Question

Find a unit vector perpendicular to each of the vectors $\overrightarrow{a}+\overrightarrow{b}\mathrm{and}\overrightarrow{a}-\overrightarrow{b},\mathrm{where}\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}\mathrm{and}\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k} \\ \overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k} \\ \therefore \overrightarrow{a}+\overrightarrow{b}=4\hat{i}+4\hat{j}+0\hat{k} \\ \overrightarrow{a}-\overrightarrow{b}=2\hat{i}+0\hat{j}+4\hat{k} \\ \left(\overrightarrow{a}+\overrightarrow{b}\right)\times \left(\overrightarrow{a}-\overrightarrow{b}\right)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right| \\ =16\hat{i}-16\hat{j}-8\hat{k} \\ \therefore \left|\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \left(\overrightarrow{a}-\overrightarrow{b}\right)\right|=\sqrt{256+256+64} \\ =\sqrt{576} \\ =24 \\ \text{Unit vector perpendicular to both}\overrightarrow{a}+\overrightarrow{b}\text{and}\overrightarrow{a}\text{-}\overrightarrow{b}\text{=}\frac{\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \left(\overrightarrow{a}-\overrightarrow{b}\right)}{\left|\left(\overrightarrow{a}+\overrightarrow{b}\right)\times \left(\overrightarrow{a}-\overrightarrow{b}\right)\right|} \\ =\frac{16\hat{i}-16\hat{j}-8\hat{k}}{24} \\ =\frac{8\left(2\hat{i}-2\hat{j}-\hat{k}\right)}{24} \\ =\frac{1}{3}\left(2\hat{i}-2\hat{j}-\hat{k}\right) \end{gathered}$$