQ and to PR , where
¡

¡

PQ i j 2k and QR i 2j k.

2
3
9 If the angle between the vectors p £ 1 ≥ and q £ x ≥ is 80º, find the
possible values of x.
1
4
10 Taking O as the origin on a cube OABCDEFG
of side 2 cm as shown, find the angle
between the diagonals OF and AG.

G

F

D

E

2 cm

C

B
2 cm

O
A

2 cm

11 A quadrilateral ABCD has coordinates A (0, 0, 1), B (1, 1, 3), C (3, 0, 6) and
D 12, 1, 42 . Show that the quadrilateral is a parallelogram.
12 A quadrilateral ABCD has coordinates A 11, 2, 12, B (2, 3, 0), C (3, 5, 3)
and D 1 2, 3, 22 . Show that the diagonals of the quadrilateral are
perpendicular. Hence state, giving a reason, whether or not the quadrilateral
is a rhombus.
13 Using the scalar product, prove that the diagonals
of this rhombus are perpendicular to one another.
q

p

R

14 In the quadrilateral PQRS shown,
prove that
¡

PR

s
S

¡ ¡
¡ ¡

QS PQ # RS QR # PS .

r

p
P

q

Q

15 If a # b a # c, show that a is perpendicular to b c.
¡

¡

16 If A, B, C, D are four points such that BC DA 0, prove that ABCD is a
parallelogram. If AB # BC 0, state with a reason whether the
¡

¡

parallelogram is a rhombus, a rectangle or a square.
17 Given that a and b are non-zero vectors show that if 冟a冟 冟b冟 then a b
and a b are perpendicular.

327

12 Vector Techniques

18 If a and b are perpendicular vectors, show that:
1a b2 # 1a b2 1a b2 # 1a b2

19 Triangle ABC is right-angled at B. Show that AB # AC 冟 AB 冟2.
¡

¡

¡

20 Given that a, b and c are non-zero vectors, a
b
c and
a # 1b c2 b # 1a c2, show that c # 1a b2 0.

Vector product
The vector product or cross product of two vectors a and b inclined at an angle u is
ˆ,, which is a vector quantity.
written as a b and equals 冟a冟冟b冟 sin un
Hence it is a vector of magnitude 冟a冟冟b冟 sin u in the direction of n where n is
perpendicular to the plane containing a and b.

Remember that nˆ is a
unit vector.

Now obviously nˆ can have one of two directions. This is decided again by using a
“right-hand screw rule” in the sense that the direction of n is the direction of a screw
turned from a to b with the right hand. This is shown in the diagram below.
n

b



If you are unsure, try
this with a screwdriver
and a screw!
a

In other words a b 冟a冟冟b冟 sin unˆ where nˆ is a unit vector perpendicular to both a
and b.
The following results are important.

Parallel vectors
If two vectors are parallel, then a b 冟a冟冟b冟 sin 0nˆ or 冟a冟冟b冟 sin pnˆ
So a b 0

The fact that the vector
product of a and b is
perpendicular to both a
and b is very important
when it comes to the
work that we will do with
planes in Chapter 13.
Remember that for parallel
vectors, a # b 冟a冟冟b冟

Perpendicular vectors
Remember that for
perpendicular vectors,
a#b 0

p
If two vectors are perpendicular, then a b 冟a冟冟b冟 sin nˆ
2
So a b 冟a冟冟b冟nˆ

Commutativity
b

We know a b 冟a冟冟b冟 sin unˆ 1 and that
b a 冟b冟冟a冟 sin unˆ 2.

nˆ 1



By thinking of the right-hand screw rule we can see that
nˆ 1 and nˆ 2 must be in opposite directions.
Hence a b b a.

328

nˆ 2

a

12 Vector Techniques

The vector product is not commutative.

Distributivity
The vector product is distributive across addition.

r

1p

q2

r

p

r

q.

