1

CHAPTER 6 : VECTORS

6.1 Lines in Space

6.1.1 Angle between Two Lines

6.1.2 Intersection of Two lines

6.1.3 Shortest Distance from a Point to a Line

6.2 Planes in Space

6.2.1 Intersection of Two Planes

6.2.2 Angle between Two Planes

6.2.3 Angle between a Line and a Plane

6.2.4 Shortest Distance from a Point to a Plane

Review:

Basic Concepts

Vectors in Space

The Dot Products

The Cross Products

2

Basic Concepts

What is scalar?

a quantity that has only magnitude

What is vector?

a quantity that has magnitude and

direction

A vector can be represented by a directed line

segment where

length of the line segment

- the magnitude of the vector

direction of the line segment

- the direction of the vector

3

2 2,Q x y

Terminal Point

1 1,P x y

Initial Point

PQ

A vector can be written as PQ , or a . The

order of the letters is important. PQ

means the vector is from P to Q or the

position vector Q relative to P, QP means

vector is from Q to P or the position vector

P relative to Q.

If P 11, yx is the initial point and

Q 22 , yx is the terminal point of a directed

line segment, PQ then component form

of vector v that represents PQ is

4

12

12

12121 ,,yy

xxyyxxvv x

and the magnitude or the length of v is

2 2

2 1 2 1

2 2

1 2

x x y y

c c

v

1 1,P x y

2 2,Q x y

1 1 1 10, 0 ,OP x y x y

2 2 2 20, 0 ,OQ x y x y

-Note-

Any vector that has magnitude of 1 unit = unit

vector.

5

Example:

Find the component form and length of the

vector v that has initial point 7,3 and

terminal point 5,2 .

Solution:

2 3,5 7 5,12v

2 2

5 12 25 144 13v

Example:

Given 5,2v and 4,3w , find each of

the following vectors:

a) v2

1

b) vw c) wv 2

Answer:

a) 1,5 2 b) 5, 1 c) 4,13

6

Theorem:

If a is a non-null vector and if a is a unit

vector having the same direction as a, then

ˆ

aa =

a

-Note-

To verify that magnitude is 1, a = 1

Example:

Find a unit vector in the direction of

5,2v and verify that it has length 1.

Solution:

2 2

2,5 12,5

292 5

v

2 2

2 29 5 29 1 1v

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Standard Unit Vectors

Three standard unit vectors are: i, j dan k

z

y

x

1

1

1

i

j

k

Vectors i, j and k can be written in

components form:

i = < 1, 0, 0 >,

j = < 0, 1, 0 > and

k = < 0, 0, 1>

and can interpreted as

a = <x, y, z>

= xi + yj + zk

8

The vector QP with initial point

111 ,, zyxP and terminal point 222 ,, zyxQ

has the standard representation

kji )()()( 121212 zzyyxxQP

or

121212 ,, zzyyxxPQ

Example:

Let u be the vector with initial point (2,-5)

and terminal point (-1,3), and let jiv 2 .

Write each of the following vectors as a linear

combination of i and j.

a) u

b) vuw 32

9

0x

y

v

v

1x

1y

-Note-

If is the angle between v and the positive

x – axis then we can write

cos and y sinx v v ;2 2

1 2v x y .

Example:

The vector v has a length of 3 and makes an

angle of 630

with the positive x-axis.

Write v as a linear combination of the unit

vectors i and j.

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Vectors in Space

Properties of Vectors in Space

Let 321 vvv ,,v and 321 www ,,w be

vectors in 3 dimensional space and k is a

constant.

1. wv if and only if

332211 wvwvwv ,, .

2. The magnitude of v is 2

3

2

2

2

1 vvv v

3. The unit vector in the direction of v is

v

,,

v

v 321 vvv

4. 332211 wvwvwv ,,wv

5. 321 kvkvkvk ,,v

6. Zero vector is denoted as 0 000 ,, .

7. + v w w v

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8. Let 1 2 3, ,u u uu ,

then + + u + v w u v w .

9. + u 0 u

10. +u -u = 0

11. +dc +d u = cu u

12. +c c cu + v u v

13. c d cdu u

14. 1 and 0u = u u = 0

15. c =cu u

Example:

Express the vector PQ if it starts at point

),,( 856P and stops at point ),,( 937Q in

components form.

