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

1 Answer
Mar 11, 2018

78^o \ 78o to nearest degree

Explanation:

The angle, thetaθ, 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