Find a vector of magnitude 4 units perpendicular to each of the vectors $2\hat{i} - \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} - \hat{k}$. |
$\pm \frac{4}{3}(\hat{j} + \hat{k})$ $\pm 2\sqrt{2}(\hat{j} + \hat{k})$ $\pm 2(\hat{j} - \hat{k})$ $\pm 4(\hat{i} + \hat{j} + \hat{k})$ |
$\pm 2\sqrt{2}(\hat{j} + \hat{k})$ |
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, $\vec{d} = 4 \left( \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \right)$ $= \frac{4}{3\sqrt{2}}(3\hat{j} + 3\hat{k}) = \frac{2\sqrt{2}}{3} \times 3(\hat{j} + \hat{k})$ $= 2\sqrt{2}(\hat{j} + \hat{k})$ |