4. 3-Dimensional Vector Geometry

specialistmathematics

specialistmathematics

mathematics for year 12

Haese & Harris Publications

Robert HaeseSandra HaeseMichael HaeseRoger DixonJon RobertsMichael TeubnerAnthony Thompson

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Color profile: DisabledComposite Default screen

SPECIALIST MATHEMATICSMATHEMATICS FOR YEAR 12

This book is copyright

Copying for educational purposes

Disclaimer

Robert Haese B.Sc.Sandra Haese B.Sc.Michael Haese B.Sc.Hons.Roger Dixon B.Ed.Jon Roberts B.Sc.Hons.(Ma.Sc.), Dip.Ed.Michel Teubner B.Sc., Ph.D.Anthony Thompson B.Sc., Dip.T, Dip.Ed., Grad.Dip.Ed.Admin.

Haese & Harris Publications3 Frank Collopy Court, Adelaide Airport SA 5950Telephone: (08) 8355 9444, Fax: (08) 8355 9471email:

National Library of Australia Card Number & ISBN 1 876543 79 5

© Haese & Harris Publications

Published by Raksar Nominees Pty Ltd, 3 Frank Collopy Court, Adelaide Airport SA 5950

First Edition 2002

Cartoon artwork by John Martin. Artwork by Piotr Poturaj, Joanna Poturaj and David PurtonCover design by Piotr Poturaj. Cover photography Piotr PoturajComputer software by David Purton and Mark Foreman

Typeset in Australia by Susan Haese (Raksar Nominees). Typeset in Times Roman 10 /11

. Except as permitted by the Copyright Act (any fair dealing for thepurposes of private study, research, criticism or review), no part of this publication may bereproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic,mechanical, photocopying, recording or otherwise, without the prior permission of the publisher.Enquiries to be made to Haese & Harris Publications.

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[email protected]

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FOREWORD

Mathematics for Year 12 Specialist Mathematics is our interpretation of the concepts outlined

in the Stage 2 Mathematics Curriculum Statement. It is not our intention to define the course,

and teachers are encouraged to use other resources.

Our package is a textbook with an interactive CD. The CD displays the contents of the

textbook plus many interactive features such as simulations, computer demonstrations, video

clips, graphing packages, spreadsheets etc to stimulate the interest of students and assist their

teachers.

The book is language rich and technology rich. Some exercises are simply designed to build

skills, but every effort has been made to contextualise problems to help students see the

everyday uses and practical application of the mathematics they are studying.

The book contains many problems, from basic to the advanced, to cater for a wide range of

student abilities and interests. Much emphasis has been placed on the gradual development of

concepts with appropriate worked examples. However, we have provided material for those

who look towards further studies or applications of mathematics for their career choices. It is

not our intention that each chapter should be worked through in full. Time constraints will

not allow for this. Consequently, teachers must select exercises carefully, according to the

abilities and prior knowledge of their students, in order to make the most efficient use of time

and give as thorough coverage of work as possible.

The extensive use of graphics calculators and computer packages throughout the book

enables students to realise the importance, application and appropriate use of technology. No

single aspect of technology has been favoured. It is as important that students work with a

pen and paper as it is that they use their calculator or graphics calculator, or use a spreadsheet

or graphing package on computer.

The interactive features of the CD allow immediate access to our own specially designed

geometry packages, graphing packages and more. Teachers are provided with a quick and

easy method of demonstrating concepts, and students can discover for themselves, and revisit

when necessary.

Instructions for graphics calculators are on the CD and can be printed for students.

In this changing world of mathematics education, we believe the contextual approach shown

in this book, with the associated use of technology, will enhance the students’ understanding,

knowledge and appreciation of mathematics.

RCH SHH

PMH RLD

JR MDT AWT

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3-dimensionalvector

geometry

3-dimensionalvector

geometry

4Chapter

This chapter covers the following · · · · · · · · · · · · · · · · · ·

· · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · · ·

�

�

�

�

�

�

�

�

�

�

�

the position of a point in space

distance, midpoints, ratio of division

3-dimensional vectors and their geometric representation

vector equality and operations with vectors

parallelism and unit vectors

the scalar product and its properties

the vector product and its properties

areas and volumes using vectors

lines in space

planes and their intersection with lines and other planes

measuring angles using vectors

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126 3-DIMENSIONAL VECTOR GEOMETRY (Chapter 4)

INTRODUCTION

Vectors are used to specify quantities which

have size (magnitude) and direction.

Such quantities include:

² velocity ² force ² acceleration ² displacement

² work ² moments ² momentum ² weight

To specify points in space (or 3-dimensional

space) we need a point of reference, O, called

the origin.

Through O we draw 3 mutually perpendicular

lines and call them the X, Y and Z-axes. The

X-axis is considered to come directly out of the

page.

In the diagram alongside the coordinate planes

divide space into 8 regions, each pair of planes

intersecting on the axes.

The positive direction of each axis is a solid line

whereas the negative direction is ‘dashed’.

The position vector of P is¡!OP = [x, y, z].

3-DIMENSIONAL COORDINATESA

y

x

(3, 4)

4

3

X

Y

Z

Any point in space can be specified byan of numberswhere and are the in theand directions from the origin O, to

P,( , , )

, ,P.

ordered triple

steps

x y zx y z X Y

Z

Z

XY

x z

y

P( , , )x y z

At Stage 1, we considered 2-dimensional coordinate and vector geometry.

Points were specified in terms of ordered pairs such as (3, 4).

(3, 4) is found by starting at the origin O(0, 0), moving 3 units along the

x-axis in the positive direction and then 4 units vertically.

Vectors were also specified using a number pair. a = [3, 4] is a vector

which is represented by a directed line segment or arrow and has x-step

3 and y-step 4 as shown.

Vectors are used in navigation, physics, engineering and a host of other applied sciences.

In 3-dimensional geometry we extend the number plane to space where there is a third

dimension, depth. Consequently, points are specified as an ordered number triple, for example,

(3, 4, 7) and vectors likewise, for example, [3, 4, 7].

It should be noted that vectors become a bridge between geometry and algebra.

For example, if a is parallel to b then a = kb for some scalar k.

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3-DIMENSIONAL VECTOR GEOMETRY (Chapter 4) 127

To help us visualise the 3-D position of a point

on our 2-D paper it is useful to complete a rect-

angular prism (or box) with the origin O as one

vertex, the axes as sides adjacent to it, and P is

at the vertex opposite O.

DISTANCE AND MIDPOINTS

) AB =p

(x2 ¡ x1)2 + (y2 ¡ y1)2 + (z2 ¡ z1)2

A simple extension from 2-D to 3-D geometry also gives the

midpoint of AB =

µx1 + x2

2,y1 + y2

2,z1 + z2

2

¶:

Illustrate the points: a A(0, 2, 0) b B(3, 0, 2) c C(¡1, 2, 3)

a b c

Example 1

Z

XY

x z

y

P( , , )x y z

3-D POINTPLOTTER

Z

Y

X

A

2

Z

Y

X

B

2

3

Z

YX

C

2

3

�1

P

A( , , )x y z1 1 1

B( , , )x y z2 2 2

Qb

a c

Z

Y

X

P( , , )a b c

a

b

c

A B

Triangle OAB is right angled at A

) OB2 = a2 + b2 ...... (1) fPythagorasgTriangle OBP is right angled at B

) OP2 = OB2 + c2 fPythagorasg) OP2 = a2 + b2 + c2 ffrom (1)g) OP =

pa2 + b2 + c2

Now for two general points A(x1, y1, z1)

and B(x2, y2, z2)

a, the x-step from A to B ===

=

x2 ¡ x1b, the y-step from A to B = y2 ¡ y1c, the z-step from A to B = z2 ¡ z1

) AB =q

(x2 ¡ x1)2 + (y2 ¡ y1)2 + (z2 ¡ z1)2

or AB =q

(¢

¢¢

¢

x

y

x

z

)2 + (¢y)2 + (¢z)2

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1 Illustrate P and find its distance to the origin O if P is:

a (0, 0, ¡3) b (0, ¡1, 2) c (3, 1, 4) d (¡1, ¡2, 3)

2 For each of the following:

i find the distance AB ii find the midpoint of AB.

a A(¡1, 2, 3) and B(0, ¡1, 1) b A(0, 0, 0) and B(2, ¡1, 3)

c A(3, ¡1, ¡1) and B(¡1, 0, 1) d A(2, 0, ¡3) and B(0, 1, 0)

3 Show that P(0, 4, 4), Q(2, 6, 5) and R(1, 4, 3) are vertices of an isosceles triangle.

4 Determine the nature of triangle ABC using distances for:

a A(2, ¡1, 7), B(3, 1, 4) and C(5, 4, 5)

b A(0, 0, 3), B(2, 8, 1) and C(¡9, 6, 18)

c A(5, 6, ¡2), B(6, 12, 9) and C(2, 4, 2).

d A(1, 0, ¡3), B(2, 2, 0) and C(4, 6, 6).

5 A sphere has centre C(¡1, 2, 4) and diameter AB where A is (¡2, 1, 3).

Find the coordinates of B and the radius of the circle.

6 a State the coordinates of any point on the Y -axis.

b Find the coordinates of two points on the Y -axis which arep

14 units from

B(¡1, ¡1, 2).

7 Find the Cartesian equation of all points P(x, y, z) which move so that:

a they are always 4 units from the origin O

b they are alwaysp

3 units from the point A(¡1, 2, 2).

8 The distance of P(x, y, z) to A(¡2, 1, 4) is equal to the distance from P to B(1, 3, 2).

Find the Cartesian equation of the locus of P.

If A(¡1, 2, 4) and B(1, 0, ¡1) are two points in space, find:

a the distance from A to B b the coordinates of the midpoint of AB.

a AB =p

(1 ¡ ¡1)2 + (0 ¡ 2)2 + (¡1 ¡ 4)2

=p

4 + 4 + 25

=p

33 units

b midpoint is

µ¡1 + 1

2,

2 + 0

2,

4 + (¡1)

2

¶i.e., (0, 1, 32 )

Example 2

EXERCISE 4A

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OPENING PROBLEM

0

INVESTIGATION 1


In this investigation we will examine more complicated sets of points

in 3-D space. Click on the icon to open the demonstration.

1 Click on f(x, y, z): x = 1, y = 2, z = 4g then

f(x, y, z): x = 1, y = 2g then f(x, y, z): x = 1g.

Write a brief statement of observations.

2 Click on f(x, y, z): 0 6 x 6 2, 1 6 y 6 3, z = 0g then

f(x, y, z): 0 6 x 6 2, 1 6 y 6 3, z = 2g then

f(x, y, z): 0 6 x 6 2, 1 6 y 6 3, 0 6 z 6 2g.

What did you notice?

3 Click on f(x, y, z): x2 + y2 = 9, z = 0gf(x, y, z): x2 + y2 = 9, z = 4gf(x, y, z): x2 + y2 = 9, 0 6 z 6 4g. What did you notice?

4 Click on f(x, y, z): x2 + y2 + z2 = 9g. Record your observations.

9 Illustrate and describe:

a f(x, y, z): z = 3g b f(x, y, z): x = 1, z = 3gc f(x, y, z): x2 + z2 = 4, y = 0g d f(x, y, z): y2 + z2 6 4, x = 2ge f(x, y, z): x = 2, 1 6 y 6 3, 2 6 z 6 4gf f(x, y, z): 0 6 x 6 2, 1 6 y 6 3, 2 6 z 6 4g

Three dimensional geometry enables us to solve problems in space provided that each point

can be specified by coordinates.

What to do:

3-DDEMO

GARAGEVIEWERA .garage is illustrated below Three dimensional coordinate

axes have been inserted for convenience.

Z

Y

X

3 m

4 m

6 m

R

Q

M

L

C

BA

PK

Y

X

O

Q L

C

M

B

K

A

R

P

plan view

floor

Z

X

1 m 1 m5m

O

right end view

M is 5 m above the floor.

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² Can you specify the coordinates of key labelled points on the diagrams?

² Can you specify the midpoint of RM?

² Can you find distances such as the distance PC?

² Can you find a point on AC which divides AC in the ratio 2 : 3?

² Can you write a vector between any two key points?

² What point on QK is nearest to L?

² If an electric light pole has base (6, 2, 0), what point on the roof is nearest to

(6, 2, 5) and what is the shortest distance?

² Can we describe lines and planes algebraically in 3-dimensional geometry?

After completing this chapter you should be able to answer these questions and many more.