Proof
Consider two vectors p and q with the third vector r which is perpendicular to both p
and q. Hence the plane containing p and q also contains p q, and r is perpendicular
to that plane.

r
p

p

r p, r q and r 1p
are perpendicular to r.

q
q

q 2 must also lie in this plane as all three are vectors that

r
p

B

A

r

q
p

q

C

Now r

And r

1p

q
(p

q)

E

G

F

D

¡

q

Similarly r

p

r
r

AE which is a vector of magnitude r q sin 90°
¡

p

rq

AG which is a vector of magnitude r p sin 90°
q2

¡

AF which is a vector of magnitude r 1p

rp

q2 sin 90°

r 1p

q2

Hence the sides of the quadrilateral AEFG are r times the lengths of the sides in
quadrilateral ABCD. Since the angles in both figures are the same ( r p is a 90° rotation
of p, r q is a 90° rotation of q, and r 1p q2 is a 90° rotation of p q ), ABCD
and AEFG are both parallelograms.
¡

Now we know that AF

¡

¡

AE

r p
Hence r 1p q 2
is perpendicular to p and q.

EF

r

q, and we have proved the distributive law when r

Now let us consider the case where r is not perpendicular to p and q. In this case we
need to form the plane perpendicular to r with vectors p and q inclined at different
angles to r. p1, q1 and 1p1 q1 2 are the projections of p, q and 1p q2 on this
plane. u is the angle between r and p.

329

12 Vector Techniques



p

q

r

p
p1

p1



q



q1

q1

Now r p 冟r冟冟p冟 sin unˆ and r p1 冟r冟冟p1冟 sin 90°nˆ , where nˆ is a vector perpendicular
to r, p and p1.
Also 冟p1冟 冟p冟 sin u and hence r p1 冟r冟冟p冟 sin unˆ r p.
Using an identical method, r q1 r q and r 1p1 q1 2 r 1p q2.
Since r is perpendicular to p1, q1 and p1 q1 we can use the distributive law that we
proved earlier, that is, r 1p1 q1 2 r p1 r q1.
Hence r 1p q2 r p r q and we have proved that the distributive law also
holds when r is not perpendicular to p and q.

Vector product of vectors in component form
Let p a1i b1j c1k and q a2i b2j c2k
˛

˛

˛

˛

˛

˛

Then p q 1a1i b1j c1k2 1a2i b2j c2k2
˛

˛

˛

˛

˛

˛

1a1a2i i b1b2j j c1c2k k2 1a1b2i j b1c2j k
c1a2k i b1a2j i c1b2k j a1c2i k2
˛

˛

˛

˛

˛

˛

˛

˛

˛

We need to look at what happens with various combinations of i, j and k:
i i j j k k 1 1 sin 0° perpendicular vector 0
For the others the answer will always be
1 1 sin

p
perpendicular vector perpendicular vector
2

From the definition of the unit vectors the perpendicular vector to i and j is k, to j and k
is i, and to i and k is j. The only issue is whether it is positive or negative, and this can be
determined by the “right-hand screw rule”. A list of results is shown below.
i j k

j i k

j k i

k j i

k i j

i k j

Hence p q a1b2k b1c2i c1a2j b1a2k c1b2i a1c2j
˛

˛

˛

˛

˛

˛

1b1c2 c1b2 2i 1c1a2 a1c2 2j 1a1b2 b1a2 2k
1b1c2 c1b2 2i 1a1c2 c1a2 2j 1a1b2 b1a2 2k
This can be written as a determinant:
i
p q 3 a1
a2

330

j
b1
b2

k
c1 3
c2

The angle between r
and p1 is 90º because
this is how we set up the
plane in the beginning.

12 Vector Techniques

So there are two ways of calculating the vector product. We normally use this
determinant form as we rarely know the angle between two vectors.
We can use the vector product to calculate the angle between two vectors, but unless
there is a good reason, we would normally use the scalar product. One possible reason
would be if we were asked to find the sine of the angle between the vectors.

