Steps for finding a missing vector:

**Write the vector equation** for the problem (drawing a diagram usually helps a lot!)**Decompose** the given vectors and **organize** known information in a table.- Write algebraic equations and
**solve.** **Find ****m****agnitude and angle** of the resultant if required.

When writing a vector equation for one of these problems, it'll be in a form like

$\overline{)\stackrel{\rightharpoonup}{A}{+}\stackrel{\rightharpoonup}{B}{+}{.}{.}{.}{=}\stackrel{\rightharpoonup}{R}}$,

where the **missing vector** is one of the vectors on the **left-hand side** of the equation. Note that if the problem is about *returning to the starting point*, $\stackrel{\rightharpoonup}{R}=\stackrel{\rightharpoonup}{0}$.

For decomposing given vectors, we'll have to __define a coordinate system__ if the problem hasn't, then use the equations

$\overline{)\begin{array}{rcl}{A}_{x}& {=}& \left|\stackrel{\rightharpoonup}{A}\right|\mathrm{cos}\theta \\ {A}_{y}& {=}& \left|\stackrel{\rightharpoonup}{A}\right|\mathrm{sin}\theta \end{array}}$

If the final answer is supposed to be in magnitude-angle notation, we'll need to use these equations to get the magnitude and direction:

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

$\overline{){\mathrm{tan}}{}{\theta}{=}\frac{{A}_{y}}{{A}_{x}}}$

**Step 1.** This problem has the plane flying three separate legs (24.0 km at 34.0° south of east, 46.0 km north, and the unknown). We'll call them $\stackrel{\rightharpoonup}{A}$, $\stackrel{\rightharpoonup}{B}$, and $\stackrel{\rightharpoonup}{C}$ in order. The resultant $\stackrel{\rightharpoonup}{R}$ is 32.0 km west. So our vector equation is

$\overline{)\stackrel{\rightharpoonup}{A}{+}\stackrel{\rightharpoonup}{B}{+}\stackrel{\rightharpoonup}{C}{=}\stackrel{\rightharpoonup}{R}}$

**Step 2.** Define a coordinate system so that north is the +*y*-direction and east is the +*x*-direction. Then we can decompose vector $\stackrel{\rightharpoonup}{A}$ into *x*- and *y*-components. "South of east" in this coordinate system is **counterclockwise from the +***x*-axis, so

$\begin{array}{lll}{A}_{x}& =& \left|\stackrel{\rightharpoonup}{A}\right|\mathrm{cos}\theta \\ & =& (24.0\mathrm{km})\mathrm{cos}(-34.0\xb0)\\ & =& 19.90\mathrm{km}\end{array}$ $\begin{array}{lll}{A}_{y}& =& \left|\stackrel{\rightharpoonup}{A}\right|\mathrm{sin}\theta \\ & =& (24.0\mathrm{km})\mathrm{sin}(-34.0\xb0)\\ & =& -13.42\mathrm{km}\end{array}$