Question

What is the angle between `<-4,8,0>` and `< 1,0,-2>`?

Answer 1

`78^o \` to nearest degree

Explanation:

The angle, `theta`, between vectors `bb(ul a)` and `bb(ul b)` is given by:

`bb(ul a) * bb(ul b) = |bb(ul a)| \ |bb(ul b)| \ cos theta`

So if we define:

`bb(ul a) = <<-4,8,0>>`
`bb(ul b) = <<1,0,-2>>`

Then we calculate the scalar (or dot product):

`bb(ul a) * bb(ul b) = <<-4,8,0>> * <<1,0,-2>>`
`\ \ \ \ \ \ \ \ \ = (-4)(1) + (8)(0) + (0)(-2)`
`\ \ \ \ \ \ \ \ \ = -4`

And we calculate the norms:

`|bb(ul a)| = sqrt((-4)^2+8^2+0^2)`
`\ \ \ \ \ = sqrt(16+64+0)`
`\ \ \ \ \ = sqrt(80)`

`|bb(ul b)| = sqrt(1^2+0^2+(-2)^2)`
`\ \ \ \ \ = sqrt(1+0+4)`
`\ \ \ \ \ = sqrt(5)`

So using `bb(ul a) * bb(ul b) = |bb(ul a)| \ |bb(ul b)| \ cos theta` we can write:

`-4 = sqrt(80)sqrt(5) \ cos theta`

`:. cos theta = -4/sqrt(400)`
`\ \ \ \ \ \ \ \ \ \ \ \ = -1/5`

`:. theta = arccos( -1/5 )`
`\ \ \ \ \ \ = 78.46304 ...`
`\ \ \ \ \ \ = 78^o \` to nearest degree