Question

(i) Find a unit vector perpendicular to both the vectors

(ii) Find a unit vector perpendicular to the plane containing the vectors

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Given}: \\ \overrightarrow{a}=4\hat{i}-\hat{j}+3\hat{k} \\ \overrightarrow{b}=-2\hat{i}+\hat{j}-2\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 3\\ -2& 1& -2\end{array}\right| \\ =\left(2-3\right)\hat{i}-\left(-8+6\right)\hat{j}+\left(4-2\right)\hat{k} \\ =-\hat{i}+2\hat{j}+2\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{1+2^2+2^2} \\ =\sqrt{9} \\ =3 \\ \text{Unit vector perpendicular to}\overrightarrow{a}\text{and}\overrightarrow{b}\text{=}\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}=\frac{-\hat{i}+2\hat{j}+2\hat{k}}{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Given}: \\ \overrightarrow{a}=2\hat{i}+\hat{j}+\hat{k} \\ \overrightarrow{b}=\hat{i}+2\hat{j}+\hat{k} \\ \stackrel{}{\therefore \overrightarrow{a}}\times \stackrel{}{\overrightarrow{b}}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& 1\\ 1& 2& 1\end{array}\right| \\ =\left(1-2\right)\hat{i}-\left(2-1\right)\hat{j}+\stackrel{}{\left(4-1\right)\stackrel{⏜}{k}} \\ =-\hat{i}-\hat{j}+3\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{1+1+9} \\ =\sqrt{11} \\ \text{Unit vector perpendicular to the plane containing vectors}\overrightarrow{a}\text{and}\overrightarrow{b}=\pm \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|} \\ \text{Unit vector perpendicular to the plane containing vectors}\overrightarrow{a}\text{and}\overrightarrow{b}=\pm \frac{1}{\sqrt{11}}\left(-\hat{i}-\hat{j}+3\hat{k}\right) \end{gathered}$$