Vectors

ba

a b

( )i a b b a Addition is commutative

( )ii a b a b

( ) ( ) ( ) Addition is associativeiii a b c a b c

( ) The negative of a vector has the same magnitude and direction

as but the opposite sense.

iv u

u( ) Subtraction of vectors can be defined as ( )v a b a b

(vi) If vector u is multiplied by a scalar k, then the product ku is a vector in the same direction as u but k times the magnitude.

If 1, then the magnitude is increasedk

If 1, then the magnitude is decreasedk

If 0, then has the same sense as k ku u

If 0, then has the opposite sense to k ku u

u

2u

( )k l u ku lu

( )k u v ku kv

Vector Directions

Hold your right hand out in front of you as if to shake hands

direction - Fingersxdirection - From Palmy

direction - From Thumbz

1 2 3Given ( , , )A a a a

1

2

3

then This is the position vector of

a

a a A

a

If and have position vectors and b respectively, thenA B a AB b a ��������������

If divides in the ration : , thenP AB m n��������������

��������������

��������������AP m

nPB

mb naP

m n

12 2 2

2 1 2 3

3

If , then (magnitude)

u

u u u u u u

u

A unit vector is a vector of magnitude 1.

1au a

a

Unit vectors in the x direction, y direction and z direction are denoted i, j and k.

1

0

0

i

0

1

0

j

0

0

1

k

Any vector can be expressed in terms of i, j and k for example;

2

3 2 3 5

5

i j k

P

P’

The projection of a point P on a plane is the point P’, at the foot of the perpendicular to the plane passing through P.

P

P’

Q’

Q

The projection of a line PQ on a plane is the line P’Q’, which joins the projections of P ands Q on the plane

The angle between a line and a plane , is the angle between the line and its projection on the plane. (degrees or radians)

The scalar product or dot product of the vectors and is defined by;a b

. cosa b a b

Where is the angle between the positive directions of and .a b

Note that cos is the magnitude of the projection of on b b a

b

cosb

. is scalara b

2. cos0a a a a a

. .a b b a

.( ) . .a b c a b a c

. 0 and 0, 0 and are perpendiculara b a b a b

1 1

2 2

3 3

and

a b

a a b b

a b

1 1 2 2 3 3.a b a b a b a b

1 1 2 2 3 3cosa b a b a b

a b

Direction Ratios

1

2

3

Given these components uniquely determine the magnitude and direction of .

a

a a a

a

Any scalar multiple of a has the same direction but not necessarily the same sense as a.

It follows that the ratio a1:a2:a3 can be used to describe uniquely the direction ratio of the vector. If two vectors have equal direction ratios then they are parallel.

Direction Cosines

2 . cos cos u u j

3 . cos cos u u k

1

2

3

If , and are the angles the vector a makes with , and

respectively and is a unit vector in the direction of a

then we know;

ox oy oz

u

u u

u

1 . cos cos u u i cos

giving cos

cos

u

These components are called the direction cosines of a

2 2 2Note: cos cos cos 1u

2 2 2cos cos cos 1

This means that , and are not independent.

Find (a) the direction ratio(b) the direction cosines

of the vector a = 6i + 8j +24k

( ) Direction ratio = 6 :8 : 24

3: 4 :12

a

2 2 2( ) 6 8 24 26b a 6 8 2426 26 26au i j k

6cos

268

cos2624

cos26

Page 44 Exercise 1 Questions 6 – 8Page 46 Exercise 2 Questions 1, 2, 9

TJ Exercise 1

The Vector Product

Two vectors can be combined by the binary operation, the scalar product, to produce a number (or scalar).

It is also possible to define an operation where two vectors, a and b, multiply to produce a vector. This is referred to as the vector product, or cross product, as denoted a × b.

Any two non parallel vectors, a and b, define a plane. Let n be a unit vector perpendicular to this plane (a normal) so that a, b and n form a right handed system.

Any two non parallel vectors, a and b, define a plane. Let n be a unit vector perpendicular to this plane (a normal) so that a, b and n form a right handed system.

a b

u

a along the fingers

b from the palm

n along the thumb.