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Solution:

2 1 2 1 2 1, ,PQ x x y y z z

7 6,3 5,9 8PQ

1, 2,1PQ

Example:

Given that 213 ,,a , 461 ,,b . Find

(a) ba 3 (b) b

(c) a unit vector which is in the direction of b.

(d) find the unit vector which has the same

direction as ba 3 .

Answer:

(a) 0,19,10 (b) 53 (c) 41 23

2, ,6 3

(d) 1 (e) 1

1,6,453

(f)

10,19,10

461

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Parallel Vector

have same slopes

1 2 ; constantsv v

So, there are multiples of each other.

Example:

Vector w has initial point (2,-1,3) and

terminal point (-4,7,5). Which of the

following vectors is parallel to w?

Solution:

4 2,7 1,5 3w

6,8,2w

One example: 2 6,8,2 12,16,4

Another is 1

6,8,2 3,4,12

.

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Example: (Collinear Points)

Determine whether the point 0,1,2,3,2,1 QP

and 6,7,4 R lie on the same line.

Solution:

2 1,1 2,0 3 1,3, 3PQ

22 21 3 3 19PQ

4 2,7 1, 6 0 2,6, 6QR

22 22 6 6 2 19QR

4 1,7 2, 6 3 3,9, 9PR

22 23 9 9 3 19PR

Thus, PR PQ QR . Since one vector is a

multiple of the other, the two vectors are

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parallel and since they share a common point

Q, they must be the same line.

The Dot Product

-Theorem-

If 321 vvv ,,v and 321 www ,,w , then the

scalar product wv is

wv 321 vvv ,, 321 www ,,

332211 wvwvwv

-Note-

The dot product is also called

- the scalar product

- the inner product

The dot product of two vectors is a scalar.

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The Angle between Vectors

Refer to the figure below, let

1 2 3, ,u u OP u u u ,

1 2 3, ,v v OQ v v v

be two vectors and let be the angle between

them, with 0 .

z

y

x

1 2 3, ,Q v v v 1 2 3, ,P u u u

vu

c

Compute the distance, c between points P and

Q in two ways.

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1) Using the Distance formula

2 2 22

1 1 2 2 3 3

2 2 2 2 2 2

1 2 3 1 2 3

1 1 2 2 3 32

c u v u v u v

u u u v v v

u v u v u v

2 2

1 1 2 2 3 32u v u v u v u v ---(1)

2) Using the Law of Cosines

2 22 2 cosc u v u v ---(2)

Equating equation (1) and (2), we get

1 1 2 2 3 3cos

u v u v u v u v

u v u v

Example:

If v = 2i-j+k, w = i+j+2k and the angle

between v and w is 60, find wv .

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Solution:

22 2 2 2 22 1 1 1 1 2 cos 3v w

6 6 cos 3 6 1 2 3

Example:

Given that 3,2,2 u , 1,8,5v and

2,3,4 w , find

(a) vu

(b) wvu

(c) wvu

(d) the angle between u and v

(e) the angle between v and w.

Answer:

(a) -3

(b) 12, 9,6

(c) -23

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(d) 94 24

(e) 87 46

Example:

Let A=(4,1,2), B=(3,4,5) and C=(5,3,1) are

the vertices of a triangle. Find the angle at

vertex A.

Answer:

79 12

-Theorem-

The nature of an angle , between two vectors

u and v.

is an acute angle if and only if 0 vu

is an obtuse angle if and only if 0 vu

= 90 if and only if 0 vu . The

Vectors u and v are orthogonal /

perpendicular.

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Example:

Show that the given vectors are perpendicular

to each other.

(a) i and j

(b) 3i-7j+2k and 10i+4j-k

-Theorem-

(Properties of Dot Product)

If u,v and w are nonzero vectors and k is a

scalar,

1. uvvu

2. wuvuwvu

3. vuvu kk

4. 2

vvv

5. 0 u00u

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The Cross Products

- The cross product (vector product) vu is

a vector perpendicular to u and v.

(illustrated in figure below)

- The direction is determined by the right

hand rule.

u v

u v

If the first two fingers of the right hand

point in the directions of andu v

respectively, then the thumb points in

the direction of u v .

