Question #63619

1 Answer
Cesareo R.
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^@#

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