Question

What is the angle between `<-8,0,6>` and `<3,2,-3 >`?

Answer 1

-156.43 degrees

Explanation:

The vectors can be rewritten as -8i +6k and 3i+2j-3k
If `theta` is the angle between these two vectors then,

`cos theta`=

`( -8*3 +0*2 +6*(-3)) /(|sqrt (8^2+0^2 +6^2)| | sqrt (3^2 +2^2 +(-3)^2|)`

=`-42/(sqrt100 sqrt22)`

=`(-4.2)/sqrt22`= -0.8945

`theta`= 156.43 degrees