Question

What is the angle between `<0,-7,6>` and `<-1,9,-3>`?

Answer 1

`theta=arccos(-81/sqrt(7735))~~2.7414`

Explanation:

We'll use the fact that the dot product `vec(u)*vec(v)` can be calculated in two ways:

`vec(u)*vec(v) = u_1v_1+u_2v_2+u_3v_3 = ||vecu||*||vecv||cos(theta)`

Dividing by `||vecu||*||vecv||`, we get

`cos(theta) = (u_1v_1+u_2v_2+u_3v_3)/(||vecu||*||vecv||)`

or

`theta = arccos((u_1v_1+u_2v_2+u_3v_3)/(||vecu||*||vecv||))`

In our case, we have `vecu = < 0, -7, 6 >` and `vecv = < -1, 9, -3 >`.

If we calculate their magnitudes, we get

`||vecu|| = sqrt(0^2+(-7)^2+6^2) = sqrt(85)`
and
`||vecv|| = sqrt((-1)^2+9^2+(-3)^2) = sqrt(91)`

So, the formula we derived gives us

`theta = arccos((0(-1)+(-7)(9)+6(-3))/(sqrt(85)*sqrt(91)))`

`=arccos(-81/sqrt(7735))`

`~~2.7414`