Question

What is the angle between `<-8,2,8>` and `< 6,-4,5>`?

Answer 1

The angle is `80.9 ^o`(3sf)

Explanation:

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

`bb(vec A * vec B) = || bb(vec A) || \ || bb(vec B) || \ cos theta`

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

So for this problem, let the angle between `bb(vecu)` and `bb(vecv)` be `theta` and let:

`bb(vec u) = <<-8, 2, 8>>`
`bb(vec v) = <<6, -4, 5>>`

The vector norm is given by;

`| bb(vec u) | = || <<-8, 2, -8 >> || = sqrt(64+4+64)=sqrt(132)`
`| bb(vec v) | = || <<6, -4, 5 >> || = sqrt(36+16+25)=sqrt(77)`

And the scaler (or "dot") product is:

`bb(vec u * vec v) = <<-8, 2, 8>> bb(*) <<6, -4, 5>>`
`\ \ \ \ \ \ \ \ \ \ \ = (-8)(6) + (2)(-4) + (8)(5)`
`\ \ \ \ \ \ \ \ \ \ \ = -48 -8 +40`
`\ \ \ \ \ \ \ \ \ \ \ = -16`

And so using `bb(vec A * vec B) = || bb(vec A) || \ || bb(vec B) || \ cos theta` we have:

`-16 = sqrt(132)sqrt(77) cos theta`
`:. cos theta = -16/sqrt(10164)`
`" " = -(8sqrt(21))/231`
`" " =-0.15870392 ...`
`=> theta = 99.1316 ... ^o`

This is not acute, so the acute angle between the vectors is:

`theta_("acute") = 180^o - 99.1316 ... ^o`
`" " = 80.8683 ...^o`