Consider a point P(x1, y1, z1).

The x, y and z-steps from the origin to P are x1, y1 and

z1 respectively.

So¡!OP = [x1, y1, z1] is the vector which emanates

from O and terminates at P.

In general, if A(x1, y1, z1) and B(x2, y2, z2) are two

points in space then:

¡!AB = [ x2 ¡ x1, y2 ¡ y1, z2 ¡ z1

x-step y-step z-step

]

¡!AB is called ‘vector AB’ or ‘the position vector of B relative to A (or from A)’

Note:¡!OA has length j ¡!

OA j or simply OA,¡!AB has length j ¡!

AB j or simply AB.

3-DIMENSIONAL VECTORSBZ

Y

X

P( , , )x y z

x

y

z

A B

If A is (3, ¡1, 2) and B is (1, 0, ¡2) find: a¡!OA b

¡!AB

a¡!OA = [3, ¡1, 2] b

¡!AB = [1 ¡ 3, 0 ¡ ¡1, ¡2 ¡ 2]

= [¡2, 1, ¡4]

Example 3

If P is (¡3, 1, 2) and Q is (1, ¡1, 3), find j ¡!PQ j :

¡!PQ = [1 ¡ ¡3, ¡1 ¡ 1, 3 ¡ 2]

= [4, ¡2, 1]

) j ¡!PQ j =

p42 + (¡2)2 + 12

=p

21 units

Example 4

= [ ¢x; ¢y; ¢z ]

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1 Consider the point T(3, ¡1, 4).

a Draw a diagram to locate the position of T in space.

b Find¡!OT. c How far is it from O to T?

2 Given A(¡3, 1, 2) and B(1, 0, ¡1) find:

a¡!AB and

¡!BA b the length of

¡!AB and

¡!BA.

3 Given A(3, 1, 0) and B(¡1, 1, 2) find¡!OA,

¡!OB, and

¡!AB:

4 Given M(4, ¡2, ¡1) and N(¡1, 2, 0) find:

a the position vector of M from N b the position vector of N from M

c the distance between M and N.

5 For A(¡1, 2, 5), B(2, 0, 3) and C(¡3, 1, 0) find the position vector of:

a A from O and the distance of A from O

b C from A and the distance of C from A

c B from C and the distance of B from C.

6 Find the distance from Q(3, 1, ¡2) to:

a the Y -axis b the origin c the Y OZ plane.

As for 2-D vectors, 3-D vectors are represented by directed line segments (called arrows).

Consider the vector represented by the line segment from O to A.

EXERCISE 4B

If A is (¡1, 3, 2) and B(2, 1, ¡4) find:

a the position vector of A from B b the distance between A and B.

a The position vector of A from B is¡!BA = [¡1 ¡ 2, 3 ¡ 1, 2 ¡ (¡4)]

= [¡3, 2, 6]b AB = j ¡!

BA j=

p9 + 4 + 36

=p

49

= 7 units

Example 5

GEOMETRIC REPRESENTATIONC

² This vector could be represented by

¡!OA or a or ea .

" "bold used used by

in text books students

A

a

O

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DISCUSSION


² The magnitude (length) could be represented by j ¡!OA j or OA or jaj or jeaj.

If a = [a1, a2, a3] then jaj =pa 21 + a 22 + a 23 :

VECTOR EQUALITY

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

So, if arrows are used to represent vectors, then

equal vectors are parallel and equal in length.

This means that equal vector arrows are transla-

tions of one another, but in space.

If a = [a1, a2, a3] and b = [b1, b2, b3], then

a = b , a1 = b1, a2 = b2, a3 = b3.

a = b implies that vector a is parallel to vector b.

Consequently, a and b are opposite sides of a

parallelogram, and certainly lie in the same plane.

1 Find a, b and c if:

a [a¡ 4, b¡ 3, c+ 2] = [1, 3, ¡4]

b [a¡ 5, b¡ 2, c+ 3] = [3 ¡ a, 2 ¡ b, 5 ¡ c]

2 Find scalars a, b and c if:

a 2[1, 0, 3a] = [b, c¡ 1, 2] b [2, a, 3] = [b, a2, a+ b]

c a[1, 1, 0] + b[2, 0, ¡1] + c[0, 1, 1] = [¡1, 3, 3]

a a

a

a

b

��

Do any three points in space define plane? What about four points? Illustrate.

What simple test(s) on four points in space enables us to deduce that the pointsare vertices of parallelogram? Consider using vectors and not using vectors.

a

a

Find a, b, and c if [a¡ 3, b¡ 2, c¡ 1] = [1 ¡ a, ¡b, ¡3 ¡ c]

As [a¡ 3, b¡ 2, c¡ 1] = [1 ¡ a, ¡b, ¡3 ¡ c] then

a¡ 3 = 1 ¡ a b¡ 2 = ¡b and c¡ 1 = ¡3 ¡ c) 2a = 4 2b = 2 and 2c = ¡2

) a = 2, b = 1 and c = ¡1

Example 6

EXERCISE 4C

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3 A(¡1, 3, 4), B(2, 5, ¡1), C(¡1, 2, ¡2) and D (r, s, t) are four points in space.

Find r, s and t if:

a¡!AC =

¡!BD b

¡!AB =

¡!DC

4 A quadrilateral has vertices A(1, 2, 3), B(3, ¡3, 2), C(7, ¡4, 5) and D(5, 1, 6).

a Find¡!AB and

¡!DC. b What can be deduced about the quadrilateral ABCD?

5 PQRS is a parallelogram. P is (¡1, 2, 3), Q(1, ¡2, 5) and R(0, 4, ¡1).

a Use vectors to find the coordinates of S.

b Use midpoints of diagonals to check your answer.

We define vector addition geometrically in the following way:

To add a and b:

Step 1: first draw a, then

Step 2: at the arrowhead end of

a draw b, and then

Step 3: join the beginning of a to

the arrowhead end of b

and this is vector a + b.

So given

we have

OPERATIONS WITH VECTORSDVECTOR ADDITION

a

b

a

a + b b

Example 7

ABCD is a parallelogram. A is (¡1, 2, 1), B is (2, 0, ¡1) and D is (3, 1, 4).

Find the coordinates of C.

First we draw an axis free sketch:

Let C be (a, b, c).

Now as AB is parallel to DC and has the same

length then¡!DC =

¡!AB,

i.e., [a¡ 3, b¡ 1, c¡ 4] = [3, ¡2, ¡2]

a¡ 3 = 3, b¡ 1 = ¡2, c¡ 4 = ¡2

) a = 6, ) b = ¡1, ) c = 2 So, C is (6, ¡1, 2).

Check: midpoint of DB isµ3 + 2

2,

1 + 0

2,

4 + ¡1

2

¶i.e.,

¡52 , 12 , 32

¢midpoint of AC isµ¡1 + 6

2,

2 + ¡1

2,

1 + 2

2

¶i.e.,

¡52 , 12 , 32

¢

A( 1, 2, 1)� � �

D(3, 1, 4)� �C( , , )a b c� �

B(2, 0, 1)� �

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¡a is the negative of a.

Notice that ¡a has the same magnitude

as a but is in the opposite direction.

The zero vector is written as 0 and for any vector a, a + (¡a) = (¡a) + a = 0.

VECTOR SUBTRACTION

To subtract one vector from another, we simply add its negative, i.e., a ¡ b = a + (¡b).

Geometrically:

For and

then

Numbers such as 2 and ¡3 are also referred to as scalars. If a is a vector, what would 2a

and ¡3a mean?

By definition 2a = a + a, 3a = a + a + a, etc.

¡3a = 3(¡a) = (¡a) + (¡a) + (¡a)

So, if a is then

So, 2a is in the direction of a but is twice as long as a

3a is in the direction of a but is three times longer than a

¡3a is oppositely directed to a and is three times longer than a.

NEGATIVE VECTORS

ZERO VECTORS

SCALAR MULTIPLICATION

a �a

a

a2a

a

a

3a

a

�a

�a

�3a

�a

ab

a

�b

b

a b�

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Note: ² If a is a vector and k is a scalar,

ka is also a vector and we are performing scalar multiplication.

If k > 0, ka and a have the same direction.

If k < 0, ka and a have opposite directions.

² When drawing the addition of three 3-D vectors on a 2-D page we must

remain aware that the vectors are generally not coplanar,

i.e., not in the same plane.

As with 2-D vector we can state vector operations in component form.

If a = [a1, a2, a3] and b = [b1, b2, b3], then:

² a + b = [a1 + b1, a2 + b2, a3 + b3] Addition

² a ¡ b = [a1 ¡ b1, a2 ¡ b2, a3 ¡ b3] Subtraction

² ka = [ka1, ka2, ka3] Scalar multiplication

SOME PROPERTIES OF VECTORS

1 For a = [2, ¡1, 1], b = [1, 2, ¡3] and c = [0, 1, ¡3] find:

a a + b b a ¡ b c b + 2c

d a ¡ 3c e a + b + c f c ¡ 12a

g a ¡ b ¡ c h 2b ¡ c + a i c ¡ 2a + 3b

ALGEBRAIC FORM

If a = [¡1, 2, ¡3] and b = [4, ¡1, 2] find:

a

a

a

a

+

+

2

2

b

b

b

b

2

2

a

a

¡

¡

3

3

b

b

= [¡1, 2, ¡3] + 2[4, ¡1, 2] = 2[¡1, 2, ¡3] ¡ 3[4, ¡1, 2]

= [¡1, 2, ¡3] + [8, ¡2, 4] = [¡2, 4, ¡6] ¡ [12, ¡3, 6]

= [7, 0, 1] = [¡14, 7, ¡12]

Example 8

² a + b = b + a

(a + b) + c = a + (b + c)

a + 0 = 0 + a = a

a + (¡a) = (¡a) + a = 0

² jkaj = jkj jaj where ka is parallel to a

length of kamodulus of k

length of a

EXERCISE 4D

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2 If a = [¡1, 1, 3], b = [1, ¡3, 2] and c = [¡2, 2, 4] find:

a jaj b jbj c jb + cj d ja ¡ cj e jajb f1

jaja

3 For and , show how to find geometrically:

a ¡a b 2b c 12a d ¡1

2b

e a + b f a ¡ b g b ¡ a h 2b ¡ a

4 Solve for x:

a 2x + a = b b b ¡ 3x = 2a c a + 2x = b ¡ x

5 If a = [¡1, 2, 3] and b = [2, ¡2, 1] find x if:

a 2a + x = b b 3x ¡ a = 2b c 2b ¡ 2x = ¡a

6 By letting a = [a1, a2, a3], prove that jkaj = jkj jaj :

Notice that if¡!OA = a and

¡!OB = b where O is the origin

then¡!AB = b ¡ a and

¡!BA = a ¡ b.

7 If¡!OA = [¡2, ¡1, 1] and

¡!OB = [1, 3, ¡1] find

¡!AB and hence find the distance

from A to B.

If a = [¡1, 3, 2]

find jaj :If a = [¡1, 3, 2] then

jaj =p

(¡1)2 + 32 + 22

=p

1 + 9 + 4

=p

14 units

Example 9

If a = [¡1, 0, 2] and b = [2, 1, ¡3] find x if:

a 2x = b b b ¡ 2x = a

a 2x = b

) 2x = [2, 1, ¡3]

) x = 12 [2, 1, ¡3]

) x = [1, 12 , ¡32 ]

b b ¡ 2x = a

) b ¡ 2x ¡ b = a ¡ b

) ¡2x = a ¡ b

) ¡2x = [¡1, 0, 2] ¡ [2, 1, ¡3]

) ¡2x = [¡3, ¡1, 5]

) x = 1¡2 [¡3, ¡1, 5]

) x = [32 , 12 , ¡52 ]

Example 10

ab

a

b

A

B

O

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8 The position vectors of A, B, C and D from O are [2, 1, ¡2], [0, 3, ¡4],

[1, ¡2, 1] and [¡2, ¡3, 2] respectively. Deduce that¡!BD = 2

¡!AC.

9 In the given figure BD is parallel to OA and

half its length. Find in terms of a and b vector

expressions for:

a¡!BD b

¡!AB c

¡!BA

d¡!OD e

¡!AD f

¡!DA

10 If¡!AB = [¡1, 3, 2],

¡!AC = [2, ¡1, 4] and

¡!BD = [0, 2, ¡3] find:

a¡!AD b

¡!CB c

¡!CD

PARALLELISM

Notice that a = [2, 6, ¡4] is parallel to b = [1, 3, ¡2] and c = [4, 12, ¡8] as

a = 2b and a = 12c. Also a = [2, 6, ¡4] is parallel to d = [¡3, ¡9, 6] as a = ¡3

2d.

a b

A

D

B

O

If¡!AB = [¡1, 3, 2],

¡!AC = [2, ¡1, 4] find

¡!BC.

