Remember that the unit-vector form is a vector equation that relates a vector to its x, y, and z components:

$\boxed{\stackrel{\rightharpoonup }{A}=A_x \hat{\mathrm{i}}+A_y \hat{\mathrm{j}}+A_z \hat{\mathrm{k}}}$

Anytime we're asked to relate the magnitude and direction of a three-dimensional vector to its components, there are four basic equations we might use:

$\boxed{\left|\stackrel{\rightharpoonup }{A}\right|=\sqrt{{A_x}^2+{A_y}^2+{A_z}^2}}$ (1)

$\boxed{A_x=\left|\stackrel{\rightharpoonup }{A}\right| \cos \alpha }$ (2)

$\boxed{A_y=\left|\stackrel{\rightharpoonup }{A}\right| \cos \beta }$   (3)

$\boxed{A_z=\left|\stackrel{\rightharpoonup }{A}\right| \cos \gamma }$   (4)

(a) For this problem we're given that $\stackrel{\rightharpoonup }{V}=20.5 \hat{\mathrm{i}}+22.5 \hat{\mathrm{j}}-13.5 \hat{\mathrm{k}}$. Plugging these magnitudes into equation (1), we get

$\begin{array}{rcl}\left|\stackrel{\rightharpoonup }{A}\right|& =& \sqrt{{(20.5)}^2+{(22.5)}^2+{(-13.5)}^2}\\ & =& \mathbf{33}\mathbf{.}\mathbf{30}\end{array}$

The magnitude of the vector is 33.3 units.

(b) For the second part of the problem, we solve equations 2-4 for the angle and substitute our known values: