A unit vector perpendicular to the plane of $\vec a=2\hat i-6\hat j-3\hat k$ and $\vec b = 4\hat i +3\hat j-\hat k$, is
Answer & explanation
Correct answer: option 3
We have,
$\vec a×\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\2&-6&-3\\4&3&-1\end{vmatrix}=5(3\hat i-2\hat j+6\hat k)$
$∴|\vec a×\vec b|=5\sqrt{9+4+36}=35$
Hence, required unit vector $\hat n$ is given by
$\hat n=\frac{5}{35}(3\hat i-2\hat j+6\hat k)=\frac{1}{7}(3\hat i -2\hat j+6\hat k)$