Question

Question #63619

Answer 1

`theta_z = 90^@`

Explanation:

Let `vec v = {v_x,v_y,v_z}` and `hat e_x,hat e_y,hat e_z` the axis unity vectors.

We have

`<< vec v, hat e_x>> = norm(vec v)cos(theta_x) = v_x`
`<< vec v, hat e_y>> = norm(vec v)cos(theta_y) = v_y`
`<< vec v, hat e_z>> = norm(vec v)cos(theta_z) = v_z`

also

`v_x^2+v_y^2+v_z^2=norm(vec v)^2 = norm(vec v)^2cos^2(theta_x)+norm(vec v)^2cos^2(theta_y)+norm(vec v)^2cos^2(theta_z)`

simplifying we have

`cos^2(theta_x)+cos^2(theta_y)+cos^2(theta_z) = 1`

so

`cos(theta_z) = pm sqrt(1-cos^2(theta_x)-cos^2(theta_y))`

having now `theta_x = 150^@` and `theta_y = 60^@`

we have `cos(theta_z)=0` so `theta_z = pm pi/2 = pm 90^@`