¡!BC =

¡!BA +

¡!AC

= ¡[¡1, 3, 2] + [2, ¡1, 4]

= [3, ¡4, 2]

Example 11

PARALLELISM AND UNIT VECTORSE

Find r and s given that a = [2, ¡1, r] is parallel to b = [s, 2, ¡3].

Since a and b are parallel, then a = kb for some scalar k

) [2, ¡1, r] = k[s, 2, ¡3]

) 2 = ks, ¡1 = 2k and r = ¡3k

Consequently, k = ¡12 and ) 2 = ¡1

2s and r = ¡3¡¡12

¢) r = 3

2 and s = ¡4

Example 12

If two vectors are parallel, then one is a scalar multiple of the other and vice versa.

Note: ² If a is parallel to b, then there exists a

scalar, k say, such that a = kb.

² If a = kb for some scalar k, then

I a is parallel to b, and

I jaj = jkj jbj :

a b

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1 a = [2, ¡1, 3] and b = [¡6, r, s] are parallel. Find r and s.

2 Find scalars a and b, given that [3, ¡1, 2] and [a, 2, b] are parallel.

3 a Find a vector of length 1 unit which is parallel to a = [2, ¡1, ¡2].

[Hint: Let the vector be ka.]

b Find a vector of length 2 units which is parallel to b = [¡2, ¡1, 2].

4 What can be deduced from the following?

a¡!AB = 3

¡!CD b

¡!RS = ¡1

2

¡!KL c

¡!AB = 2

¡!BC d

¡!BC = 1

3

¡!AC

5 The position vectors of P, Q, R and S from O are [3, 2, ¡1], [1, 4, ¡3], [2, ¡1, 2] and

[¡1, ¡2, 3] respectively.

a Deduce that PR and QS are parallel.

b What is the relationship between the lengths of PR and QS?

6 Given that r1a + s1b = r2a + s2b where a and b are not parallel, prove that r1 = r2and s1 = s2.

[Hint: Suppose that s1 6= s2 and derive a contradiction.]

7 Prove that ja + bj 6 jaj + jbj using a geometrical argument.

[Hint: Consider a a b b a and b parallel c any other cases.]

A unit vector is any vector which has a length of one unit.

For example,

² [1, 0, 0] is a unit vector as its length isp

12 + 02 + 02 = 1

² [ 1p2

, 0, ¡ 1p2

] is a unit vector as its length is

r³1p2

´2+ 02 +

³¡ 1p

2

´2= 1

i = [1, 0, 0], j = [0, 1, 0] and k = [0, 0, 1]

the X, Y and Z-axes respectively.

Notice that a = [a1, a2, a3] , a = a1i + a2j + a3k.

component form unit vector form

Thus, a = [2, 3, ¡5] can be written as a = 2i + 3j ¡ 5k and vice versa.

The question arises: “How do we find quickly the unit vector in the direction of another

vector?”

EXERCISE 4E.1

UNIT VECTORS

The unit vector form, sometimes called the algebraic form, is useful as order does not have

to be preserved during addition.

For example, a + b = ( i + 32 2j ¡ 5k) + (i ¡ j + k) = 2i ¡ 2j + k + 3j + i ¡ 5k

not parallel to

are special unit vectors in the direction of

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The answer is:

and this vector is obtained by dividing each of the components of the original vector by the

length of the original vector.

Proof: The unit vector in the direction of a = [a1, a2, a3] has form ka where k is a

positive scalar.

Since jkaj = 1, then jkj jaj = 1 fusing jkaj = jkj jajg) jkj =

1

jaji.e., k = § 1

jaj) k =

1

jaj as k > 0

) the unit vector is1

jaj [a1, a2, a3] or .

So, for a = [2, 3, ¡5], jaj =p

4 + 9 + 25 =p

38

) [ 2p38

, 3p38

, ¡5p38

] is the unit vector in the direction of a.

Note: [ 2p38

, 3p38

, ¡5p38

] and [ ¡2p38

, ¡3p38

, 5p38

] are the unit vectors which are

parallel to [2, 3, ¡5].

1 Express the following vectors in component form and find their length:

a i ¡ j + k b 3i ¡ j + k c i ¡ 5k d 12 (j + k)

2 Express in terms of i, j and k:

a [2, 1, 3] b [2, 0, ¡5] c [0, 0, ¡3] d [¡3, 0, 1]

3 Find the length of:

a [1, 0, ¡1] b [12 , 12 , 1p2] c i ¡ j + k d 2i ¡ j

4 For a = ¡i + j + k and b = 3i ¡ j + 2k find in terms of i, j and k:

a a + b b b ¡ a c 2a + 5b d 3a ¡ 2b

5 Find a unit vector in the direction of:

a [1, 2, ¡3] b [1, 0, ¡1] c i ¡ k d 2i ¡ j + 5k

6 Find unit vectors which are parallel to:

a [2, 1, 0] b i ¡ j + 2k

·a1

jaj ;a2

jaj ;a3

jaj

is the unit vector in the direction of a = [a1, a2, a3]

·a1

jaj ;a2

jaj ;a3

jaj¸

¸

EXERCISE 4E.2

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7 Find a vector that is in the direction of:

a [2, 1, ¡2] and is 6 units in length

b [3, ¡1, 2] and is 2 units in length

c [1, 2, ¡4] and is 3 units in length

d i + j ¡ k and is 3 units in length.

8 a A is (2, 3, 4) and B is 14 units from A in the direction of ¡3i + 2j + 6k.

Find the coordinates of B.

b P is (¡1, 2, 1) and Q is 3 units from P in the direction of i ¡ j + k.

Find the coordinates of Q.

c R is (¡3, 1, 4) and S is 9 units from R in the direction of [2, ¡1, ¡2].

Find the coordinates of S.

COLLINEAR POINTS AND RATIO OF DIVISION

Three or more points are said to be collinear if

they lie on the same straight line.

Notice that,

A, B and C are collinear if¡!AB = k

¡!BC for

some scalar k.

Find a vector in the direction of [4, ¡1, 1] which is 5 units long.

If a = [4, ¡1, 1], then jaj =p

16 + 1 + 1 =p

18 units

) [ 4p18

, ¡1p18

, 1p18

] is a unit vector in the direction of a

) 5[ 4p18

, ¡1p18

, 1p18

] or [ 20p18

, ¡5p18

, 5p18

] is the required vector.

Example 13

C

B

A

A is (¡1, 3, 2) and B is 2p

3 units from A in the direction i ¡ j + k.

Find the coordinates of B.

¡!OB =

¡!OA +

¡!AB

= [¡1, 3, 2] + 2p

3h1p3

, ¡ 1p3

, 1p3

i

unit vector in direction of¡!AB

= [¡1, 3, 2] + [2, ¡2, 2]

= [1, 1, 4]) B is the point (1, 1, 4).

Example 14

A( 1, 3, 2)� � � � �1, 1, 1]

B

O

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Note: If the ratio of AB : BC = 5 : 3 we say

that B divides AC in the ratio 5 : 3

or B divides CA in the ratio 3 : 5:

B divides AC means the same as¡!AB :

¡!BC:

In this case¡!AB :

¡!BC = 8 : ¡3 = ¡8 : 3

i.e., B divides AC externally in the ratio 8 : 3. fB is external to the line segment ACg

If A is (¡1, 4, 7) and B(3, 0, 5) find:

a P if P divides AB in the ratio 3 : 1

b Q if Q divides BA externally in the ratio 3 : 1.

a¡!AP :

¡!PB = 3 : 1 b

¡!BQ :

¡!QA = ¡3 : 1

)¡!OP =

¡!OA +

¡!AP =

¡!OA + 3

4

¡!AB

= [¡1, 4, 7] + 34 [4, ¡4, ¡2]

= [¡1, 4, 7] + [3, ¡3, ¡32 ]

= [2, 1, 512 ]

)¡!OQ =

¡!OA +

¡!AQ =

¡!OA + 1

2

¡!BA

= [¡1, 4, 7] + 12 [¡4, 4, 2]

= [¡1, 4, 7] + [¡2, 2, 1]

= [¡3, 6, 8]

) P is (2, 1, 512 ) ) Q is (¡3, 6, 8)

1

2

C

B

A

3

5

B

C

A

3

5

Prove that A(¡1, 2, 3), B(4, 0, ¡1) and C(14, ¡4, ¡9) are collinear and hence

find the ratio in which B divides CA.

¡!AB = [5, ¡2, ¡4]¡!BC = [10, ¡4, ¡8] = 2[5, ¡2, ¡4]

)¡!BC = 2

¡!AB

) BC is parallel to AB and since B is common to both, A, B and C are collinear.

To find the ratio in which B divides CA, we find¡!CB :

¡!BA = ¡2[5, ¡2, ¡4] : ¡[5, ¡2, ¡4] = 2 : 1

) B divides CA internally in the ratio 2 : 1.

Example 15

Example 16

A

BP31

Q

BA 2

1

Note: a could also be done using¡!AP = 3

¡!PB

) [x+ 1, y ¡ 4, z ¡ 7] = 3[3 ¡ x, 0 ¡ y, 5 ¡ z]and then solving x+ 1 = 9 ¡ 3x, y ¡ 4 = ¡3y, z ¡ 7 = 15 ¡ 3z

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1 a Prove that A(¡2, 1, 4), B(4, 3, 0) and C(19, 8, ¡10) are collinear and hence

find the ratio in which A divides CB.

b Prove that P(2, 1, 1), Q(5, ¡5, ¡2) and R(¡1, 7, 4) are collinear and hence

find the ratio in which Q divides PR.

2 a A(2, ¡3, 4), B(11, ¡9, 7) and C(¡13, a, b) are collinear. Find a and b.

b K(1, ¡1, 0), L(4, ¡3, 7) and M(a, 2, b) are collinear. Find a and b.

3 P, Q and R have position vectors from O of p, q and r respectively.

If r = kp + 23q, and P, Q and R are collinear, find k.

4 For A(3, 1, 1), B(¡1, 2, 0), C(1, ¡1, 4) and D(3, ¡2, 4) find:

a Q if Q divides BC internally in the ratio 1 : 2

b R if R divides CA externally in the ratio 3 : 4

c S if S divides BA internally in the ratio 3 : 1

d T if T divides CB externally in the ratio 2 : 5

e X if X divides AD internally in the ratio 2 : 7

f Y if Y divides DB externally in the ratio 5 : 3

5 For P(¡1, 3, 2) and Q(4, 1, ¡5) find:

a X if X divides QP internally in the ratio 3 : 7

b Y if Y divides PQ externally in the ratio 3 : 5

6 For A(a1, a2, a3) and B(b1, b2, b3), find P if P divides AB in the ratio r : s.

With 2-D vectors a = [a1, a2] and b = [b1, b2] we define

a ² b = a1b1 + a2b2 as the scalar product or dot product of the vectors.

a ² b = jaj jbj cos µ is the geometric definition of scalar product where µ is the angle

between a and b.

It is only natural that we define the scalar product of 3-D vectors in the same way.

THE SCALAR PRODUCT OF 3-DIMENSIONAL VECTORS

If a = [a1, a2, a3] and b = [b1, b2, b3],

the scalar product of a and b (also known as the dot product) is defined as

a ² b = a1b1 + a2b2 + a3b3.

Once again, this definition is simply an extension of the 2-dimensional definition, adding the

Z-component.

EXERCISE 4E.3

THE SCALAR PRODUCTF

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ALGEBRAIC PROPERTIES OF THE SCALAR PRODUCT

Dot product has the same algebraic properties for 3-D vectors as it has for its 2-D counterparts.

I a ² b = b ² a

I a ² a = jaj2

I a ² (b + c) = a ² b + a ² c and

(a + b) ² (c + d) = a ² c + a ² d + b ² c + b ² d

These properties are proven in general by using vectors such as

a = [a1, a2, a3], b = [b1, b2, b3], etc.

GEOMETRIC PROPERTIES OF THE SCALAR PRODUCT

I If µ is the angle between vectors a and b then:

a ² b = jaj jbj cos µ

Note: One vector is translated so that both vectors emanate from the same point.