The angle between a and b need NOT be 900

The vector product is defined as sina b n a b

where is the angle between the positive directions of a and b.

If either a = 0, or b = 0, then n is not defined and a × b is defined as 0.

The following properties follow directly from the definition.

is a vector with the same sense and direction as a b n

= a b sin is the area of the parallelogram defined by and a b a b

=n a a sin 0a a

=n a a sin 0 : parallel vectors have a vector product of zeroa ka k

0, and 0 and a 0 and b are parallel vectorsa b b a

= ( ) : The vector product is not commutative.a b b a

= i j sin90oi j k k

i j k

Similarly:

k i j

i

jk

= i j sin90oi j k k

j k i

i k j

k j i

j i k

The vector product is distributive over addition.

a b c a b a c and a b c a c b c

b

a3

3

600

03 3sin 60a b

92

03 3sin 60b a

92

Same magnitude but different sense. (see page 48)

Component form of the Vector Product

1 1

2 2

3 3

Suppose and then

a b

a a b b

a b

1 2 3 1 2 3( ) ( )a b a i a j a k b i b j b k Using the distributive law:

1 1 1 2 1 3

2 1 2 2 2 3

3 1 3 2 3 3

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( )

a b i i a b i j a b i k

a b j i a b j j a b j k

a b k i a b k j a b k k

1 2 1 3 2 1 2 3 3 1 3 20 ( ) ( ) ( ) ( ) ( ) ( ) 0a b k a b j a b k o a b i a b j a b i

2 3 3 2 3 1 1 3 1 2 2 1( ) ( ) ( )a b a b i a b a b j a b a b k

2 3 3 2 3 1 1 3 1 2 2 1( ) ( ) ( )a b a b a b i a b a b j a b a b k

This is normally written as:

2 3 3 2 1 3 3 1 1 2 2 1( ) ( ) ( )a b a b a b i a b a b j a b a b k

We do this so as to use the following shortcut notation

1 2 3

1 2 3

i j k

a b a a a

b b b

Calculate the area of a triangle whose vertices areA(2, -1, 3), B(5, -1, 2), C(2, 3, 4)

A

B

C

1Area = sin

2AB AC ����������������������������

12AB AC ����������������������������

3 01

0 42

1 1

1(0 4) (3 0) (12 0)

2i j k

14 3 12

2i j k

21 1316 9 144

2 2units

3 0 1

0 4 1

i j k

a b

Calculate the shortest distance from the point P(1, 2, 3) to the straight line passing through the points A(1, 3, -2) and B(2, 2, -1)

AB��������������

AP��������������

A

P

BP’

' sinPP AP ��������������

sinAB AP

AB

����������������������������

��������������AB AP

AB

����������������������������

��������������

1 0

1 , 1

1 5

AB b a AP p a

����������������������������1 1 1

0 1 5

i j k

a b

( 5 1) (5 0) ( 1 0)

3

i j k

4 5 1

3

i j k 16 25 1

3

14 units

The Scalar Triple Product

A parallelopiped is a region of 3D space bounded by three pairs of parallel lines.

Its volume can be calculated by multiplying the area of one plane in a pair and the perpendicular distance between the planes.

V = Ah

h

c

a

b

Suppose that the parallelopiped is defined by three vectors a, b and c which form a right handed system!Area of Base = b c

Distance between the planes = cosh a

a cosV b c

Now any of the 3 pairs of parallel planes could have been used.

.( )a b c

a cosV b c

Now any of the 3 pairs of parallel planes could have been used.

