🤓 Based on our data, we think this question is relevant for Professor Munsell & Lewicki's class at PURDUE.

For the vectors shown in the figure, determine **B** - **2****A**. Vector magnitudes are given in arbitrary units.

Whenever we're adding and subtracting vectors in 2D, we follow these steps:

- Draw a
**diagram**and resolve the vectors into**components**. **Organize**your known information.**Add or subtract**vectors as required.- Convert back to
**magnitude-angle notation**(unless the problem only asks for components).

For step 1, we'll use these equations to find the *x-* and *y*-components of our *given* vectors:

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

For step 4, we'll use these equations to calculate the magnitude and angle from the +*x*-axis:

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

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

Remember that if we solve the last equation for *θ*, we always have to check whether the result we get makes sense based on the components. We might need to add or subtract 180°, because the inverse tangent function only returns angles between −90° and 90°.

**Step 1.** For this problem, we're *given a diagram*, so we don't need to draw one from scratch. Unfortunately one of the given angles is measured **clockwise** from the **negative** *x*-axis, and we need *θ*_{B} as a **counterclockwise** angle from the **positive** *x*-axis for our formulas to work:

${\theta}_{B}=180\xb0-56.0\xb0=\mathbf{124}\mathbf{\xb0}$

Now we can plug in known information for $\stackrel{\rightharpoonup}{B}$ and get its components, keeping an extra significant figure to reduce rounding error:

$\begin{array}{rcl}{B}_{x}& =& \left|\stackrel{\rightharpoonup}{B}\right|\mathrm{cos}{\theta}_{B}\\ & =& (26.5)\mathrm{cos}(124\xb0)\\ & =& -14.82\end{array}$ $\begin{array}{rcl}{B}_{y}& =& \left|\stackrel{\rightharpoonup}{B}\right|\mathrm{sin}{\theta}_{B}\\ & =& (26.5)\mathrm{sin}(124\xb0)\\ & =& 21.97\end{array}$

Since we're subtracting $2\stackrel{\rightharpoonup}{A}$, let's just go ahead and find 2*A _{x}* and 2

$\begin{array}{rcl}2{A}_{x}& =& \left|2\stackrel{\rightharpoonup}{A}\right|\mathrm{cos}{\theta}_{A}\\ & =& (88.0)\mathrm{cos}(28.0\xb0)\\ & =& 77.70\end{array}$ $\begin{array}{rcl}2{A}_{y}& =& \left|2\stackrel{\rightharpoonup}{A}\right|\mathrm{sin}{\theta}_{A}\\ & =& (88.0)\mathrm{sin}(28.0\xb0)\\ & =& 41.31\end{array}$

Vector Addition & Subtraction

Vector Addition & Subtraction

Vector Addition & Subtraction

Vector Addition & Subtraction