Question

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

Answer 1

`26.54^@`

Explanation:

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

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

`C=A-B=[(3),(8),(-1)]-[(4),(-3),(-1)]=[(3-4),(8-(-3)),(-1-(-1))]=[(-1),(11),(0)]`

We can find the angle between vectors using the Dot Product

The dot product states that for vectors a and b:

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

The dot product is sometimes called the inner product, because of the way the vectors a and b are multiplied and summed.

In algebra we are used to multiplying brackets in the following way.

`(a+b)(c+d)=ac+ad+bc+bd`

In the case of the dot product we multiply the vectors in the following way.

`(a+b+c) * (d+e+f)=ad+be+cf`

So we are just multiplying corresponding components and then adding them together.

Let `a = [(x),(y),(z)]`

Magnitude of `a=||a||`

`||a||=sqrt(x^2+y^2+z^2)`

Now to our example:

First find the product of:

`A*C`

`[(3),(8),(-1)]*[(-1),(11),(0)]=[(3xx-1),(8xx11),(-1xx0)]`

`=[(-3),(88),(0)]=-3+88+0=85`

Now find the magnitudes of A and C:

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

`||C||=sqrt((-1)^2+(11)^2+(0)^2)=sqrt(122)`

So we have for:

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

`85=sqrt(74)*sqrt(122)*cos(theta)`

`cos(theta)=85/(sqrt(74)*sqrt(122))`

`theta=arccos(cos(theta))=arccos(85/(sqrt(74)*sqrt(122)))=26.54^@`
( 2 .d.p.)

So the angle between vectors A and C is `26.54^@`

Notice from the diagram that the angle found by the dot product, is the angle between the vectors where they are pointing in the same relative direction.