Question

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

Answer 1

We have,
$\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$
$\therefore \overrightarrow{a}+\overrightarrow{b}=4\hat{i}+4\hat{j},\overrightarrow{a}-\overrightarrow{b}=2\hat{i}+4\hat{k}$
$(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& 4& 0\\ 2& 0& 4\end{array}\right|=\hat{i}\left(16\right)-\hat{j}\left(16\right)+\hat{k}(-18)=16\hat{i}-16\hat{j}-8\hat{k}$
$\therefore \left|\right(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b}\left)\right|=\sqrt{{16}^2+(-16{)}^2+(-8{)}^2}$
$=\sqrt{2^2\times 8^2+2^2\times 8^2+8^2}$
$=8\sqrt{2^2+2^2+1}=8\sqrt{9}=8\times 3=24$
Hence, the unit vector perpendicular to each of the vectors $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ is given by the relation.
$=\pm \frac{(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})}{|(\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-\overrightarrow{b})|}=\pm \frac{16\hat{i}-16\hat{j}-8\hat{k}}{24}$
$=\pm \frac{2\hat{i}-2\hat{j}-\hat{k}}{3}=\pm \frac{2}{3}\hat{i}\pm \frac{2}{3}\hat{j}\pm \frac{1}{3}\hat{k}$