Ex: i j k

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- The length is determined by the lengths of

u and v and the angle between them.

- If we change the order informing the cross

product, then we change the direction.

Ex:

v u u v

-Theorem-

If kjiu 321 uuu and kjiv 321 vvv ,

then,

1 2 3

1 2 3

2 3 1 3 1 2

2 3 1 3 1 2

2 3 3 2 1 3 3 1 1 2 2 1

u u u

v v v

u u u u u u

v v v v v v

u v u v u v u v u v u v

i j k

u v

i j k

i j k

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Properties of Cross Product

(a) 0uu

(b) uvvu

(c) wuvuwvu

(d) vuvuvu kkk

(e) u // v if and only if 0u v

(f) u 0u00

Example:

1) Given that 4,0,3u and 2,5,1 v ,

find

(a) vu

(b) uv

2) Find two unit vectors that are

perpendicular to the vectors u 2i+2j-3k

and v = i+3j+k.

24

Answer:

1) (a) -20i+10j+15k (b) 20i-10j-15k

2)

111, 5,4

162

(The unit vector in the

opposite direction is also a unit vector

perpendicular to both u and v )

Further geometry interpretation of the cross

product comes from computing its magnitude.

2 2 2 2

2 3 3 2 3 1 1 3 1 2 2 1u v u v u v u v u v u v u v

2 2 2 2 2 2 2

1 2 3 1 2 3

2

1 1 2 2 3 3

2 2 2

2 2 2 2 2

2 2 2

2 2 2

cos

1 cos

sin

u v u u u v v v

u v u v u v

u v u v

u v u v

u v

u v

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with is the angle between u and v .

Therefore, sin .u v u v

A D

B C

u

v

sinu

From the figure above, we can see that the

magnitude of the cross product is the area of

the parallelogram of which arrows

representing the two vectors are adjacent

sides.

Area of a parallelogram vuvu sin

Area of triangle = vu2

1

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Example:

(a) Find an area of a parallelogram that is

formed from vectors u = i + j - 3k and

v = -6j + 5k.

(b) Find an area of a triangle that is formed

from vectors u = i + j - 3k and

v = -6j + 5k.

Answer:

(a) 230 (b) 230

2

Scalar Triple Product

-Theorem-

If 111 ,, zyxa , 222 ,, zyxb and

333 ,, zyxc ,

then

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333

222

111

zyx

zyx

zyx

cba

Properties of The Scalar Triple Product

1) cbacba

2) a b c a c b b c a

3) a c b a b c

4) 0 a a b

5) a d b c a b c d b c

Example:

If a = 3i + 4j - k, b = -6j+5k and c = i+j-k,

evaluate

(a) cba (b) cba

(c) bac (d) cab

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6.1 Lines in Space

In this section we use vectors to study lines in

three-dimensional space.

HOW LINES CAN BE DEFINED USING VECTORS?

The most convenient way to describe a line in

space is to give a point on it and a nonzero

vector parallel to it.

Suppose L is a straight line that passes

through ),,( 000 zyxP and is parallel to the

vector kjiv cba .

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Thus, a point ),,( zyxQ also lies on the line if

vPQ t .

Let,

0r OP and r OQ ,

Then

0r rPQ .

vrr t 0

vrr t 0

cbatzyxzyx ,,,,,, 000

-Theorem-

(Parametric Equations for a Line)

The line through the point ),,( 000 zyxP and

parallel to the nonzero vector cba ,,A has

the parametric equations,

30

atxx 0 , btyy 0 , ctzz 0 .

If we let 000 ,, zyx0R denote the position

vector of ),,( 000 zyxP and zyx ,,R the

position vector of the arbitrary point Q(x,y,z)

on the line, then we write equation (1) in the

vector form,

ARR t 0 .

Example:

Give the parametric equations for the line

through the point (6,4,3) and parallel to the

vector 7,0,2 .

-Theorem-

(Symmetric Equations for a line)

31

The line through the point ),,( 000 zyxP and

parallel to the nonzero vector cba ,,A has

the symmetrical equations,

c

zz

b

yy

a

xx 000

.

Example:

Given that the symmetrical equations of a line

in space is 2

4

4

3

3

12

zyx.

