Find the unit vector in the direction of the sum of vectors $\mathbf{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\mathbf{b} = 2\hat{j} + \hat{k}$. |
$\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$ $\frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})$ $\frac{2}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$ $2\hat{i} + \hat{j} + 2\hat{k}$ |
$\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$ |
The correct answer is Option (1) → $\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$ ## Let $\mathbf{c}$ denote the sum of $\mathbf{a}$ and $\mathbf{b}$. We have, $\mathbf{c} = \mathbf{a} + \mathbf{b}$ $= 2\hat{i} - \hat{j} + \hat{k} + 2\hat{j} + \hat{k} = 2\hat{i} + \hat{j} + 2\hat{k}$ $∴$ Unit vector in the direction of $\mathbf{c} = \frac{\mathbf{c}}{|\mathbf{c}|} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{2^2 + 1^2 + 2^2}} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{9}}$$ $\hat{c} = \frac{2\hat{i} + \hat{j} + 2\hat{k}}{3}$ |