Question

Find the unit vector perpendicular to each of the vectors $6\hat{i}+2\hat{j}+3\hat{\mathit{k}}$ and $3\hat{i}-2\hat{k}$

Answer 1

Step 1: Find the normal vector

Let, vector $\overrightarrow{a}$ be perpendicular to each of the vectors $6\hat{i}+2\hat{j}+3\hat{\mathit{k}}$ and $3\hat{i}-2\hat{k}$

Then,

$$\begin{gathered} \overrightarrow{a}=\left(6\hat{i}+2\hat{j}+3\hat{k}\right)\times \left(3\hat{i}-2\hat{k}\right) \\ =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 6& 2& 3\\ 3& 0& -2\end{array}\right| \\ =\hat{i}\left(-4\right)-\hat{j}\left(-12-9\right)+\hat{k}\left(0-6\right) \\ =-4\hat{i}+21\hat{j}-6\hat{k} \end{gathered}$$

Step 2: Calculate the magnitude of the normal vector

To find the unit vector, divide the vector by its magnitude

For a vector $\overrightarrow{v}=a\hat{i}+b\hat{j}+c\hat{k}$, the magnitude $\left|\overrightarrow{v}\right|=\sqrt{a^2+b^2+c^2}$

Thus, the magnitude of vector $\overrightarrow{a}$ is,

$$\begin{gathered} \left|\overrightarrow{a}\right|=\sqrt{{(-4)}^2+{21}^2+{(-6)}^2} \\ \Rightarrow \left|\overrightarrow{a}\right|=\sqrt{16+441+36} \\ \Rightarrow \left|\overrightarrow{a}\right|=\sqrt{493} \end{gathered}$$

Step 3: Find the unit vector

The unit vector $\hat{a}$ of vector $\overrightarrow{a}$ is given as,

$\begin{array}{rcl}\hat{a}& =& \frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}\\ & \Rightarrow & \hat{a}=\frac{-4\hat{i}+21\hat{j}-6\hat{k}}{\sqrt{493}}\end{array}$

Therefore, the unit vector perpendicular to the given vectors is $\frac{-4\hat{i}+21\hat{j}-6\hat{k}}{\sqrt{493}}$ .