Find,

(a) a point on the line.

(b) a vector that is parallel to the line.

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6.1.1 Angle between Two Lines

Consider two straight lines

c

zz

b

yy

a

xxl 1111 :

and

f

zz

e

yy

d

xxl 2222 :

The line 1l parallel to the vector kjiu cba

and the line 2l parallel to the vector

kjiv fed . Since the lines 1l and 2l are

parallel to the vectors u and v respectively,

then the angle, between the two lines is

given by

vu

vu cos

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Example:

Find an acute angle between line

1l = i + 2j + t (2i – j + 2k)

and line

2l = 2i – j + k + s (3i – 6j + 2k).

34

6.1.2 Intersection of Two lines

In three dimensional coordinates (space),

two line can be

in one of the

three cases as

shown below,

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Let 1l and 2l are given by:

)2(:

(1)and:

2222

1111

f

zz

e

yy

d

xxl

c

zz

b

yy

a

xxl

From (1), we have cba ,,1v

From (2), we have fed ,,2v

Two lines are parallel if we can write

21 vv

The parametric equations of 1l and 2l are:

tczz

tbyy

taxxl

1

1

11 :

)3(

:

2

2

22

fszz

esyy

dsxxl

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Two lines are intersect if there exist unique

values of t and s such that:

fsztcz

esytby

dsxtax

21

21

21

Substitute the value of t and s in (3) to get x, y

and z. The point of intersection = (x, y, z)

Two lines are skewed if they are neither

parallel nor intersect.

Example:

Determine whether 1l and 2l are parallel,

intersect or skewed.

a) tztytxl 74,41,33:1

szsysxl 73,45,32:2

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b) zyx

l

4

2

1

1:1

3

23

1

4:2

zy

xl

Solutions:

a) for 1l :

point on the line, P = (3, 1, - 4)

vector that parallel to line, 7,4,31v

for 2l :

point on the line, Q = (2, 5, 3)

vector that parallel to line, 7,4,32v

?21 vv

1where21 vv

Therefore, lines 1l and 2l are parallel.

b) Symmetrical equations of 1l and 2l can be

rewrite as:

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1

0

4

2

1

1:1

zyxl

3

)2(

1

3

1

4:2

zyxl

Therefore:

for 1l : P = (1, 2, 0) , 1,4,11v

for 2l : Q = (4, 3, -2) , 3,1,12v

?21 vv

21 vv not parallel.

In parametric eq’s:

1

2

: 1 , 2 4 ,

: 4 , 3 , 2 3

l x t y t z t

l x s y s z s

1 4 (1)

2 4 3 (2)

2 3 (3)

t s

t s

t s

39

Solve the simultaneous equations (1), (2), and

(3) to get t and s.

5 7and

4 4s t

The value of t and s must satisfy (1), (2) and

(3). Clearly they are not satisfying (2) i.e

7 5 1 172 3 ?

4 4 4 4

Therefore, lines 1l and 2l are not intersect.

This implies the lines are skewed!

Example:

Let L1 and L2 be the lines

tztytxL 51,45,41:1

tztytxL 5,34,82:2

(a) Are the lines parallel?

(b) Do the lines intersect?

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6.1.3 Shortest Distance from a Point to a Line

Q

sinQP

P

V

Distance from a point Q to a line that passes

through point P parallel to vector v is equal to

the length of the component of PQ

perpendicular to the line.

sinQPd

where is the angle between v and vector

QP

.

41

Since

sinQPQP

vv ,

so we have the shortest distance of Q from L

as

v

v QPd

Example:

Find the distance from the point (0, 0, 0) to

the line,

5

3

4

5

3

5

zyx.

Example:

Find the distance from the point (2, 1, 3) to

the line,

3,61,22 ztytx

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Example:

Find the shortest path from the point Q(2, 0, -

2) to the line

2

1

1

1

3

2:

zyxl

6.2 Planes in Space

Suppose that is a plane. Point ),,( 000 zyxP

and ),,( zyxQ lie on it. If kji cbaN is a

non-null vector perpendicular (ortoghonal) to

, then N is perpendicular to PQ .

