Block 3

Dot ProductCalculating Angle

What is to be learned?

• How to use dot product to calculate the angle between vectors

From Before

a.b = |a| |b| cosθ

cosθ =

cosθ =

a.b |a| |b|

also a.b = x1x2 + y1y2 + z1z2

x1x2 + y1y2 + z1z2

|a| |b|

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

Find angle between a and b214( ) 7

26( )

Numerator: 2(7) + 1(2) + 4(6) = 40

Denom: |a| = √(22 + 12 + 42) = √21

|b| = √(72 + 22 + 62) = √89

cosθ = 40

√21√89= 0.925 θ = 22.30

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

Find angle between a and b213( ) -1

5 2( )

Numerator: 2(-1) + 1(5) + 3(2) = 9

Denom: |a| = √(22 + 12 + 32) = √14

|b| = √((-1)2 + 52 + 22) = √30

cosθ = 9

√14√30= 0.439 θ = 64.00

Calculating Angles with Dot Product

Rearranging formula

cosθ = a.b |a| |b|

x1x2 + y1y2 + z1z2

|a| |b|=

a = 3i + 4k ,b = 4i + 6j + 2k

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

Calculate angle between vectors

Numerator: 3(4) + 0(6) + 4(2) = 20

Denom: |a| = √(32 + 02 + 42) = √25 = 5

|b| = √(42 + 62 + 22) = √56

cosθ = 20

5√56= 0.5345 θ = 57.70

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

Find angle between a and b412( ) -3

4 5( )

Numerator: 4(-3) + 1(4) + 2(5) = 2

Denom: |a| = √(42 + 12 + 22) = √21

|b| = √((-3)2 + 42 + 52) = √50

cosθ = 2

√21√50= 0.0617 θ = 86.50

Key Question

If vectors are perpendicular cosθ = cos900 =

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

0

x1x2 + y1y2 + z1z2 = 0

Numerator

Denominator

Perpendicular Vectors

Numerator = 0

θ

Prove a = 3i – 2j is perpendicular to

b = 4i + 6j – 5k

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

If perpendicular cosθ = cos900 = 0i.e. num = 0

num = 3(4) + (-2)6 + 0(-5)

= 0 as required

must equal zero

4 -1 -2 ( )32

5( )Prove a and b are perpendicular

a = b =

If perpendicular cosθ = cos900 = 0i.e. num = 0

num = 4(3) + (-1)2 + (-2)(5)

= 0 as required

2 -1 -2( )

42 k( )

If a and b are perpendicular, find value of k

a = b =

If perpendicular cosθ = cos900 = 0i.e. num = 0

num = 2(4) + (-1)2 + (-2)(k)

= 6 – 2k

6 – 2k = 0 k = 3

If vectors are perpendicular cosθ = cos900 =

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

0

x1x2 + y1y2 + z1z2 = 0

Numerator

Denominator

Perpendicular Vectors

Numerator = 0

θ

2 -1 -3

( ) 34 k( )

If a and b are perpendicular, find value of k

a = b =

If perpendicular cosθ = cos900 = 0i.e. num = 0

num = 2(3) + (-1)4 + (-3)(k)

= 2 – 3k

2 – 3k = 0 k = 2/3

7 g -9

( ) 0g 4( )

If a and b are perpendicular, find possible values of g

a = b =

If perpendicular cosθ = cos900 = 0i.e. num = 0num = 7(0) + g(g) + (-9)(4)

= g2 – 36

g2 – 36 = 0 g2 = 36 g = 6 or -6

Key Question

A (7,5 ,7) B (3 ,4 ,6) C (5,6,9)Calculate LABC

Need BA and BC

u = BA = a – b

A

B

C

θu

vu v

757 ( ) 3

46 ( )= –

411= ( )

v = BC = c – b

569 ( ) 3

46 ( )= –

223=

( )

cosθ = x1x2 + y1y2 + z1z2

|a| |b|

Angle between u and v411( ) 2

2 3 ( )

Numerator: 4(2) + 1(2) + 1(3) = 13

Denom: |a| = √(42 + 12 + 12) = √18

|b| = √(22 + 22 + 32) = √17

cosθ = 13

√18√17θ = 420