Example
If a i 3j 2k and b 2i 4j k, find
a) the unit vector perpendicular to both a and b
b) the sine of the angle between a and b.
i
a) a b 3 1
2

j
3
4

k
23
1

i33 1 82 4 j31 44 k3 1 42 64
11i 3j 10k
Hence the unit vector is
1
1
111i 3j 10k2
111i 3j 10k2
2
2
2
211 3 1 102
2230
b) We know that a b 冟a冟冟b冟 sin unˆ
1 11i 3j 10k 212 32 22 222 1 42 2 12
1
sin u
111i 3j 10k2
2230
1
1 1 214221 sin u
2230
1 sin u

2230
214221



230
115

B 294 B 147

Example
A, B and C are the points (2, 5, 6), (3, 8, 9), and (1, 1, 0) respectively. Find the
unit vector that is perpendicular to the plane ABC.
¡

¡

The plane ABC must contain the vectors AB and BC . Hence we need a
vector perpendicular to two other vectors. This is the definition of the cross
product.
3 2
1
1 3
2
¡
¡
Now AB £8 5≥ £3≥ and BC £1 8≥ £ 7≥
9 6
3
0 9
9
Therefore the required vector is
i
AB BC 3 1
2

¡

¡

j
3
7

k
3 3
9

331

12 Vector Techniques

i3 1 272 1 212 4 j3 1 92 1 62 4 k3 1 72 1 62 4
6i 3j k
Hence the unit vector is
1
21 62 3 1 12
2

2

2

1 6i 3j k2

1
246

1 6i 3j k2

Example
2
6
4
If a £ 3 ≥, b £ 3≥ and c £ 3 ≥, find c # a b.
1
2
1
In an example like this it is important to remember that we have to do the vector
product first, because if we calculated the scalar product first we would end up
trying to find the vector product of a scalar and a vector, which is not possible.
i
j
k
a b 32
3
1 3
6 3
2
i36 34 j34 1 62 4 k3 1 62 184
3
£ 10≥
24
4
3
Therefore c # a b £ 3 ≥ # £ 10≥ 12 30 24 6
1
24

Example
Show that 1a b2 1a b2 21b a2.
Consider the left-hand side.
1a b2 1a b2 1a a2 1b a2 1a b2 1b b2
Now a a b b 0
Hence 1a b2 1a b2 1b a2 1a b2
Now we know a b and b a are the same in magnitude but in opposite
directions.
Therefore 1a b2 1a b2 21b a2

Application of vector product
We will see that a very important use of vector products is in the representation of planes,
which will be dealt with in Chapter 13. However, there are two other applications that are
useful to know.

332

12 Vector Techniques

Area of a parallelogram
B

C
h
D

A

The area of a parallelogram

base

height

AD

h

1AD 2 1AB sin u2
¡

¡

AD

AB

The area of a parallelogram is the magnitude of the vector product of two
adjacent sides.

Example
Find the area of the parallelogram ABCD where A has coordinates
12, 3, 12, B 13, 2, 12, C 14, 5, 12 and D (3, 0, 1).
We first need to find the vectors representing a pair of adjacent sides.
˛

¡

AB

3
£ 2≥
1
¡

Now AB

˛

2
£ 3 ≥
1
i
¡
AD
31
1

1
¡
£ 5≥ and AD
0
j
k
5 03
3 2

i3 1 102
10i

04
2j

j32

3
£0≥
1

04

2
£ 3 ≥
1

1
£ 3≥
2

k3 1 32

1 52 4

2k

Now the area of the parallelogram ABCD is
¡

AB

¡

AD

21 102 2

1 22 2

22

2108 units2

Area of a triangle
B

h
A

The area of triangle ABC

1
1
1
1

C

2

base

2

1AC 2 1h2

2

1AC 2 1AB sin u2
¡

2

height

AC

¡

AB

The area of a triangle is half the magnitude of the vector product of two sides.

333

12 Vector Techniques

This is consistent with the idea that the area of a triangle is half the area of a parallelogram.

