A force of 40 N acts at point P(2, 1, 3) m. The force makes angles of 60° and 45° with the X and Y axes respectively, and has a positive Z component. Express the force in vector form.

Answer: \(\vec F=20\,\hat i+28.28\,\hat j+20\,\hat k\) N (\(\theta_z=60^\circ\)).

Direction cosines satisfy \(\cos^2\theta_x+\cos^2\theta_y+\cos^2\theta_z=1\).

\(\cos^2\theta_z=1-\cos^260^\circ-\cos^245^\circ=1-0.25-0.5=0.25\Rightarrow\cos\theta_z=+0.5\) (Z component positive), so \(\theta_z=60^\circ\).

\[\vec F=F(\cos\theta_x\hat i+\cos\theta_y\hat j+\cos\theta_z\hat k)=40(0.5\hat i+0.7071\hat j+0.5\hat k)\]

\[\vec F=20\,\hat i+28.28\,\hat j+20\,\hat k\ \text{N}\]

(The point of application P only fixes where the force acts; it does not change its components.)