Proof: Let¡!OA = a and

¡!OB = b

)¡!AB =

¡!AO +

¡!OB

= ¡a + b

= b ¡ a

Now by the cosine rule, (AB)2 = (OA)2 + (OB)2 ¡ 2 (OA)(OB)cos µ

jb ¡ aj2 = jaj2 + jbj2 ¡ 2 jaj jbj cos µ

) (b ¡ a) ² (b ¡ a) = a ² a + b ² b ¡ 2 jaj jbj cos µ

) b ² b ¡ b ² a ¡ a ² b + a ² a = a ² a + b ² b ¡ 2 jaj jbj cos µ

) ¡2 a ² b = ¡2 jaj jbj cos µ

) a ² b = jaj jbj cos µ

I For non-zero vectors a and b:

a ² b = 0 , a and b are perpendicular.

Proof: a ² b = 0 , jaj jbj cos µ = 0

, cos µ = 0 as jaj 6= 0, jbj 6= 0

, µ = 90o

, a and b are perpendicular.

a

b

b a�

A

BO

�

I If a and b are two vectors then the:

projection vector of a on b is

µ ja ² bjjbj

¶1

jbjb

the length of

the projection

vector

unit vector in

the direction

of b

b

a

�P

R

Q

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Proof: In triangle PQR, cos µ =PQ

PR=

PQ

jaj) PQ = jaj cos µ

) PQ =jaj jbj cos µ

jbj

) PQ =a ² b

jbjHowever cos µ may be positive if µ is acute or negative if µ is obtuse and as

PQ is a length which must be positive,PQ =

ja ² bjjbj .

Sinceb

jbj is a unit vector in the direction of b then

the projection vector of a on b is

µa ² b

jj j

bj¶

1

jbjb.

1 For a = [2, 1, 3], b = [¡1, 1, 1] and c = [0, ¡1, 1] find:

a a ² b b b ² a c jaj2d a ² a e a ² (b + c) f a ² b + a ² c

If a = [2, 3, ¡1] and b = [¡1, 0, 2], find:

a a ² b b the angle between a and b c the projection vector of b on a.

a a ² b

= [2, 3, ¡1] ² [¡1, 0, 2]

= 2(¡1) + 3(0) + (¡1)2

= ¡2 + 0 ¡ 2

= ¡4

b a ² b = jaj jbj cos µ

) ¡4 =p

4 + 9 + 1p

1 + 0 + 4 cos µ

) ¡4 =p

14p

5 cos µ

) ¡4 =p

70 cos µ

) cos µ = ¡ 4p70

) µ = cos¡1³

¡ 4p70

´+ 118:56o

c the projection vector of b on a =

µb ² a

jj j

aj¶

1

jaja

=³

4p14

´1p14

a

= 27a

= [47 , 67 , ¡27 ]

Example 17

EXERCISE 4F

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2 Find:

a (i + j ¡ k) ² (2j + k) b i ² i c i ² j

3 Using a = [a1, a2, a3], b = [b1, b2, b3] and c = [c1, c2, c3] prove that

a ² (b + c) = a ² b + a ² c.

Hence, prove that (a + b) ²(c + d) = a ² c + a ² d + b ² c + b ² d.

4 Find t if [3, ¡1, t] and [2t, ¡3, ¡4] are perpendicular.

5 Show that a = [3, 1, 2], b = [¡1, 1, 1] and c = [1, 5, ¡4] are mutually perpendicular.

6 a Show that [1, 1, 5] and [2, 3, ¡1] are perpendicular.

b Find t if [3, t, ¡2] is perpendicular to [1 ¡ t, ¡3, 4].

7 Consider triangle ABC in which A is (5, 1, 2), B(6, ¡1, 0) and C(3, 2, 0). Using

scalar product only, show that the triangle is right angled.

8 A(2, 4, 2), B(¡1, 2, 3), C(¡3, 3, 6) and D(0, 5, 5) are vertices of a quadrilateral.

a Prove that ABCD is a parallelogram.

b Find j ¡!AB j and j ¡!

BC j. What can be said about ABCD?

c Find¡!AC ² ¡!

BD. What property of figure ABCD has been found to be valid?

9 For a = ¡i ¡ j + k and b = i + j + k find:

a a ² b

b the angle between a and b

c the projection vector of a on b

d the length of the projection vector of a on b.

10 Find the angle ABC of triangle ABC for A(3, 0, 1), B(¡3, 1, 2) and C(¡2, 1, ¡1).

Reminder:

To find the angle at B,¡!BA and

¡!BC are used.

What angle is found if¡!BA and

¡!CB are used?

11 Find the measure of angle PQR for

a P(1, 2, 3), Q(0, 2, ¡1) and R(2, ¡1, 2)

b P(2, 2, 1), Q(1, 2, 4) and R(3, 1, ¡1)

c P(4, 3, 0), Q(0, 3, ¡2) and R(¡3, 0, 1)

d P(6, 2, 1), Q(4, 3, 1) and R(3, 1, ¡3)

B

C

A

�

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12 For the cube alongside with sides of length 2 cm,

find using vector methods:

a the measure of angle ABS

b the measure of angle RBP

c the measure of angle PBS.

13 KL, LM and LX are 8, 5 and 3 units long

respectively

.

P is the midpoint of KL.

Find, using vector methods:

a the measure of angle YNX

b the measure of angle YNP.

14 For the tetrahedron ABCD:

a find the coordinates of M

b find the measure of angle DMA.

Use vector methods to determine the

measure of angle ABC.

Placing the coordinate axes as illustrated,

A is (2, 0, 0), B is (0, 4, 3) and C is (1, 4, 0)

)¡!BA is [2, ¡4, ¡3] and

¡!BC is [1, 0, ¡3]

and cos]ABC =

¡!BA ² ¡!

BC

j ¡!BA j j ¡!

BC j

=2(1) + (¡4)(0) + (¡3)(¡3)p

4 + 16 + 9:p

1 + 0 + 9

=2 + 0 + 9p

29:p

10

= 11p290

) ]ABC = cos¡1³

11p290

´+ 49:76o

Example 18

3 cm

2 cm

4 cm

C

B

A

3

2

4

C

B

A

X

Z

Y

A

P

QR

S

B C

D

W

KP L

M

YX

N

Z

M

A(2, 1, 1)� �B(1, , 1)� �

C(2, , )��

D(3, , )��

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H:\...\SA12SP04\146SP204.CDRMon Dec 02 14:32:18 2002



VECTOR PRODUCTG

15 a Find t if 2i + tj + (t¡ 2)k and ti + 3j + tk are perpendicular.

b Find r, s and t if a = [1, 2, 3], b = [2, 2, r] and c = [s, t, 1] are mutually

perpendicular.

16 Find the angle made by:

a the X-axis and the vector [1, 2, 3]

b a line parallel to the Y -axis and the vector [¡1, 1, 3].

17 Find three vectors a, b and c such that a 6= 0 and a ² b = a ² c, but b 6= c.

18

19 Given that a and b are the position vectors of two distinct points A and B (neither of

which is the origin) show that if ja + bj = ja ¡ bj then a is perpendicular to b using:

a a vector algebraic method b a geometric argument.

20 If jaj = 3 and jb j = 4, find .(a + b) ² (a ¡ b)

21 Explain why a ² b ² c is meaningless.

When scalar products are found the result is a scalar. However, there is another useful form

of vector multiplication where a vector results and so is called vector product.

Vector product arises out of an attempt to find

known vectors. Following is such an attempt.

Suppose X = [x, y, z] is perpendicular to both a = [a1, a2, a3] and b = [b1, b2, b3]

)

½a1x+ a2y + a3z = 0

b1x+ b2y + b3z = 0fas dot products are zerog

)

½a1x+ a2y = ¡a3z ...... (1)

b1x+ b2y = ¡b3z ...... (2)

We will now try to solve these two equations to get expressions for x and y in terms of z.

To eliminate x, we multiply (1) by ¡b1 and (2) by a1

¡a1b1x¡ a2b1y = a3b1z

a1b1x+ a1b2y = ¡a1b3zAdding these gives (a1b2 ¡ a2b1)y = (a3b1 ¡ a1b3)z

)y

z=a3b1 ¡ a1b3a1b2 ¡ a2b1

) y = (a3b1 ¡ a1b3)t and z = (a1b2 ¡ a2b1)t for all non-zero t

Now in (1) a1x = ¡a3(a1b2 ¡ a2b1)t¡ a2(a3b1 ¡ a1b3)t) a1x = (¡a1a3b2 + a2a3b1 ¡ a2a3b1 + a1a2b3)t

)

)

a1x

x

==a1((a

a2

2

b

b3

3

¡¡

a

a3

3

b

b2

2

))t

t

Show, using jx j2 = x ² x , that:

a ja + bj2 + ja ¡ bj2 = 2 jaj2 + 2 jbj2 b ja + bj2 ¡ ja ¡ bj2 = 4 a ² b

perpendicular to tw otheroa vector which is

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The simplest vector perpendicular to both [a1, a2, a3] and [b1, b2, b3] is obtained by

letting t = 1 and is [a2b3 ¡ a3b2, a3b1 ¡ a1b3, a1b2 ¡ a2b1].

We call this vector the cross product of a and b, and it is written as a £ b.

That is, a £ b = [a2b3 ¡ a3b2, a3b1 ¡ a1b3, a1b2 ¡ a2b1].

So that we do not have to memorise this awful looking vector we convert it to one containing

2 £ 2 determinants.

We let a1b2 ¡ a2b1 be represented by

¯̄̄¯ a1 a2b1 b2

¯̄̄¯.

So, for example,

¯̄̄¯ 2 3

4 5

¯̄̄¯ = 2 £ 5 ¡ 3 £ 4.

Consequently a2b3 ¡ a3b2 =

¯̄̄¯ a2 a3b2 b3

¯̄̄¯ and a3b1 ¡ a1b3 =

¯̄̄¯ a3 a1b3 b1

¯̄̄¯

) a £ b =

·¯̄̄¯ a2 a3b2 b3

¯̄̄¯ ,

¯̄̄¯ a3 a1b3 b1

¯̄̄¯ ,

¯̄̄¯ a1 a2b1 b2

¯̄̄¯¸

Notice the structure:

Notice also that: a £ b =

¯̄̄¯ a2 a3b2 b3

¯̄̄¯i +

¯̄̄¯ a3 a1b3 b1

¯̄̄¯j +

¯̄̄¯ a1 a2b1 b2

¯̄̄¯k

which is conveniently written as

a £ b =

¯̄̄¯̄̄ i j k

a1 a2 a3b1 b2 b3

¯̄̄¯̄̄ Ã This form is known as

a 3 £ 3 determinant.

same column same column

same column

If a = [2, 3, ¡1] and b = [¡1, 2, 4], find a £ b.

a £ b

= [2, 3, ¡1] £ [¡1, 2, 4]

=

·¯̄̄¯ 3 ¡1

2 4

¯̄̄¯ ,¯̄̄¯ ¡1 2

4 ¡1

¯̄̄¯ ,

¯̄̄¯ 2 3

¡1 2

¯̄̄¯¸

= [(12 ¡ ¡2), (1 ¡ 8), (4 ¡ ¡3)]

= [14, ¡7, 7]or

a £ b

=

¯̄̄¯̄̄ i j k

2 3 ¡1¡1 2 4

¯̄̄¯̄̄

=

¯̄̄¯ 3 ¡1

2 4

¯̄̄¯ i +

¯̄̄¯ ¡1 2

4 ¡1

¯̄̄¯ j +

¯̄̄¯ 2 3

¡1 2

¯̄̄¯ k

= 14i ¡ 7j + 7k

Example 19

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1 Calculate:

a [2, ¡3, 1] £ [1, 4, ¡2] b [¡1, 0, 2] £ [3, ¡1, ¡2]

c (i + j ¡2k) £ (i ¡ k) d (2i ¡ k) £ (j + 3k)

2 Given a = [1, 2, 3] and b = [¡1, 3, ¡1] , find a £ b and hence determine

a ² (a £ b) and b ² (a £ b). What has been verified from these results?

3 If i, j and k are the unit vectors parallel to the coordinate axes:

a find i £ i, j £ j, and k £ k

b find i £ j and j £ i, j £ k and k £ j, and i £ k and k £ i.

What do you suspect a £ a to simplify to, where a is any space vector?

What do you suspect is the relationship between a £ b and b £ a?

4 Using a = [a1, a2, a3] and b = [b1, b2, b3], prove that:

a a £ a = 0 for all space vectors a

b a £ b = ¡b £ a for all space vectors a and b.