.( ) (Dot Product)a b c

.( ) .( ) .( )V a b c b c a c a b

NOTE:

.( ) is a number NOT a vectora b c

.( ) ( ). In this expression the dot and cross are interchangeable.a b c a b c

.( ) is called the scalar triple product denoted [ , , ]a b c a b c

If , or 0, then .( ) is defined as 0.a b c a b c

Component form of the scalar product

1 1 1

2 2 2

3 3 3

a b c

a a b b c c

a b c

2 3 3 2

3 1 1 3

1 2 2 1

b c b c

b c b c b c

b c b c

1 2 3 3 2

2 3 1 1 3

3 1 2 2 1

.( )

a b c b c

a b c a b c b c

a b c b c

1 2 3 3 2 2 3 1 1 3 3 1 2 2 1( ) ( ) ( )a b c b c a b c b c a b c b c

1 2 3

1 2 3

1 2 3

.( )

a a a

a b c b b b

c c c

1 2 3 3 2 2 1 3 3 1 3 1 2 2 1( ) ( ) ( )a b c b c a b c b c a b c b c

Page 52 Exercise 4

TJ Exercise 2

Equation of a line in 3 dimensionsA line in space is completely determined when we know the direction in which it runs and we know a point on the line. Its direction is unambiguously described by stating a vector parallel to the line. Such a vector is known as a position vector.

Consider the line L which passes through the point A(x1, y1, z1) with direction vector u = ai + bj + ck. Let P(x, y, z) be any point on the line L.

L

O

u

A

P

a

p

for some scalar t.AP tu��������������

p a tu p a tu 1

In Component Form this becomes

Consider the line L which passes through the point A(x1, y1, z1) with direction vector u = ai + bj + ck. Let P(x, y, z) be any point on the line L.

L

O

u

A

P

a

p

for some scalar t.AP tu��������������

p a tu p a tu 1

In Component Form this becomes

1 1

1 1

1 1

x x a x at

y y t b y bt

z z c z ct

Giving

1 1 1, ,x x at y y bt z z ct 2

Also: 1 1 1x x y y z zt

a b c

3

We have three form of the equation of a line in space.

1. p a tu

1 1 12. , ,x x at y y bt z z ct

1 1 13.x x y y z z

ta b c

Equation 1 is known as the vector equation.

The system of equations (2) is the parametric form. (t being the parameter).

The system of equations (3) is the symmetric form. (also referred to as the standard or canonical form)

•‘t’ is often omitted in the symmetric form but has to be inserted to convert to other forms.•If any of the components of the direction vector is zero then some parts of the symmetric form will be undefined in which case the parametric form is better.•Each point on L is uniquely associated with a value of the parameter t.•The equation of a particular line is NOT unique.

1 2 4 2 1 and

2 4 6 1 2 3x y z x y z

Both equations represent a line that passes through (3, 2, 2) and is parallel to i + 2j + 3k.

(a) Write down the symmetric form of the equations of the line which passes through (1, -2, 8) and is parallel to 3i + 5j + 11k.

(b) Does the point (-2, -7, -3) lie on the line?

(a) Direct substitution gives: 1 2 8

3 5 11x y z

(b) Substituting (-2, -7, -3) we get:

2 11

3

7 21

5

3 81

11

Since the results are consistent (-1), the point lies on the line.

Find the equations of the line passing through A(2, 1, 3) and B(3, 4, 5)

The line MUST be parallel to AB��������������

3 2 1

4 1 3

5 3 2

AB b a

��������������

Since the line passes through A we get:2 1 3

1 3 2x y z

t

Page 66 Exercise 9A Questions 1(a), (b), 2(a), 3(a), (c), (e) and 5Page 67 Exercise 9B Question 2

TJ Exercise 3

3 4 5NOTE: we could have chosen point B.

1 3 2x y z

t

The Equation of a Plane

A plane in space can be uniquely identified if:

•3 points on the plane are known

•2 Lines on the plane are known

•1 point on the plane and a normal to the plane are known

Suppose that relative to a right handed set of axes, we have a

plane called . Let ( , , ) be a typical point on the plane and

let be a normal to the plane passing through the plane at

P x y z

a

a b

c

Q.

a

Q

P

Since is perpendicular to ,QP a��������������

. 0a QP ��������������

.( ) 0a p q

. . 0a p a q

Since both a and Q are fixed a.q is a constant on the plane.

Let k = a . q then

. 0a p k

.a p k If we know the normal a and any point P we can easily compute k

.

a x

b y k

c z

ax by cz k

This is the equation of the plane

To find the equation of a plane we must be able to reduce given data to(i) The components of a suitable normal vector(ii) A point on the plane

P, Q, R and S are the points (1, 2, 3), (2, 1, -4), (1, 1, 1) and (7, -6, 5) respectively.

