Question

If `A = <3 ,-1 ,5 >`, `B = <5 ,2 ,-9 >` and `C=A-B`, what is the angle between A and C?

Answer 1

`38.43 °` (2dp)

Explanation:

The angle `theta` between two vectors `vec A` and `vec B` is related to the modulus (or magnitude) and scaler (or dot) product of `vec A` and `vec A` by the relationship:

`vec A * vec B = ||A|| \ ||B|| \ cos theta`

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem,

`vec A=<<3,-1,5>>`
`vec B=<<5,2,-9>>`

And so;

`vec C = vec A - vec B`
`\ \ \ \ = <<3,-1,5>> - <<5,2,-9>>`
`\ \ \ \ = <<-2,-3,14>>`

The moduli are given by;

`|| vec A || = || <<3,-1,5>> ||`
`\ \ \ \ \ \ \ \ = sqrt( (3)^2+(-1)^2+(5)^2 )`
`\ \ \ \ \ \ \ \ = sqrt( 9 + 1 + 25 )`
`\ \ \ \ \ \ \ \ = sqrt( 35 )`

`|| vec C || = || <<-2,-3,14>> ||`
`\ \ \ \ \ \ \ \ = sqrt( (-2)^2+(-3)^2+(14)^2 )`
`\ \ \ \ \ \ \ \ = sqrt( 4 + 9 + 196 )`
`\ \ \ \ \ \ \ \ = sqrt( 209 )`

And the scaler product is:

`vec A * vec C = || <<3,-1,5>> || * || <<-2,-3,14>> ||`
`\ \ \ \ \ \ \ \ \ \ = (3)(-2) + (-1)(-3) + (5)(14)`
`\ \ \ \ \ \ \ \ \ \ = -6 + 3 + 70`
`\ \ \ \ \ \ \ \ \ \ = 67`

And so using `vec A * vec C = || A || \ || C || \ cos theta` we have:

`67 = sqrt(35) * sqrt(209) * cos theta`
`:. cos theta = (67)/(sqrt(7315))`
`:. cos theta = 0.783371 ...`
`:. theta = 38.43 °` (2dp)

So the acute angle between the vectors is `38.43 °` (2dp)