5 For a = [1, 3, 2] , b = [2, ¡1, 1] and c = [0, 1, ¡2] find:

a b £ c b a ² (b £ c) c

¯̄̄¯̄̄ 1 3 2

2 ¡1 10 1 ¡2

¯̄̄¯̄̄

EXERCISE 4G.1

For a = [2, 1, ¡1], b = [1, 2, 3] and c = [2, 0, 4] find:

a b £ c b a ² (b £ c) c

¯̄̄¯̄̄ 2 1 ¡1

1 2 32 0 4

¯̄̄¯̄̄

a b £ c = [1, 2, 3] £ [2, 0, 4]

=

¯̄̄¯ 2 3

0 4

¯̄̄¯ i +

¯̄̄¯ 3 1

4 2

¯̄̄¯ j +

¯̄̄¯ 1 2

2 0

¯̄̄¯ k

= 8i + 2j ¡ 4k

b a ² (b £ c) = [2, 1, ¡1] ² [8, 2, ¡4]

= 16 + 2 + 4

= 22

c

¯̄̄¯̄̄ 2 1 ¡1

1 2 32 0 4

¯̄̄¯̄̄ =

¯̄̄¯ 2 3

0 4

¯̄̄¯ (2) +

¯̄̄¯ 3 1

4 2

¯̄̄¯ (1) +

¯̄̄¯ 1 2

2 0

¯̄̄¯ (¡1)

= 8(2) + 2(1) + (¡4)(¡1)

= 16 + 2 + 4

= 22

Example 20

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6 Repeat 5 for vectors of your choosing.

7 If a = i + 2k, b = ¡j + k and c = 2i ¡ k, determine:

a a £ b b a £ c c (a £ b) + (a £ c) d a £ (b + c)

8 What do you suspect to be true from 7?

Check with vectors a, b and c of your choosing.

9 Prove that a £ (b + c) = a £ b + a £ c using a = [a1, a2, a3], b = [b1, b2, b3]

and c = [c1, c2, c3].

10 Use a £ (b + c) = (a £ b) + (a £ c) to prove that

(a + b) £ (c + d) = (a £ c) + (a £ d) + (b £ c) + (b £ d).

[Notice that the order of the vectors must be maintained as x £ y = ¡y £ x.]

11 Simplify:

a a £ (a + b) b (a + b) £ (a + b)

c (2a + b) £ (a ¡ 2b) d 2a ² (a £ b)

12 Find all vectors perpendicular to both:

a [2, ¡1, 3] and [1, 1, 1] b [¡1, 3, 4] and [5, 0, 2]

c i + j and i ¡ j ¡ k d i ¡ j ¡ k and 2i + 2j ¡ 3k

13 Find all vectors perpendicular to both a = [2, 3, ¡1] and b = [1, ¡2, 2] and hence

find a vector of length 5 units which is perpendicular to both a and b.

Simplify (a + b) £ (a ¡ b).

(a + b) £ (a ¡ b) = (a £ a) ¡ (a £ b) + (b £ a) ¡ (b £ b)

= 0 + (b £ a) + (b £ a) ¡ 0

= 2(b £ a)

Example 21

Find all vectors which are perpendicular to both a = [1, 2, ¡1] and

b = [1, 0, ¡3].

a £ b =

·¯̄̄¯ 2 ¡1

0 ¡3

¯̄̄¯ ,

¯̄̄¯ ¡1 1

¡3 1

¯̄̄¯ ,

¯̄̄¯ 1 2

1 0

¯̄̄¯¸

= [¡6, 2, ¡2]

= ¡2[3, ¡1, 1]

) the vectors have form k[3, ¡1, 1] where k is any non-zero real number.

Example 22

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14 Find a direction vector of a normal to the plane passing through the points:

a A(1, 3, 2), B(0, 2, ¡5) and C(3, 1, ¡4)

b P(2, 0, ¡1), Q(0, 1, 3) and R(1, ¡1, 1).

In the exercise just concluded you should have observed the following properties of cross

product.

I a £ b is a vector which is perpendicular to both a and b.

I a £ a = 0 for all space vectors a.

I a £ b = ¡b £ a for all space vectors a and b,

i.e., a £ b and b £ a have the same length, but in opposite directions.

I a ² (b £ c) =

¯̄̄¯̄̄ a1 a2 a3b1 b2 b3c1 c2 c3

¯̄̄¯̄̄ and is called the scalar triple product.

I a £ (b + c) = (a £ b) + (a £ c) and hence

(a + b) £ (c + d) = (a £ c) + (a £ d) + (b £ c) + (b £ d).

DIRECTION OF a b�£�

We have already observed that as a £ b = ¡b £ a

then a £ b and b £ a are oppositely directed.

But, what is the direction of each?

Consider i £ j and j £ i. In the last Exercise,

we saw that i £ j = k and j £ i = ¡k.

Find a direction vector of a normal to the plane passing through the points

A(1, ¡1, 2), B(3, 1, 0) and C(¡1, 2, ¡3).

¡!AB = [2, 2, ¡2]¡!AC = [¡2, 3, ¡5]

Now n is perpendicular to both¡!AB

and¡!AC.

fA normal is perpendicular to every line in the plane.g

Thus n =

·¯̄̄¯ 2 ¡2

3 ¡5

¯̄̄¯ ,

¯̄̄¯ ¡2 2

¡5 ¡2

¯̄̄¯ ,¯̄̄¯ 2 2

¡2 3

¯̄̄¯¸

= [¡4, 14, 10]

= ¡2[2, ¡7, ¡5]

Thus any non-zero multiple of [2, ¡7, ¡5] will do.

Example 23

B

CA

n

X

Z

Y

k

ji

X

Z

Y

�k

ji

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In general, the direction

of a £ b is determined

by the right hand rule.

To determine the direction of x £ y use the

right hand, where the fingers turn from x to y

and the thumb points in the direction of x £ y.

THE LENGTH OF a b�£�

As a £ b = [a2b3 ¡ a3b2, a3b1 ¡ a1b3, a1b2 ¡ a2b1] then

ja £ bj =p

(a2b3 ¡ a3b2)2 + (a3b1 ¡ a1b3)2 + (a1b2 ¡ a2b1)2

However, another very useful form of the length of a £ b exists. This is:

ja £ bj = jaj jbj sin µ where µ is the angle between a and b.

Proof: We start with jaj2 jbj2 sin2 µ

= jaj2 jbj2 (1 ¡ cos2 µ)

= jaj2 jbj2 ¡ jaj2 jbj2 cos2 µ

= jaj2 jbj2 ¡ (a ² b)2

= (a 21 + a 22 + a 23 )(b 21 + b 22 + b3)2 ¡ (a1b1 + a2b2 + a3b3)

2

which on expanding and then factorising

= (a2b3 ¡ a3b2)2 + (a3b1 ¡ a1b3)2 + (a1b2 ¡ a2b1)2= ja £ bj2

and so ja £ bj = jaj jbj sin µ fas sin µ > 0g

Immediate consequences are:

I If u is a unit vector in the direction of a £ b then a £ b = jaj jbj sin µ u

[In some texts this is the geometric definition of a £ b.]

I If a and b are non-zero vectors, then a £ b = 0 , a is parallel to b.

1 a Find i £ k and k £ i using the original definition

of a £ b.

b Check that the right-hand rule correctly gives

the direction of i £ k and k £ i.

c Check that a £ b = jaj jbj sin µ u could be used

to find i £ k and k £ i.

a b! b

a

b a!

b

a

EXERCISE 4G.2

X

Z

Y

k

ji

x y!

x y

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2 Consider a = [2, ¡1, 3] and b = [1, 0, ¡1].

a Find a ² b and a £ b.

b Find cos µ using a ² b = jaj jbj cos µ.

c From b, find sin µ using sin2 µ + cos2 µ = 1.

d Find sin µ using ja £ bj = jaj jbj sin µ.

3 Prove the property:

“If a and b are non-zero vectors then a £ b = 0 , a is parallel to b.”

4 O is the origin. Find:

a¡!OA and

¡!OB b

¡!OA £ ¡!

OB and j ¡!OA £ ¡!

OB j :c Explain why the area of triangle OAB is

12 j ¡!

OA £ ¡!OB j :

5 A, B and C are 3 distinct points with non-zero position vectors a, b and c respectively.

a If a £ c = b £ c, what can be deduced about¡!OC and

¡!AB?

b If a + b + c = 0, what relationship exists between a £ b and b £ c?

c If c 6= 0 and b £ c = c £ a, prove that a £ b = kc for some scalar k.

TRIANGLES

Proof: Area = 12£ product of two sides £ sine of included angle

= 12 £ jaj jbj sin µ

= 12 ja £ bj

Find the area of ¢ABC given A(¡1, 2, 3), B(2, 1, 4) and C(0, 5, ¡1).

Area = 12 j ¡!

AB £ ¡!AC j

= 12 j[3, ¡ 1, 1] £ [1, 3, ¡ 4]j

= 12

·¯̄̄¯ ¡1 1

3 ¡4

¯̄̄¯ ,¯̄̄¯ 1 3

¡4 1

¯̄̄¯ ,¯̄̄¯ 3 ¡1

1 3

¯̄̄¯¸

= 12 j[1, 13, 10]j

= 12

p1 + 169 + 100

= 12

p270 units2

AREAS AND VOLUMESH

O

B( 1, 1, 2)� � �

A(2, 3, 1)� �

If a triangle has defining vectors a and b

then its area is 12 ja £ bj units2. �

a

b

Example 24

A( 1, , )� ��

C(0, , 1)��

B(2, , )��

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PARALLELOGRAMS

If a parallelogram has defining vectors a and b then

its area is ja £ bj units2.

PARALLELEPIPED

If a parallelepiped has defining vectors a, b and c then

its volume is

ja ² (b £ c)j =

¯̄̄¯̄̄ a1 a2 a3b1 b2 b3c1 c2 c3

¯̄̄¯̄̄ units3.

Proof: Volume = (area of base) £ (perp. height)

= jb £ cj £ AN

= jb £ cj £ jaj sin µ fas sin µ =AN

jaj g= jaj jb £ cj sin µ= jaj jb £ cj cosÁ

where Á is the angle between a and b £ c

= ja ² (b £ c)j as cosÁ > 0.

TETRAHEDRON

If a tetrahedron has defining vectors a, b and c then

its volume is

16 ja ² (b £ c)j = 1

6

¯̄̄̄¯̄¯̄̄¯̄̄ a1 a2 a3b1 b2 b3c1 c2 c3

¯̄̄¯̄̄¯̄̄̄¯̄ units3:

Proof: Volume = 13 (area of base) £ (perp. height)

= 13 £ 1

2 jb £ cj £ AN

= 16 jb £ cj jaj sin µ fas sin µ =

AN

jaj g= 1

6 jaj jb £ cj cosÁ

where Á is the angle between a and b £ c

= 16 ja ² (b £ c)j as cosÁ > 0.

a

b

a

c

b

modulus

modulus

determinant

b c!

a

a

c

b�

�

�

N

O

A

B

C

The proof follows directly from that of triangle as the parallelogram consists of two congru-ent triangles with defining vectors and

a.a b

a

c

b

a

a

c

b

� N

O

AB

C

�b c!

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Find the volume of the tetrahedron with vertices P(0, 0, 1), Q(2, 3, 0),

R(¡1, 2, 1) and S(1, ¡2, 4).

¡!PQ = [2, 3, ¡1],¡!PR = [¡1, 2, 0],¡!PS = [1, ¡2, 3]

9>>=>>; are the defining vectors from P

Thus the volume = 16

¯̄̄¯̄̄¯̄̄¯̄̄ 2 3 ¡1

¡1 2 01 ¡2 3

¯̄̄¯̄̄¯̄̄¯̄̄

= 16

¯̄̄¯2¯̄̄¯ 2 0

¡2 3

¯̄̄¯+ 3

¯̄̄¯ 0 ¡1

3 1

¯̄̄¯¡ 1

¯̄̄¯ ¡1 2

1 ¡2

¯̄̄¯¯̄̄¯

= 16 j2(6) + 3(3) ¡ 1(0)j

= 16 j12 + 9j

= 216 = 312 units3

1 Calculate the area of triangle ABC for:

a A(2, 1, 1), B(4, 3, 0), C(1, 3, ¡2)

b A(0, 0, 0), B(¡1, 2, 3) and C(1, 2, 6)

c A(1, 3, 2), B(2, ¡1, 0) and C(1, 10, 6)

2 Calculate the area of parallelogram ABCD for A(¡1, 2, 2), B(2, ¡1, 4) and

C(0, 1, 0).