(a) Find the equation of the plane perpendicular to PQ which contains the point P

(b) which of the other points lie on the plane?

2 1 1

( ) 1 2 1

4 3 7

a PQ q p

��������������1 1

1 . 2 1 2 21 22

7 3

k

Equation of the plane = 7 22x y z

( ) (1,1,1) does not satisfy the equation

S(7,-6,5) does satisfy the equation so lies on the plane.

b R

Find the equation of the plane which passes through the points A(-2, 1, 2)B(0, 2, 5) ans C(2, -1, 3).

Strategy: Find a Normal to the plane.

Two vectors that lie on the plane are and .

is a normal vector to the plane.

AB AC

AB AC

����������������������������

����������������������������

2 4

1 2

3 1

AB AC

����������������������������2 1 3

4 2 1

i j k

AB AC

����������������������������1 6 7

(2 12) 10

4 4 8

Using Point A to find k 14 10 16 20k

Required Equation of the plane = 7 10 8 20x y z

Vector Equation of a plane

AC��������������

uAC��������������

t AB��������������

A plane can be defined by any two non zero non parallel vectors which lie upon it.

Let and be such a pair defining the plane . Let R be any point on the plane.

Then and a

AB AC

AB AC

����������������������������

����������������������������nd are coplanar.AR

��������������

AR t AB uAC ������������������������������������������

AB��������������

AR��������������

R

( ) ( )r a t b a u c a

( ) ( )r a t b a u c a (1 )r t u a tb uc

This is known as the vector equation of a plane.

( ) ( ) can be written as r a t b a u c a r a tb uc

Where A is a point on the plane, and b and c are vectors parallel to the plane.

(a) Find the equation of the plane, in vector form, which contains the points A(1, 2, -1), B(-2, 3, 2) and C(4, 5, 2).

(b) Find the point on the plane corresponding to the parameter values t=2, u=3

1 2 4

( ) 2 , 3 , 5

1 2 2

a a b c

1 2 4

(1 ) (1 ) 2 3 5

1 2 2

r t u a tb uc t u t u

1 2 4

2 2 2 3 5

1 2 2

t u t u

t u t u

t u t u

1 3 3

2 3

1 3 3

t u

t u

t u

4

( ) When 2, 3 13 . So the required point is R(4, 13, 14)

14

b t u r

Find the cartesian equation of the plane whose vector equation is

(2 3 ) (1 3 ) (3 7 )r t u i t u j t u k

2 3 3 3 7r i ti ui j t j u j k tk uk

2 3 ( ) (3 3 7 )i j k t i j k u i j k

Expressed in component form

2 1 3

1 1 3

3 1 7

r t u

comparing with r a tb uc

We see that the point (2, 1, 3) lies in the plane and

1 3

1 and -3

1 -7

are vectors lying in the plane.

Hence a vector normal to the plane is

1 3

1 3

1 7

1 1 1

3 3 7

i j k

10

4

6

Using the point (2, 1, 3) to find k,

10 2

4 . 1 34

6 3

10 4 6 34x y z or 5 2 3 17x y z

Page 57 Exercise 6 Questions 1, a, b, 2a, 3, 4a, c, 5a, 9, 10Page 63 Exercise 8 As many as you can manage (A/B)

TJ Exercise 4

Intersection of 2 lines

Two lines in space can be

•Parallel•Intersect at a point•Skew. (not parallel but never intersect)

Strategy for finding the point of Intersection of two lines

•Express the equations of the lines in parametric form using parameters t1 and t2

•Equate the corresponding expressions for x, y and z producing 3 equations in two unknowns•Use two of the equations to find the values of t1 and t2 •Substitute these values into the third equation. If they satisfy then the point of intersection has been found. If not, then the lines do not intersect.

Show that the lines with equations 5 ( 2) and

12 3 5 intersect and find the point of intersection.

