Let **V** = 20.5 **i** + 22.5 **j** − 13.5 **k**.

(a) Find the magnitude of **V**.

(b) What angle does this vector make with the *x*-, *y-*, and *z*-axes?

Solution: Let V = 20.5 i + 22.5 j − 13.5 k.(a) Find the magnitude of V.(b) What angle does this vector make with the x-, y-, and z-axes?

Let **V** = 20.5 **i** + 22.5 **j** − 13.5 **k**.

(a) Find the magnitude of **V**.

(b) What angle does this vector make with the *x*-, *y-*, and *z*-axes?

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

$\overline{)\stackrel{\rightharpoonup}{A}{=}{{A}}_{{x}}{}\hat{\mathrm{i}}{+}{{A}}_{{y}}{}\hat{\mathrm{j}}{+}{{A}}_{{z}}{}\hat{\mathrm{k}}}$

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:

$\overline{)\left|\stackrel{\rightharpoonup}{A}\right|{=}\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}+{{A}_{z}}^{2}}}$ (1)

$\overline{){{A}}_{{x}}{=}\left|\stackrel{\rightharpoonup}{A}\right|{}{\mathrm{cos}}{}{\alpha}}$ (2)

$\overline{){{A}}_{{y}}{=}\left|\stackrel{\rightharpoonup}{A}\right|{}{\mathrm{cos}}{}{\beta}}$ (3)

$\overline{){{A}}_{{z}}{=}{\left|\stackrel{\rightharpoonup}{A}\right|}{}{\mathrm{cos}}{}{\gamma}}$ (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: