Vectors (5)Vectors (5)•Scaler ProductScaler Product

•Angle between linesAngle between lines

Angle between VectorsThe angle can be measured if they are placed ….

“Head-to-Head” “Tail-to-Tail”

-1 7[ ] 4

3[ ]Angle between Vectors - example

a = b =

|a| = ((-1)2 + 72)

|a| = 50

|b| = (42 + 32)= 5

|b - a| = (52 + -42)= 41

b - a = 5 -4[ ]

-1 7[ ] 4

3[ ]Angle between Vectors - example (2)

a = b =

|a| = 50|b| = 5

|b - a| = 41

How can you find the angle now?

a2 = b2 + c2 - 2bc cos A

The Cosine Rule

A B

C

ab

c

anglessides

… is used for working out angles and sides in non-right angled triangles

It is ….

a2 = b2 + c2 - 2bc cos AUsing the Cosine Rule ...

A

B C

anglessides 505

41

41 = 50 + 25 - 2 x 50 x 5 cos A

= 61.3o

3

2

1

a

a

a

a

3

2

1

b

b

b

b

Angle between Vectors - general case

33

22

11

3

2

1

3

2

1

ab

ab

ab

a

a

a

b

b

b

abc

|a| = (a12 + a2

2 + a32)

|b| = (b12 + b2

2 + b32)

|c| = ((b1-a1)2 + (b2-a2)2 +(b3-a3)2)

|c|2 = (b1-a1)2 + (b2-a2)2 +(b3-a3)2

|c|2 = a12 + a2

2 + a32 + b1

2 + b22 + b3

2 -2(a1 b1+ a2 b2+

a3 b3)

Expand and rearrange

|c|2 = |a|2 + |b|2 -2(a1 b1+ a2 b2+ a3 b3)

a2 = b2 + c2 - 2bc cos A

Generalizing

anglessides |a||b|

|c|=|b - a|

|c|2 = |a|2 + |b|2 - 2|a||b| cos

c=b - a

Cosine Rule

|c|2 = |a|2 + |b|2 -2(a1 b1+ a2 b2+ a3 b3)|a|2 + |b|2 -2(a1 b1+ a2 b2+ a3 b3) = |a|2 + |b|2 - 2|a||b| cos -2(a1 b1+ a2 b2+ a3 b3) = - 2|a||b| cos

a1 b1+ a2 b2+ a3 b3 = |a||b| cos

Generalizing (cont.)

a1 b1+ a2 b2+ a3 b3 = |a||b| cos cos = a1 b1+ a2 b2+ a3 b3

|a||b|

The Scaler Product

a1 b1+ a2 b2+ a3 b3 = |a||b| cos

The scaler product is defined as ...

Previously,

… was proved

the value of … a1 b1+ a2 b2+ a3 b3

or |a||b| cos

The scaler product is written as ...a.b… it’s also known as the dot product

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

a.b = |a||b| cos

Scaler Product (cont.)

cos = a1 b1+ a2 b2+ a3 b3

|a||b|

becomes

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

cos = a.b|a||b|

Parallel Vectors

cos = a.b|a||b|

Occur

…when cos = 1… so = cos-1(1) = 0

degreesi.e. the lines are Parallel

Perpendicular Vectors

cos = a.b|a||b|

a.bIf = 0,

…then cos = 0… so = cos-1(0) = 90

degreesi.e. the lines are Perpendicular

So, if a.b = 0 then the lines are perpendicular

Example (2D) - angle between vectors

Given: a = 3i + 4j and b = i - 3j

The scaler product is written as ...a.b

a.b = (3 x 1) + (4 x -3)

The j componentsThe i components

cos = a.b|a||b|

|a| = (32 + 42) = 25 = 5

|b| = (12 + (-3)2) = 10

= 4 - 12 = -8

cos = -8 = 0.506 510

= cos-1(0.506) = 120.4o

Angle between 3D VectorsAngle between 3D Vectors

7

3

2

a

5

2

1

b

The scaler product is written as ...a.b

a.b = (2 x 1) + (3 x -2) + (7 x 5)

cos = a.b|a||b|

|a| = (22 + 32 + 72) = 62

|b| = (12 + (-2)2 + 52) = 30

= 2 - 6 + 35 = 31

cos = 31 = 0.719 6230 = cos-1(0.719) = 44.0o