5 2 4

x y z

x y z

Using parameters t1 and t2;

1 2

1 2

1 2

5 5 12

2 2 3

4 5

x t x t

y t y t

z t z t

Equating:

1 2

1 2

1 2

5 7

2 1

4 5

t t

t t

t t

1

2

3

2 21 2 3 6 0 2t t 1 3t

Substituting these values into equation 3 shows that they satisfy the equation. Hence the lines intersect.

2, 1, 3x y z (2, 1, -3)

Find the acute angle between the lines with equations

2 1 11 and 3, 4, 8

1 2 1x y z

x t y t z

3 4 8The second equation can be rewritten as

1 1 0x y z

1 -1

This gives the vectors -2 and 1

-1 0

01 2 0 3cos 150

26 2

0Hence the acute angle is 30

Angle between two planes

P

π1

π2 A

B

C

The angle between two planes can be constructed by picking any point B on the line of intersection of the planes and drawing perpendiculars to the line BA and BC on both planes. The angle ABC is the required angle.

Viewing from P along the line of intersection we can see that the angle ABC is the same as the angle between the normal.

Find the angle between the planes with equations, x + 2y + z = 5 and x + y = 0.

1 1

By in spection the normals are: 2 and 1

1 0

a b

. 1 1 0 3cos

26 2

a ba b

030

Page 70 Exercise 11 Questions 1 and 2Page 59 Exercise 7A Questions 1, 2 and 3

TJ Exercise 5 questions 1 and 2

Angle between a line and a plane

Is the compliment of the angle between the line and the normal to the plane

u

a

90-

sin cos(90 )

.a ua u

Use modulus since <90

Strategy•Change the equation of the line to parametric form if necessary.•Substitute x y and z into the equation of the plane.•Solve this subsequent equation to obtain a value of the parameter, t, and hence the coordinates of the point of intersection.

7 11 24Given the line and the plane

3 4 136 4 5 28, find

(a) The point of intersection

(b) The angle the line makes with the plane.

x y z

x y z

( ) 3 7 4 11 13 24a x t y t z t (3 7,4 11,13 24)t t t

Substitute this into the equation of the plane.

6(3 7) 4(4 11) 5(13 24) 28t t t 31 62 0t

2t

point of intersection (1, 3, -2)

( ) (6 4 5 ) (3 4 13 )b a i j k u i j k

( ) (6 4 5 ) (3 4 13 )b a i j k u i j k

.sin cos(90 )

a ua u

6 3 4 4 5 13

36 16 25 9 16 169

31

77 194

0.254

014.7 to 3 significant figures

Page 68 Exercise 10 Questions 1, 2, 3, 4

TJ Exercise 5 Question 3

Line of intersection of two planes

Two planes must either be parallel or intersect in a line.

To determine the equations of the line of intersection we need to know its direction vector and a point on the line.

To find a point on the line:•The line must either cross the (x,y) plane (which has equation z = 0) or be parallel to it.•If it crosses it, set z = 0 in the equations of both planes to obtain a pair of simultaneous equations in x and y. Solving them will provide the required point on the line (x1, y1, 0)•If the line is parallel to the (x,y) plane then a similar point can be found on the (x,z) plane by a similar strategy. (set y = 0)

To find the direction vector:•The line of intersection lies in both planes.•Its direction vector is therefore perpendicular to the normal vector in each plane.•Thus the direction vector is parallel to the vector product of these normal vectors.

Find the equations of the line of intersection of the planes with equations

2 3 1 and 2 3x y z x y z

Let 0z 2 1 and 2 3x y x y

2(2 1) 3y y 5 2 3y 1y 1x

( 1, 1,0) lies on the line of intersection.

Normal Vectors are: 2 3u i j k 2v i j k

5

1 2 3 5

2 1 1 5

i j k

u v

1 1 Equation of the line is

5 5 5x y z

1 1

1 1 1x y z

also

For the intersection of 3 planes look at page 76 and 77.

Page 72 Exercise 12 Questions 1, 2Page 78 Exercise 15 Questions 1a, c, 2a, c

TJ Exercise 5 Questions 5 to 12

END OF TOPIC