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: |
$\pm\left[\frac{-5 \hat{i}+\hat{j}+2 \hat{k}}{\sqrt{30}}\right]$ $\pm\left[\frac{5 \hat{i}+\hat{j}-2 \hat{k}}{\sqrt{30}}\right]$ $\pm\left[\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\right]$ $\pm\left[\frac{-5 \hat{i}-\hat{j}-2 \hat{k}}{\sqrt{30}}\right]$ |
$\pm\left[\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\right]$ |
The correct answer is Option (3) - $\pm\left[\frac{5 \hat{i}-\hat{j}+2 \hat{k}}{\sqrt{30}}\right]$ $\vec v ⊥\vec a$ $\vec v ⊥\vec b$ $\vec v=±(\vec a×\vec b)$ $\vec v=±\begin{vmatrix}\hat i&\hat j&\hat k\\1&-1&-3\\1&3&-1\end{vmatrix}$ $\vec v=±(10\hat i-2\hat j+4\hat k)$ $\hat v=\frac{\vec v}{|\vec v|}=\frac{\vec v}{\sqrt{10^2+(-2)^2+4^2}}$ $=\frac{±(10\hat i-2\hat j+4\hat k)}{2\sqrt{30}}$ $=±\frac{(5\hat i-\hat j+2\hat k)}{\sqrt{30}}$ |