3 ABCD is a parallelogram where A is (¡1, 3, 2), B(2, 0, 4) and C(¡1, ¡2, 5). Find

the

a coordinates of D b area of ABCD.

4 ABCD is a tetrahedron with A(1, ¡1, 0), B(2, 1, ¡1), C(0, 1, ¡3) and

D(¡1, 1, 2). Find the:

a volume of the tetrahedron b total surface area of the tetrahedron.

5 A(3, 0, 0), B(0, 1, 0) and C(1, 2, 3) are vertices of a parallelepiped, adjacent to vertex

O(0, 0, 0). Find the:

a coordinates of the other four vertices b measure of ]ABC

c volume of the parallelepiped.

6 If A(¡1, 1, 2), B(2, 0, 1) and C(k, 2, ¡1) are three points in space, find k if the

area of triangle ABC isp

88 units2.

Example 25

P

Q

R

S

EXERCISE 4H

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7 A, B and C are three points with position vectors a, b and c respectively. Find a formula

for S, the total surface area of the tetrahedron OABC.

8 Three distinct points, A, B and C, have position vectors a, b and c respectively. Prove

that A, B and C are collinear , (b ¡ a) £ (c ¡ b) = 0.

TEST FOR COPLANAR POINTS

Four points in space are either coplanar or form the vertices of a tetrahedron. If they are

coplanar, the volume of the tetrahedron is zero. So,

if four points A, B, C and D have position vectors a, b, c and d respectively

then A, B, C and D are coplanar , (b ¡ a) ² (c ¡ a) £ (d ¡ a) = 0:

9 Are these points coplanar?

a A(1, 1, 2), B(2, 4, 0), C(3, 1, 1) and D(4, 0, 1)

b P(2, 0, 5), Q(0, ¡1, 4), R(2, 1, 0), S(1, 1, 1)

10 Find k given that A(2, 1, 3), B(4, 0, 1), C(0, k, 2), D(1, 2, ¡1) are coplanar.

A line in space is a straight line which continues indefinitely in both directions and

contains a continuous infinite set of points.

Suppose M(x, y, z) is a point which is free to move

on a line containing a fixed point F(x1, y1, z1).

If u = [a, b, c] is a direction vector of the line, then

Are the points A(1, 2, ¡4), B(3, 2, 0), C(2, 5, 1) and D(5, ¡3, ¡1)

coplanar?

b ¡ a =¡!AB = [3 ¡ 1, 2 ¡ 2, 0 ¡ ¡4] = [2, 0, 4]

c ¡ a =¡!AC = [2 ¡ 1, 5 ¡ 2, 1 ¡ ¡4] = [1, 3, 5]

d ¡ a =¡!AD = [5 ¡ 1, ¡3 ¡ 2, ¡1 ¡ ¡4] = [4, ¡5, 3]

and (b ¡ a) ² (c ¡ a) £ (d ¡ a) =

¯̄̄¯̄̄ 2 0 4

1 3 54 ¡5 3

¯̄̄¯̄̄

= 2(9 + 25) + 4(¡5 ¡ 12)

= 0

) A, B, C and D are coplanar.

Example 26

LINES IN SPACEI

M( , , )x y z

F( , , )x y zz z zu

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¡!FM = tu for some scalar t.

) [x¡ x1, y ¡ y1, z ¡ z1] = t[a, b, c] is the vector equation of the line.

Thus x¡ x1 = at, y ¡ y1 = bt and z ¡ z1 = ct.

) x = x1 + at, y = y1 + bt and z = z1 + ct.

Summary:

If a straight line passes through (x1, y1, z1) and has direction vector [a, b, c]:

² Its vector equation is [x¡ x1, y ¡ y1, z ¡ z1] = t[a, b, c] ,

or [x, y, z] = [x1, y1, z1] + t[a, b, c].

² Its parametric equations are x = x1 + at, y = y1 + bt and z = z1 + ct.

t is called the parameter and can take any real values.

² Its Cartesian equations arex¡ x1a

=y ¡ y1b

=z ¡ z1c

, (= t).

1 Find the vector equation of the line:

a parallel to [2, 1, 3] and through the point (1, 3, ¡7)

b through (0, 1, 2) and with direction vector i + j ¡ 2k

c parallel to the X-axis and through the point (¡2, 2, 1).

2 Find the parametric equations of the line:

a parallel to [¡1, 2, 6] and through the point (5, 2, ¡1)

b parallel to 2i ¡ j + 3k and through the point (0, 2, ¡1)

c perpendicular to the XOY plane and through (3, 2, ¡1).

3 Find the Cartesian equations of the line:

a parallel to [3, ¡1, 2] and through (1, ¡2, ¡3)

b parallel to the Z-axis and through (2, 2, ¡1)

c through (2, 6, ¡1) and parallel to i ¡ 2k.

Find the vector equation, the parametric equations and the Cartesian equations of

the line through (1, ¡2, 3) in the direction 4i + 5j ¡ 6k.

The vector equation is

[x, y, z] = [1, ¡2, 3] + t[4, 5, ¡6] , t in R.

The parametric equations are

x = 1 + 4t, y = ¡2 + 5t, z = 3 ¡ 6t, t in R.

The Cartesian equations are

x¡ 1

4=y + 2

5=z ¡ 3

¡6(= t).

Example 27

EXERCISE 4I

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4 Write down the parametric equations of the line given byx¡ 1

2=y + 5

¡3= 2 ¡ z

and find the point on the line with y-coordinate 4.

5 Given the line with equations x¡ 1 = 2(y ¡ 2) = 4z, find the equations of the line

in: a parametric form b vector form.

Find the parametric equations of the line through A(2, ¡1, 4) and B(¡1, 0, 2).

We require a direction vector for the line (¡!AB or

¡!BA)

¡!AB = [¡1 ¡ 2, 0 ¡ ¡1, 2 ¡ 4] = [¡3, 1, ¡2]

Using the point A, the equations are: x = 2 ¡ 3t, y = ¡1 + t, z = 4 ¡ 2t (t in R)

[Using the point B, the equations are: x = ¡1 ¡ 3s, y = s, z = 2 ¡ 2s, (s in R)]

Note: Both sets of equations generate the same set of points and s = t¡ 1.

6 Find the parametric equations of the line through:

a A(1, 2, 1) and B(¡1, 3, 2) b C(0, 1, 3) and D(3, 1, ¡1)

c E(1, 2, 5) and F(1, ¡1, 5) d G(0, 1, ¡1) and H(5, ¡1, 3)

Example 29

Example 28

Consider the line defined by x¡ 2 =1 ¡ y

2=z + 3

¡1:

a Write down the parametric equations of the line.

b

c Find the point on the line with z-coordinate 5.

a Let x¡ 2 =1 ¡ y

2=z + 3

¡1= t

) x = 2 + t, 1 ¡ y = 2t and z + 3 = ¡ti.e., x = 2 + t, y = 1 ¡ 2t, z = ¡3 ¡ t (t any real number).

b l = [1, ¡2, ¡1] f tgc We could use the original equations with z = 5

i.e., x¡ 2 =1 ¡ y

2=

5 + 3

¡1= ¡8

) x = ¡6, y = 17, z = 5 So the point is (¡6, 17, 5).

or Use the parametric equations, where ¡3 ¡ t = 5

) ¡8 = t

) x = 2 ¡ 8 = ¡6 y = 1 ¡ 2(¡8) = 17

i.e., the point is (¡6, 17, 5).

Write down the direction vector of the line.

the coefficients of

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8 Find points on the line with parametric equations x = 2 ¡ t, y = 3 + 2t and

z = 1 + t which are 5p

3 units from the point (1, 0, ¡2).

9 Consider the pair of lines 3(x¡ 1) = 2(y + 2) = z ¡ 3 and x = y = z.

a Write down a and b, the direction vectors of the lines.

b Find the parametric equations of the line through (1, 2, 3) that is perpendicular to

both of the original lines.

10 Find the coordinates of the foot of the perpendicular:

a from (1, 1, 2) to the line with equations x = 1 + t, y = 2 ¡ t, z = 3 + t

b from (2, 1, 3) to the line with vector equation [x, y, z] = [1, 2, 0] + s[1, ¡1, 2]:

11 Find the distance from:

a (3, 0, ¡1) to the line with equations x = 2 + 3t, y = ¡1 + 2t, z = 4 + t

b (1, 1, 3) to the line with vector equation [x, y, z] = [1, ¡1, 2] + t[2, 3, 1]:

Find the coordinates of the foot of the perpendicular from P(¡1, 2, 3) to the line

with parametric equations x = 1 + 2t, y = ¡4 + 3t, z = 3 + t.

A(1 + 2t, ¡4 + 3t, 3 + t) is any point on

¡!PA = [1 + 2t¡ ¡1, ¡4 + 3t¡ 2, 3 + t¡ 3]

= [2 + 2t, ¡6 + 3t, t]

and v = [2, 3, 1] is the direction vector of the line.

Now as¡!PA and v are perpendicular,

¡!PA ² v = 0

) [2 + 2t, ¡6 + 3t, t] ² [2, 3, 1] = 0

) 2(2 + 2t) + 3(¡6 + 3t) + 1(t) = 0

) 4 + 4t¡ 18 + 9t+ t = 0

) 14t = 14

) t = 1

and substituting t = 1 into the parametric equations we obtain the foot of the

perpendicular (3, ¡1, 4).

Note: As t = 1,¡!PA = [4, ¡3, 1] and j ¡!

PA j=

p16 + 9 + 1

=p

26 units

) the shortest distance from P to the line isp

26 units.

Example 30

v

A�� t t t

P��

the given line.

7 Find the coordinates of the point where the line with parametric equations

x = 1 ¡ t, y = 3 + t and z = 3 ¡ 2t meets:

a the XOY plane b the Y OZ plane c the XOZ plane.

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SHORTEST DISTANCE FROM POINT TO LINE

If a line has direction vector v and a known point A,

the shortest distance d from P to the line is given by

d =j ¡!

AP £ v jjvj

Proof: sin µ =d

j ¡!AP j

and so d =j ¡!AP j sin µ

=j ¡!

AP j jvj sin µjvj

=j ¡!

AP £ v jjvj

12 Check your answers to question 11 using the distance from a point to a line formula.

13 x2 + y2 + z2 = 26 is the equation of a sphere, centre (0, 0, 0) and radiusp

26 units.

Find the point(s) where the line through (3, ¡1, ¡2) and (5, 3, ¡4) meets the given

sphere.

vA

N

P

d line

AN

P

d line

�

Use the distance formula to find the shortest distance from P(¡1, 2, 3) to the line

x = 1 + 2t, y = ¡4 + 3t, z = 3 + t (as in Example 30).

The direction vector of the line is v = [2, 3, 1]

and if t = 0, say, A on the line is (1, ¡4, 3)

and¡!AP = [¡1 ¡ 1, 2 ¡ ¡4, 3 ¡ 3]

= [¡2, 6, 0]

and¡!AP £ v =

·¯̄̄¯ 6 0

3 1

¯̄̄¯ ,¯̄̄¯ 0 ¡2

1 2

¯̄̄¯ ,¯̄̄¯ ¡2 6

2 3

¯̄̄¯¸

= [6, 2, ¡18]

Now d =j ¡!

AP £ v jjvj

=

p62 + 22 + (¡18)2p

22 + 32 + 12

=

p364p14

=p

26 units

Example 31

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Two lines in space are either parallel, intersecting or skew,

i.e., or

Skew lines are any lines which are neither parallel nor intersecting.

² If the lines are parallel, the angle between them is 0o.

² If the lines are intersecting, the angle between them is µo, as shown.

² If the lines are skew, there is still an angle that one line makes with the other and if

we translate one line to intersect the other, the angle between the original lines is

defined as the angle between the intersecting lines,

i.e.,

LINE CLASSIFICATION

�

point of intersection

line 1line 2

�

Line 1 has equations x = ¡1 + 2s, y = 1 ¡ 2s and z = 1 + 4s.

Line 2 has equations x = 1 ¡ t, y = t and z = 3 ¡ 2t.