Example
Find the area of the triangle ABC with coordinates A 11, 3, 12, B 1 2, 1, 42
and C 14, 3, 32 .
˛

˛

˛

¡

¡

We begin by finding AB and BC .
2 1
3
4 1 22
6
¡
Now AB £ 1 3 ≥ £ 2≥ and BC £ 3 1 ≥ £2≥
4 1 12
3
3 1 42
1
i
j
k
¡
¡
Hence AB BC 3 3 2 3 3
6
2
1
¡

i3 1 22 1 62 4 j3 1 32 1 18 2 4 k3 1 62 1 12 2 4
4i 15j 6k
Now the area of triangle ABC is
¡
1 ¡
2277
1
冟 AB BC 冟 242 1 15 2 2 62
units2
2
2
2

Exercise 4
1 If a i 2j 3k and b 2i 4j k, find:
a a b
b a 1a b2
c b 1a b2
d a 13a 2b2
g a # 1a b2

e 12a b2 a

f 1a 2b2 12a b2

2 Find the value of 冟a b冟 for the given modulus of a, modulus of b and angle
between the vectors a and b.
a 冟a冟 3, 冟b冟 7, u 60°
b 冟a冟 9, 冟b冟 213, u 120°
c 冟a冟 218, 冟b冟 2, u 135°
3 If OPQ is a triangle, show that
¡

OP

¡

OQ

¡

OP

¡

PQ .

4 If a b 0, show that a kb where k is a scalar.
5 Given that b c c a show that a b is parallel to c.
6 Consider two vectors a and b. If a # a b 0 write down the angle
between a and b.
3
4
7 If a £ 4≥ and b £2≥, find
3
3
a the unit vector nˆ perpendicular to both a and b
b the sine of the angle u between a and b.
8 If a i 3j k and b i 2k, find
a the unit vector nˆ perpendicular to both a and b
b the sine of the angle u between a and b.

334

12 Vector Techniques

9 P, Q and R are the points (0, 0, 3), (3, 4, 6) and 10, 1, 02 respectively. Find
the unit vector that is perpendicular to the plane PQR.
10 Two sides of a triangle are represented by the vectors 1i j k2 and
15i 2j 2k2. Find the area of the triangle.
11 Relative to the origin the points A, B and C have position vectors
0
2
4
£ 1 ≥, £ 2≥ and £ 1≥ respectively. Find the area of the triangle ABC.
2
1
3
12 The triangle ABC has its vertices at the points A (0, 1, 2), B (0, 0, 1) and
C (2, 6, 3). Find the area of the triangle ABC.
¡

¡

¡

¡

¡

¡

13 Given that AB i 4j, AC 3i j 2k, AP 2 AB and AQ 4 AC ,
find the area of triangle APQ.
14 A parallelogram OABC has one vertex O at the origin and the vertices A
and B at the points (3, 4, 0) and (0, 5, 5) respectively. Find the area of the
parallelogram OABC.
15 A parallelogram PQRS has vertices at P 10, 2, 12, Q 12, 3, 72 and
R 1 1, 0, 42 . Find the area of the parallelogram PQRS.
˛

˛

˛

¡

16 A parallelogram PQRS is such that PX 5i j 2k and
¡

¡

¡

¡

PY 3i 7j k, where PQ 5 PX and Y is the midpoint of PS .
¡

¡

Find the vectors representing the sides PQ and PS and hence calculate
the area of the parallelogram.
1
4
3
17 If a £6≥, b £ 7 ≥ and c £ 4 ≥, determine whether or not
0
1
1
a 1b c2 1a b 2 c.
18 If a b a c, show that the vector c b is parallel to a.

