Find a vector of magnitude 4 units perpendicular to each of the vecto…

The correct answer is Option (2) → $\pm 2\sqrt{2}(\hat{j} + \hat{k})$ ##

Given vectors are:

$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$

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

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix}$

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

Now, $|\vec{a} \times \vec{b}| = \sqrt{(3)^2 + (3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$

Therefore required vector is, $\vec{d} = 4 \left( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \right)$

$= \frac{4}{3\sqrt{2}}(3\hat{j} + 3\hat{k}) = \frac{2\sqrt{2}}{3} \times 3(\hat{j} + \hat{k})$

$= 2\sqrt{2}(\hat{j} + \hat{k})$