Line 3 has equations = ¡y ¡ 1 =zx ¡¡ 41

32:

a Show that line 1 and line 2 are parallel.

b Show that line 2 and line 3 intersect and find the angle between them.

c Show that line 1 and line 3 are skew.

a Line 1 has direction vector [2, ¡2, 4] = 2[1, ¡1, 2]

Line 2 has direction vector [¡1, 1, ¡2] = ¡[1, ¡1, 2]

) both line 1 and line 2 are parallel to [1, ¡1, 2] and so must be parallel

to each other.

b Substituting line 2’s equations into line 3’s, gives

1 ¡ t¡ 1

2= ¡t¡ 1 =

3 ¡ 2t¡ 4

3

¡ t2

= ¡t¡ 1 =¡1 ¡ 2t

3

) ¡t = 2(¡t¡ 1) and 3(¡t¡ 1) = ¡1 ¡ 2t

) ¡t = ¡2t¡ 2 and ¡3t¡ 3 = ¡1 ¡ 2t

) t = ¡2 and ¡2 = t

i.e., t = ¡2 (common solution)

Example 32

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14 Classify the following line pairs as either parallel, intersecting or skew and in each case

determine the measure of the angle between each pair:

a x = 1 + 2t, y = 2 ¡ t, z = 3 + t andx+ 2

3= 3 ¡ y =

z ¡ 1

2

b x = ¡1 + 2t, y = 2 ¡ 12t, z = 4 + 12t andx+ 3

4=y ¡ 2

3=z + 1

¡1

c x = 6t, y = 3 + 8t, z = ¡1 + 2t andx¡ 2

3=y

4= z ¡ 1

) the lines intersect at t = ¡2, i.e., at (3, ¡2, 7)

x¡ 1

2=

¡y ¡ 1

1=z ¡ 4

3

)x¡ 1

2=y + 1

¡1=z ¡ 4

3

) l3 = [2, ¡1, 3]

The acute angle between lines 2 and 3 is found using

cos µ =jl2 ² l3jjl2j jl3j

=j(¡2) + (¡1) ¡ 6jp

1 + 1 + 4 £ p4 + 1 + 9

=9p

6 £ p14

) µ = cos¡1³

9p84

´+ 10:89o

c Substituting line 1’s equations into line 3’s, gives

¡1 + 2s¡ 1

2= ¡1 + 2s¡ 1 =

1 + 4s¡ 4

3

i.e.,2s¡ 2

2= 2s¡ 2 =

4s¡ 3

3

i.e., s¡ 1 = 2s¡ 2 =4s¡ 3

3

) 1 = s and 3(2s¡ 2) = 4s¡ 3

i.e., 6s¡ 6 = 4s¡ 3

i.e., 2s = 3

s = 32

This is absurd as s cannot be 1 and 32 simultaneously.

Thus the lines cannot meet and are not parallel, ) are skew.

We could now find the angle between the lines using the method in b.

lx� �� "

lc� ��"

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d x = 2 ¡ y = z + 2 andx¡ 1

3=y + 2

¡1=

2z ¡ 1

4

e x = 1 + t, y = 2 ¡ t, z = 3 + 2t andx¡ 2

3=

3 ¡ y2

= z + 5

f x = 1 ¡ 2t, y = 8 + t, z = 5 and x = 2 + 4s, y = ¡1 ¡ 2s, z = 3.

THE SHORTEST DISTANCE BETWEEN SKEW LINES

Suppose line 1 contains point A and has direction

vector u and line 2 contains point B and has direction

vector v.

Line 2 is translated to line 20 so that line 20 and line

1 intersect.

u £ v is a vector which is perpendicular to both

u and v and line 2 is parallel to the shaded plane

containing lines 1 and 20.

So, the shortest distance d is the length of the projection vector of¡!AB on u £ v

i.e., d =

¯̄̄¡!AB ² (u £ v)

¯̄̄ju £ vj .

Find the shortest distance between the skew lines x = t, y = 1 ¡ t, z = 2 + t and

x = 3 ¡ s, y = ¡1 + 2s, z = 4 ¡ s.

line 1 contains A(0, 1, 2) and has direction u = [1, ¡1, 1]

line 2 contains B(3, ¡1, 4) and has direction v = [¡1, 2, ¡1]

So¡¡!AB = [3 ¡ 0;¡1 ¡ 1; 4 ¡ 2] = [3, ¡2, 2]

and u £ v =

·¯̄̄¯ ¡1 1

2 ¡1

¯̄̄¯ ,

¯̄̄¯ 1 1

¡1 ¡1

¯̄̄¯ ,¯̄̄¯ 1 ¡1

¡1 2

¯̄̄¯¸

= [¡1, 0, 1]

Now d =

¯̄̄¡!AB ² (u £ v)

¯̄̄ju £ vj

=j(3)(¡1) + (¡2)(0) + (2)(1)jp

(¡1)2 + 02 + 12

=j¡1jp

2

= 1p2

units

Example 33

v

v

u

u v!

line 2

line 1

line 2'

A

B

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15 Find the shortest distance between the skew lines:

a x = 1 + 2t, y = ¡t, z = 2 + 3t and x = y = z

b x = 1 ¡ t, y = 1 + t, z = 3 ¡ t and x = 2 + s, y = 1 ¡ 2s, z = s

Suppose a plane in space has normal vector n = [A, B, C]

and that it passes through the fixed point F(x1, y1, z1).

M(x, y, z) moves anywhere in the plane.

Now¡!FM is perpendicular to n

) n ² ¡!FM = 0

) [A, B, C] ² [x¡ x1, y ¡ y1, z ¡ z1] = 0

) A(x¡ x1) +B(y ¡ y1) + C(z ¡ z1) = 0

) Ax+By +Cz = Ax1 +By1 +Cz1

Note: u ² ¡!FM = 0 is the vector equation of the plane. It could also be written as

n ² (m ¡ f) = 0.

So, if a plane has normal vector [A, B, C] and passes through the point (x1, y1, z1)

then it has equation Ax+By +Cz = Ax1 +By1 +Cz1 = D, say,

where D is a constant.

1 Find the equation of the plane:

a with normal vector [2, ¡1, 3] and through (¡1, 2, 4)

b perpendicular to the line through A(2, 3, 1) and B(5, 7, 2) and through A

c perpendicular to the line connecting A(1, 4, 2) and B(4, 1, ¡4) and containing

P such that AP : PB = 1 : 2

d containing A(3, 2, 1) and the line x = 1 + t, y = 2 ¡ t, z = 3 + 2t.

2 State the normal vector to the plane with equation:

a 2x+ 3y ¡ z = 8 b 3x¡ y = 11

c z = 2 d x = 0

PLANESJ

Find the equation of the plane with normal vector [1, 2, 3] and containing the

point (¡1, 2, 4).

Since n = [1, 2, 3] and (¡1, 2, 4) lies on the plane, the equation is

x+ 2y + 3z = (¡1) + 2(2) + 3(4)

i.e., x+ 2y + 3z = 15

Example 34

EXERCISE 4J

n� "A B C, ,

F( , , )x y zz z z

M( , , )x y z

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3 Find the equation of the:

a XOZ-plane

b plane perpendicular to the Z-axis and through (2, ¡1, 4).

4 Find the equation of the plane through:

a A(0, 2, 6), B(1, 3, 2) and C(¡1, 2, 4),

b A(3, 1, 2), B(0, 4, 0) and C(0, 0, 1),

c A(2, 0, 3), B(0, ¡1, 2) and C(4, ¡3, 0).

5 Find the equations of the following lines:

a through (1, ¡2, 0) and normal to the plane x¡ 3y + 4z = 8

b through (3, 4, ¡1) and normal to the plane x¡ y ¡ 2z = 11.

Find the equation of the plane through A(¡1, 2, 0), B(3, 1, 1) and C(1, 0, 3).

If n is the normal vector, then n =¡!AB £ ¡!

AC

= [4, ¡1, 1] £ [2, ¡2, 3]

=

·¯̄̄¯ ¡1 1

¡2 3

¯̄̄¯ ,¯̄̄¯ 1 4

3 2

¯̄̄¯ ,

¯̄̄¯ 4 ¡1

2 ¡2

¯̄̄¯¸

= [¡1, ¡10, ¡6]

= ¡[1, 10, 6]

Thus the plane has equation x+ 10y + 6z = (¡1) + 10(2) + 6(0)

i.e., x+ 10y + 6z = 19

[Note: Check that all 3 points satisfy this equation.]

Example 35

A

C

B

n

Find the parametric equations of the line through A(¡1, 2, 3) and B(2, 0, ¡3)

and hence find where this line meets the plane with equation x¡ 2y + 3z = 26:

¡!AB = [3, ¡2, ¡6]

) line AB has parametric equations

x = ¡1 + 3t, y = 2 ¡ 2t, z = 3 ¡ 6t ..... (¤)

and this line meets the plane x¡ 2y + 3z = 26 where

¡1 + 3t¡ 2(2 ¡ 2t) + 3(3 ¡ 6t) = 26

) 4 ¡ 11t = 26

) ¡11t = 22

) t = ¡2

) meets plane at (¡7, 6, 15) fsubstitute t = ¡2 into ¤g

Example 36

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6 Find the parametric equations of the line through A(2, ¡1, 3) and B(1, 2, 0) and hence

find where this line meets the plane with equation x+ 2y ¡ z = 5.

7 Find the parametric equations of the line through P(1, ¡2, 4) and Q(2, 0, ¡1) and

hence find where this line meets:

a the Y OZ-plane

b the plane with equation y + z = 2

c the line with equationsx¡ 3

2=y + 2

3=z ¡ 30

¡1.

8 In the following, find the foot of the normal from A to the given plane and hence find

the shortest distance from A to the plane:

a A(1, 0, 2); 2x+ y ¡ 2z + 11 = 0

b A(2, ¡1, 3); x¡ y + 3z = ¡10

c A(1, ¡4, ¡3); 4x¡ y ¡ 2z = 8

9 Find the coordinates of the mirror image of A(3, 1, 2) when reflected in the plane

x+ 2y + z = 1.

10 Does the line through (3, 4, ¡1) and normal to x + 4y ¡ z = ¡2 intersect any of

the coordinate axes?

11 Find the equations of the plane through A(1, 2, 3) and B(0, ¡1, 2) which is parallel

to:

a the X-axis b the Y -axis c the Z-axis.

Find the coordinates of the foot of the normal from A(2, ¡1, 3) to the plane

x¡ y + 2z = 27. Hence find the shortest distance from A to the plane.

x¡ y + 2z = 27 has normal vector [1, ¡1, 2]

) the parametric equations of AN are

x = 2 + t, y = ¡1 ¡ t, z = 3 + 2t

and this line meets the plane x¡ y + 2z = 27 where

2 + t¡ (¡1 ¡ t) + 2(3 + 2t) = 27

i.e., 2 + t+ 1 + t+ 6 + 4t = 27

i.e., 6t+ 9 = 27

) 6t = 18

) t = 3 ) N is (5, ¡4, 9).

The shortest distance AN =p

(5 ¡ 2)2 + (¡4 ¡ ¡1)2 + (9 ¡ 3)2

=p

9 + 9 + 36 =p

54 units.

A��

N

Example 37

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12 Show that the lines x ¡ 1 =y ¡ 2

2= z + 3 and x + 1 = y ¡ 3 = 2z + 5 are

coplanar and find the equation of the plane which contains them.

13 A(1, 2, k) lies on the plane x+ 2y ¡ 2z = 8. Find:

a the value of k

b the coordinates of B such that AB is normal to the plane and 6 units from it.

14 Q is any point in the plane

Ax+By +Cz = D.

d is the distance from P(x1, y1, z1)

to the given plane.

a Explain why d =j ¡!

QP ² n jjnj .

b Hence, show that d =jAx1 +By1 +Cz1 +Djp

A2 +B2 +C2.

c Check your answers to question 8 using the formula of b.

15 Find the distance from:

a (0, 0, 0) to x+ 2y ¡ z = 10 b (1, ¡3, 2) to x+ y ¡ z = 2.

16 Find the distance between the parallel planes:

a x+ y + 2z = 4 and 2x+ 2y + 4z + 11 = 0

b ax+ by + cz + d1 = 0 and ax+ by + cz + d2 = 0.

17 Show that the line x = 2 + t, y = ¡1 + 2t, z = ¡3t is parallel to the plane

11x¡ 4y + z = 0, and find its distance from the plane.

18 Find the equations of the two planes which are parallel to 2x¡ y + 2z = 5 and

2 units from it.

THE ANGLE BETWEEN A LINE AND A PLANE

Suppose a line has direction vector l and

a plane has normal vector n and allow n

to intersect the line making an angle of µo

o

with it.