Review exercise
M
C
7
4
1

M–

M+

CE

%

8

9



5

6

÷

2

3
+

0

ON
X

=


M
C

7

4

1

0

M–

M+

CE

%

8

9



5

6

÷

2

3

+

ON
X

=

1 The points P, Q, R, S have position vectors p, q, r, s given by
p i 2k
q 1.2j 1.4k
r 5i 6j 8k
s j 7k
respectively. The point X lies on PQ produced and is such that PX 5PQ,
and the point Y is the midpoint of PR.
a Show that XY is not perpendicular to PY.
b Find the area of the triangle PXY.
c Find a vector perpendicular to the plane PQR.
d Find the cosine of the acute angle between PS and RS.
2
1
2
2 Let a £1≥, b £ p ≥ and c £ 4≥
0
6
3
a Find a b
b Find the value of p, given that a b is parallel to c. [IB May 06 P1 Q11]

335

12 Vector Techniques


M

M–

M+

ON

C

CE

%

X

7

8

9



5

6

÷

2

3

4

1

=

+

0

1 m
1 2m
3 The point A is given by the vector £2 m≥ and the point B by £2 2m≥,
3 m
3 2m
¡

¡

relative to O. Show that there is no value of m for which OA and OB are
perpendicular.

✗ 4 If a and b are unit vectors and u is the angle between them, express 冟a b冟 in
M
C

M–

M+

CE

%

8

9



5

6

÷

2

3

7

4

1

ON
X

=

+

0

terms of u.

[IB May 93 P1 Q12]

✗ 5 A circle has a radius of 5 units with a centre at (3, 2). A point P on the circle
M
C

M–

M+

CE

%

8

9



5

6

÷

2

3

7

4

1

ON
X

=

+

0

has coordinates (x, y). The angle that this radius makes with the horizontal is
¡
u. Give a vector expression for OP .

vectors a and b such that 冟a b冟 冟a b冟, find the
✗ 6 Given two non-zero
#
M
C

M–

M+

CE

%

8

9



5

6

÷

2

3

7

4

1

ON
X

=

+

0

value of a b.

[IB Nov 02 P1 Q18]

that the points A (0, 1, 3), B (5, 3, 2) and C (15, 7, 0) are collinear (that
✗ 7 Show
is, they lie on the same line).
M
C

M–

M+

CE

%

8

9



5

6

÷

2

3

7

4

1

M
C

ON
X

=

+

0

M–

M+

CE

%

8

9



5

6

÷

2

3

7
4
1

ON
X

=

+

0

8 Find the angle between the vectors v i j 2k and w 2i 3j k.
Give your answer in radians.
[IB May 02 P1 Q5]

circle shown has centre D, and the points
✗ 9 The
A, B and C lie on the circumference of the
M
C

7

4

1

M–

M+

CE

%

8

9



5

6

÷

2

3

+

0

ON
X

C

=

circle. The radius of the circle is 1 unit.
¡

¡

b

Given that DB a and DC b, show that
ACˆ B 90°.

A

✗ 10 Let a be the angle between a and b, where
M
C

M–

M+

CE

%

8

9



5

6

÷

2

3

7

4

1

+

0

D

B

a

ON
X

=

p
. Express
4
[IB Nov 00 P1 Q11]

a 1cos u2 i 1sin u2j, b 1sin u 2i 1cos u2j and 0 6 u 6
a in terms of u.

M
C

M–

M+

CE

%

8

9



5

6

÷

2

3

7
4
1

+

0

ON
X

=

11 The points X and Y have coordinates (1, 2, 3) and 12, 1, 02 respectively.
¡

¡

¡

¡

OA 2 OX and OB 3 OY . OABC is a parallelogram.
a Find the coordinates of A, B and C.
b Find the area of the parallelogram OABC.
¡

¡

c Find the position vector of the point of intersection of OB and AC .
¡

¡

d The point E has position vector k. Find the angle between AE and BE .
e Find the area of the triangle ABE.

✗ 12 Given u 3i 2j 5k and v i 4j mk, and that the vector
M
C

7

4

1

0

M–

M+

CE

%

8

9



5

6

÷

2

3

+

ON
X

=

12u 3v2 has magnitude 2265, find the value of m. [IB Nov 93 P1 Q8]

336




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