Question

Find a unit vector normal to the plane containing

$$\begin{gathered} \overrightarrow{v}=2\hat{i}+3\hat{j}-\hat{k} \\ \overrightarrow{w}=-2\hat{i}-4\hat{j}+3\hat{k} \end{gathered}$$

Short answer

$\hat{a}=\frac{-13}{\sqrt{237}}\hat{i}+\frac{8}{\sqrt{237}}\hat{j}-\frac{2}{\sqrt{237}}\hat{k}$

Step-by-step solution

Step 1. Given vectors in the plane 

$$\begin{gathered} \overrightarrow{v}=2\hat{i}+3\hat{j}-\hat{k} \\ \overrightarrow{w}=-2\hat{i}-4\hat{j}+3\hat{k} \end{gathered}$$

Step 2. The cross product

We know that the unit vector perpendicular to

is:

$\hat{a}=\pm \frac{\overrightarrow{v}\times \overrightarrow{w}}{\left|\overrightarrow{v}\times \overrightarrow{w}\right|}$

So;

$$\begin{gathered} \overrightarrow{v}=2\hat{i}+3\hat{j}-\hat{k} \\ \overrightarrow{w}=-2\hat{i}-4\hat{j}-3\hat{k} \\ \overrightarrow{v}\times \overrightarrow{w}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& -1\\ -2& -4& -3\end{array}\right| \\ \overrightarrow{v}\times \overrightarrow{w}=\hat{i}(-9-4)-\hat{j}(-6-2)+\hat{k}(-8+6) \\ \overrightarrow{v}\times \overrightarrow{w}=-13\hat{i}+8\hat{j}-2\hat{k} \\ |\overrightarrow{v}\times \overrightarrow{w}|=\sqrt{169+4+64} \\ |\overrightarrow{v}\times \overrightarrow{w}|=\sqrt{237} \end{gathered}$$

Step 3. The unit vector is :

$\hat{a}=\frac{-13}{\sqrt{237}}\hat{i}+\frac{8}{\sqrt{237}}\hat{j}-\frac{2}{\sqrt{237}}\hat{k}$