The required angle is Á and

sinÁ = cos µ =jn ² ljjnj jlj

) Á = sin¡1µ jn ² lj

jnj jlj¶

P( , , )x y zz z z

N

d

Q

Ax By Cz D� � �

n

line

l��

��

��

plane

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19 Find the acute angle between:

a the plane x¡ y + z = 5 and the linex¡ 1

4=y + 1

3= z + 2

b the plane 2x¡ y + z = 8 and the line x = t+ 1, y = ¡1 + 3t, z = t

c the plane 3x+ 4y ¡ z = ¡4 and the line x¡ 4 = 3 ¡ y = 2(z + 1).

cos µ =jn1 ² n2jjn1j jn2j

So, if two planes have normal vectors n1 and n2 and µ is the

acute angle between them then µ = cos¡1µ jn1 ² n2j

jn1j jn2j¶

.

Find the acute angle between the plane x+ 2y ¡ z = 8 and the line with equations

x = t, y = 1 ¡ t, z = 3 + 2t.

n = [1, 2, ¡1] and l = [1, ¡1, 2]

Á = sin¡1µ j1 ¡ 2 ¡ 2jp

1 + 4 + 1p

1 + 1 + 4

¶

= sin¡1µ

3p6p

6

¶= sin¡1

¡12

¢= 30o

�

�

�

nl

Example 38

THE ANGLE BETWEEN TWO PLANES

is the cosine of the acute angle between two planes.

nx

nz

plane 2

plane 1

view here

nxnz

�

�

�

�� plane 1

plane 2

Find the acute angle between the planes with equations x+ y ¡ z = 8and 2x¡ y + 3z = ¡1.

x+ y ¡ z = 8 has normal vector

n1 = [1, 1, ¡1]

and 2x¡ y + 3z = ¡1 has normal

vector n2 = [2, ¡1, 3]

Example 39

nznx

PxPz

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If µ is the acute angle between the planes then

µ = cos¡1µ jn1 ² n2j

jn1j jn2j¶

= cos¡1µ j2 + ¡1 + ¡3jp

1 + 1 + 1p

4 + 1 + 9

¶

= cos¡1j¡2jp3p

14= cos¡1

µ2p42

¶+ 72:02o

20 Find the acute angle between the planes with equations:

a 2x¡ y + z = 3 and

x+ 3y + 2z = 8b x¡ y + 3z = 2 and

3x+ y ¡ z = ¡5c 3x¡ y + z = ¡11 and

2x+ 4y ¡ z = 2:

² Two planes in space could be

(1) intersecting (2) parallel (3) coincident

² Three planes in space could be

(1) all coincident (2) two coincident and (3) two coincident and

one other one parallel

(4) two parallel and one (5) all 3 parallel (6) all meet at the one

other point

(7) all 3 meet in a (8)

the line of intersection

common line

of any 2 is parallel to

the third plane.

THE INTERSECTION OF TWO

OR MORE PLANES

K

P Pz= x

P Pz= x

PcP P Pz= x= c P Pz= xPc

Peanuts Software (Winplot) displays these cases and should be visited.

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The following Exercise should be completed after the work on linear row operations of

Mathematics Studies has been considered.

This work is an extension of that material with geometric interpretation if planes are under

consideration.

1 a How many solutions are possible when solving simultaneously

a1x+ b1y + c1z = d1

a2x+ b2y + c2z = d2?

b Under what conditions will the planes in a be:

i parallel ii coincident?

c Solve the following using elementary row operations and interpret each system of

equations geometrically:

i x¡ 3y + 2z = 8

3x¡ 9y + 2z = 4

ii 2x+ y + z = 5

x¡ y + z = 3

iii x+ 2y + z = 5

2x+ 4y + 2z = 16

iv x+ 2y ¡ 3z = 6

3x+ 6y ¡ 9z = 18

2 Discuss the possible solution of the following systems where k is a real number, inter-

preting geometrically:

a x+ 2y ¡ z = 62x+ 4y + kz = 12

b x¡ y + 3z = 8

2x¡ 2y + 6z = k

3 For the eight possible geometric solutions of three planes in space, comment on the

possible solutions in each case.

For example, has infinitely many solutions where x, y and z

are in terms of two parameters, s and t, say.

4 Solve the following systems using elementary row operations and in each case state the

geometric meaning of your solution:

a x+ y ¡ z = ¡5

x¡ y + 2z = 11

4x+ y ¡ 5z = ¡18

b x¡ y + 2z = 1

2x+ y ¡ z = 8

5x¡ 2y + 5z = 11

c x+ 2y ¡ z = 8

2x¡ y ¡ z = 5

3x¡ 4y ¡ z = 2

d x¡ y + z = 8

2x¡ 2y + 2z = 11

x+ 3y ¡ z = ¡2

e x+ y ¡ 2z = 1

x¡ y + z = 4

3x+ 3y ¡ 6z = 3

f x¡ y ¡ z = 5

x+ y + z = 1

5x¡ y + 2z = 17

5 Solve the system of equations x¡ y + 3z = 1

2x¡ 3y ¡ z = 3

3x¡ 5y ¡ 5z = k where k takes all real values.

State the geometrical meaning of the solution in each case assuming that the equations

represent planes in space.

EXERCISE 4K

P P Pz= x= c

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REVIEW SET 4A


6 Find all values of m for which x+ 2y +mz = ¡1

2x+ y ¡ z = 3

mx¡ 2y + z = 1 has a unique solution.

In the cases where there is no unique solution, solve the system. Give geometrical

meaning to all possible solutions. Illustrate each case.

1 If a = [2, ¡3, 1], b = [¡1, 2, 3], find:

a 2a ¡ 3b b x if a ¡ 3x = b c the projection vector of a on b.

2 Show that A(1, 0, 4), B(3, 1, 12), C(¡1, 2, 2) and D(¡2, 0, ¡5) are coplanar. Find:

a the equation of the plane

b the coordinates of the nearest point on the plane to E(3, 3, 2).

3 A is (3, 2, ¡1) and B(¡1, 2, 4).

a Write down the vector equation of the line through A and B.

b Find the equation of the plane through B with normal AB.

c Find two points on the line AB which are 2p

41 units from A.

4 P1 is the plane 2x¡ y ¡ 2z = 9 and P2 is the plane x+ y + 2z = 1.

L is the line with parametric equations x = t, y = 2t¡ 1, z = 3 ¡ t.Find the acute angle:

a that L makes with P1 b between P1 and P2.

5 If jaj = 3, jbj =p

7 and a £ b = i + 2j ¡ 3k find:

a a ² b

b the area of triangle OAB given that¡!OA = a and

¡!OB = b

c the volume of tetrahedron OABC if C is the point (1, ¡1, 2).

6

Determine the measure of angle QDM

given that M is the midpoint of PS of the

rectangular prism.

7 Show that the length of ON is given by ON =ja ² bj

jbj .a

bNO

A

B

D

S

C

P

QR

10 cm

4 cm

M

7 cm

REVIEWL

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REVIEW SET 4B

REVIEW SET 4C


1 For A(3, ¡1, 1) and B(0, 2, ¡1), find the:

a vector equation of the line passing through A and B

b the coordinates of P which divides BA in the ratio 2 : 5.

2 If a = 3i ¡ j + 2k and b = i ¡ 2k find:

a 3a ¡ 2b b ja ¡ bj c the length of the projection vector of b on a.

3 If k is a scalar and a is any 3-dimensional vector, prove that jkaj = jkj jaj :4 P(¡1, 2, 3) and Q(4, 0, ¡1) are two points in space. Find:

a¡!PQ

b the angle that¡!PQ makes with the X-axis

c the coordinates of R if R divides QP in the ratio 2 : 1.

5 For C(¡3, 2, ¡1) and D(0, 1, ¡4) find the coordinates of the point(s) where the line

passing through C and D meets the plane with equation 2x¡ y + z = 3.

6 Given the linesx¡ 8

3=y + 9

¡16=z ¡ 10

7and x = 15+3t, y = 29+8t, z = 5¡5t:

a show that they are skew

b find the acute angle between them

c find the shortest distance between them.

7 a How far is X(¡1, 1, 3) from the plane x¡ 2y ¡ 2z = 8?

b Find the coordinates of the foot of the perpendicular from Q(¡1, 2, 3) to the line

2 ¡ x = y ¡ 3 = ¡12z.

1 P(2, 0, 1), Q(3, 4, ¡2) and R(¡1, 3, 2) are three points in space. Find:

a¡!PQ and j ¡!

PQ jb the parametric equations of the line through P and Q

c the equation of the plane through P, Q and R.

2 Given the point A(¡1, 3, 2), the plane 2x ¡ y + 2z = 8 and the line defined by

x = 7 ¡ 2t, y = ¡6 + t, z = 1 + 5t, find:

a the distance from A to the plane

b the coordinates of the point on the plane nearest to A

c the shortest distance from A to the line.

3 a Find the equation of the plane through A(¡1, 0, 2), B(0, ¡1, 1) and

C(1, 2, ¡1).

b Find the equation of the line, in parametric form, which passes through the origin

and is normal to the plane in a.

c Find the point where the line of b intersects the plane of a.

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REVIEW SET 4D


4 The triangle with vertices P(¡1, 2, 1), Q(0, 1, 4) and R(a, ¡1, ¡2) has an area ofp118 units2. Find a.

5 M is (¡1, 3, 4) and N(2, 0, 1). Find:

a the coordinates of two points on MN such that their distance from N isp

3 units

b a vector in the direction of¡!MN with length 2 units.

6 Solve the systemx¡ y + z = 5

2x+ y ¡ z = ¡1

7x+ 2y + kz = ¡k for any real number k,

using elementary row operations. Give geometric interpretations of your results.

1 Show that A(1, ¡3, 2), B(2, 0, 1) and C(¡1, ¡9, 4) are collinear and hence find the

ratio in which C divides BA.

2 For A(¡1, 2, 3), B(2, 0, ¡1) and C(¡3, 2, ¡4) find:

a the equation of the plane defined by A, B and C

b the measure of angle CAB

c r, given that D(r, 1, ¡r) is a point such that angle BDC is a right angle.

3 a Find where the line through L(1, 0, 1) and M(¡1, 2, ¡1) meets the plane with

equation x¡ 2y ¡ 3z = 14.

b Find the shortest distance from L to the plane.

4 Find the coordinates of the point which divides the line segment joining A(¡2, 3, 5) to

B(3, ¡1, 1) externally in the ratio 2 : 5.

5 Given A(¡1, 2, 3), B(1, 0, ¡1) and C(1, 3, 0), find:

a the normal vector to the plane containing A, B and C

b D, the fourth vertex of parallelogram ACBD

c the coordinates of the foot of the perpendicular from C to the line AB.

6 Show that the line x¡1 =y + 2

2=z ¡ 3

4is parallel to the plane 6x+7y¡5z = 8

and find the distance between them.

7 Use elementary row operations to solve the system:

x+ 3y ¡ z = 0

3x+ 5y ¡ z = 0

x¡ 5y + [2 ¡m]z = 9 ¡m2 for any real number m.

Give geometric interpretations of your results.

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REVIEW SET 4E


1 Find the volume of tetrahedron ABCD given A(3, 1, 2), B(¡1, 2, 1), C(¡2, 0, 3) and

D(4, 3, ¡1).

2 Consider the lines with equationsx¡ 3

2= y ¡ 4 =

z + 1

¡2and x = ¡1 + 3t,

y = 2 + 2t, z = 3 ¡ t.a Are the lines parallel, intersecting or skew?

b Determine the cosine of the acute angle between the lines.

3 S divides AB externally in the ratio 3 : 5 and T divides CS internally in the ratio 1 : 2.

If A, B and C have position vectors a, b and c respectively, deduce the position vector

of T.

4 For A(2, ¡1, 3) and B(0, 1, ¡1), find:

a the vector equation of the line through A and B, and hence

b the coordinates of C on AB which is 2 units from A.

5 Find the equation of the plane through A(¡1, 2, 3), B(1, 0, ¡1) and C(0, ¡1, 5).

If X is (3, 2, 4), find the angle that AX makes with this plane.

6 a Find all vectors of length 3 units which are normal to the plane x¡ y + z = 6.

b Find a unit vector parallel to i + rj + 3k and perpendicular to 2i ¡ j + 2k.

7 The distance from A(¡1, 2, 3) to the plane with equation 2x¡ y + 2z = k is

3 units. Find k.

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4. 3-Dimensional Vector Geometry

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Transcript of 4. 3-Dimensional Vector Geometry