P 0 0 0( , , )x y z Q ( , , )x y z

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Thus,

0NPQ

0,,,, 000 cbazzyyxx

0)()()( 000 zzcyybxxa

-Conclusion-

The equation of a plane can be determined if a

point on the plane and a vector orthogonal to

the plane are known.

-Theorem-

(Equation of a Plane)

The plane through the point ),,( 000 zyxP and

with the nonzero normal vector cba ,,N

has the equation

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Point-normal form:

0000 zzcyybxxa

Standard form:

ax by cz d with 0 0 0d ax by cz

Example:

1. Give an equation for the plane through the

point (2, 3, 4) and perpendicular to the

vector 4,5,6 .

2. Give an equation for the plane through the

point (4, 5, 1) and parallel to the vectors

A= 1,0,2 and B 4,1,0 .

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3. Give parametric equations for the line

through the point (5, -3, 2) and

perpendicular to the plane 5726 zyx .

6.2.1 Intersection of Two Planes

Intersection of two planes is a line. (L)

To obtain the equation of the intersecting line,

we need

a point on the line L which is given by

solving the equations of the two planes.

a vector parallel to the line L which is

46

21 NN

If cbaNN ,,21 , then the equation of

the line L in symmetrical form is

c

zz

b

yy

a

xx 000

Example:

Find the equation of the line passing through

P(2,3,1) and parallel to the line of intersection

of the planes x + 2y - 3z = 4 and x - 2y + z = 0.

Answer:

2 3 1

4 4 4

x y z

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6.2.2 Angle between Two Planes

-Properties of Two Planes-

An angle between the crossing planes is an

angle between their normal vectors.

1 2

1 2

cosN N

N N

Two planes are parallel if and only if their

normal vectors are parallel,

1 2N N .

Two planes are orthogonal if and only if

1 2 0N N .

Example:

Find the angle between plane 043 yx and

plane 522 zyx .

48

6.2.3 Angle between a Line and a Plane

Let be the angle between the normal vector

N to a plane and the line L. Then we have

Nv

Nv cos

where v is vector parallel to L. Furthermore, if

the angle between the line L and the plane ,

then

2

2

49

cos2

sinsin

Nv

Nv sin

Example:

Find the angle between the plane

523 zyx and the line

3

3

1

2

2

3

zyx.

50

6.2.4 Shortest Distance from a Point to a

Plane

(a) From a Point to a Plane

-Theorem-

The distance D between a point ),,( 111 zyxQ

and the plane dczbyax is

222

111

N

N

cba

dczbyaxPQD

Where ),,( 000 zyxP is any point on the plane.

),,( 111 zyxQ

),,( 000 zyxP

D

N

D

51

Example:

Find the distance D between the point

(1, -4, -3) and the plane 1632 zyx .

Example:

1) Show that the line

1

1

23

1

zyx

is parallel to the plane 123 zyx .

2) Find the distance from the line to the plane

in part (a).

(b) Between two parallel planes

The distance between two parallel planes

1dczbyax and 2dczbyax is given by

222

21

cba

ddD

52

Example:

Find the distance between two parallel planes

322 zyx and 7442 zyx .

-Note-

Both formulas can also be used to compute

the distance between 2 skewed lines.

(c) Between two skewed lines

N u v

v

2L

u

P

Q

d

1L

53

Assume L1 and L2 are skew lines in space

containing the points P and Q and are parallel

to vectors u and v respectively.

Then the shortest distance between L1 and L2

is the perpendicular distance between the two

lines and its direction is given by a vector

normal to both lines.

So, the distance between the two lines is

absolute value of the scalar projection of PQ

on the normal vector.

cosd

PQ

u v PQN PQ

N u v

54

Example:

Find the shortest distance from P(1, -1, 2) to

the plane 3x – 7y + z = 5.

Example:

Find the shortest distance between the skewed

lines.

l1 : x = 1+2t, y = -1+ t , z = 2 + 4t

l2 : x = -2+4s, y = -3s , z = -1+s

Example:

Find the distance between the lines

)(32:1 kitkjiL

tztyxL 3,21,0:2

55

Example:

Find the distance between the lines L1 through

the points A(1, 0, -1 ) and B(-1, 1, 0) and the

line L2 through the points C(3, 1, -1) and

D(4, 5, -2) .