Question

Let $\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$ Then a unit vector perpendicular to both $\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{a}+\overrightarrow{b}$ is

Answer 1

$\overrightarrow{a}=3\hat{i}+2\hat{j}+2\hat{k}$

$\overrightarrow{b}=\hat{i}+2\hat{j}-2\hat{k}$

$veca+\overrightarrow{b}=4\hat{i}+4\hat{j}$

$veca-\overrightarrow{b}=2\hat{i}+4\hat{k}$

unit vector perpendicular to both $\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{a}+\overrightarrow{b}$

$\overrightarrow{c}=(\overrightarrow{a}-\overrightarrow{b})\times (\overrightarrow{a}+\overrightarrow{b})$

$\therefore \overrightarrow{c}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 0& 4\\ 4& 4& 0\end{array}\right|=\hat{i}(-16)-\hat{j}(-16)+\hat{k}\left(8\right)$

$=-16\hat{i}+16\hat{j}+8\hat{k}$

Now

unit vector indirection of $\overrightarrow{c}$ is

$\frac{\overrightarrow{c}}{|\overrightarrow{c}|}$

$|\overrightarrow{c}|=\sqrt{{16}^2+{16}^2+8^2}=24$

$\therefore$ Desired unit vector = $\frac{1}{24}(-16\hat{i}+16\hat{j}+8\hat{k})$

$=\frac{1}{3}(-2\hat{i}+2\hat{j}+\hat{k})$