The unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ an $\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$, is:

$-\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}$

The correct answer is Option (4) → $-\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}$

$\vec{a}=\hat{i}+\hat{j}+\hat{k}$

$\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$

$\vec p =\vec a+\vec b=2\hat i+3\hat j+4\hat k$

$\vec q=\vec a-\vec b=-\hat j-2\hat k$

$\vec p×\vec q=-2\hat i+4\hat j-2\hat k$

$\frac{\vec p×\vec q}{|\vec p×\vec q|}=\frac{-\hat i+2\hat j-\hat k}{\sqrt{6}}$

Unit vector perpendicular to $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$