Question #63619

1 Answer
Jun 6, 2016

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^@