Unit vectors perpendiculars to the plane of vectors $\vec{a}=\hat{i}-\hat{j}-3 \hat{k}, \vec{b}=\hat{i}+3 \hat{j}-\hat{k}$ are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) - $\pm\left[\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\right]$
$\vec v ⊥\vec a$
$\vec v ⊥\vec b$
$\vec v=±(\vec a×\vec b)$
$\vec v=±\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&-3\\1&3&-1\end{vmatrix}$
$\vec v=±(10\hat i-2\hat j+4\hat k)$
$\hat v=\frac{\vec v}{|\vec v|}=\frac{\vec v}{\sqrt{10^2+(-2)^2+4^2}}$
$=\frac{±(10\hat i-2\hat j+4\hat k)}{2\sqrt{30}}$
$=±\frac{(5\hat i-\hat j+2\hat k)}{\sqrt{30}}$