Question

If `A = <3 ,-1 ,8 >`, `B = <4 ,-3 ,-1 >` and `C=A-B`, what is the angle between A and C?

Answer 1

`36.22^o` **(2,d.p.)

Explanation:

A= `((3),(-1),(8))`

B= `((4),(-3),(-1))`

C= A - B: C= `((3),(-1),(8))-((4),(-3),(-1))=((-1),(2),(7))`

Angle between A and C:

`((3),(-1),(8))*((-1),(2),(7))`

This is called the Dot product, Scaler product or Inner product and is defined as:

`a*b= |a|*|b|cos(theta)`

`|A|= sqrt(3^2+(-1)^2+8^2)= sqrt(74)`

`|C|= sqrt((-1)^2+2^2+7^2)= sqrt(54)=3sqrt(6)`

`A*C= ((3),(-1),(8))*((-1),(2),(7))=((-3),(-2),(56))`

We need to sum the components in the product vector, so:

`-2-3+56=51`

So now we have:

`51 = sqrt(74)*3sqrt(6)cos(theta)=51/(sqrt(74)*3sqrt(6))= cos(theta)`

`-> 51/(3sqrt(444))=cos(theta)`

`theta= arccos(51/(3sqrt(444))) =36.22^o` (2,d.p.)

The angle is formed where the vectors are moving in the same relative direction. i.e where both arrow heads are pointing.