Question

Let $\mathbf{a} = \mathbf{i} - \mathbf{j},\mathbf{b} = \mathbf{j} - \mathbf{k},\mathbf{c} = \mathbf{k} - \mathbf{i}$. If $\widehat{\mathbf{d}}$ is a unit vector such that $\mathbf{a}.\mathbf{d} = 0 = \lbrack\mathbf{bcd}\rbrack$, then $\widehat{\mathbf{d}}$ is equal to

• A. $\pm \frac{\mathbf{i} + \mathbf{j} - \mathbf{k}}{\sqrt{3}}$
• B. $\pm \frac{\mathbf{i} + \mathbf{j} + \mathbf{k}}{\sqrt{3}}$
• C. $\pm \frac{\mathbf{i} + \mathbf{j} - 2\mathbf{k}}{\sqrt{16}}$
• D. $\pm \mathbf{k}$

Answer 1

Answer: (C)

$\pm \frac{\mathbf{i} + \mathbf{j} - 2\mathbf{k}}{\sqrt{16}}$

Answer 2

Let $\widehat{\mathbf{d}} = \alpha\mathbf{i} + \beta\mathbf{j} + \gamma\mathbf{k}$

$\mathbf{a}.\widehat{\mathbf{d}} = 0$ ⇒ $(\mathbf{i} - \mathbf{j}).(\alpha\mathbf{i} + \beta\mathbf{j} + \gamma\mathbf{k}) = 0$ ⇒ $\alpha - \beta = 0$ ⇒ $\alpha = \beta\lbrack\mathbf{bcd}\rbrack = 0$ ⇒ $(\mathbf{b} \times \mathbf{c}).\mathbf{d} = 0$ ⇒ $\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 1 & - 1 \

• 1 & 0 & 1 \end{matrix} \right|$ .

$(\alpha\mathbf{i} + \beta\mathbf{j} + \gamma\mspace{6mu}\mathbf{k}) = 0$ ⇒ $(\mathbf{i} + \mathbf{j} + \mathbf{k})\mspace{6mu}.\mspace{6mu}(\alpha\mspace{6mu}\mathbf{i} + \beta\mspace{6mu}\mathbf{j} + \gamma\mspace{6mu}\mathbf{k}) = 0$

⇒ $\alpha + \beta + \gamma = 0$ ⇒ $\gamma = - (\alpha + \beta) = - 2\alpha$; $(\beta = \alpha)$

$|\widehat{\mathbf{d}}| = 1$ ⇒ $\alpha^{2} + \beta^{2} + \gamma^{2} = 1$ ⇒ $\alpha^{2} + \alpha^{2} + 4\alpha^{2} = 1$

⇒ $\alpha = \pm \frac{1}{\sqrt{6}} = \beta$ and $\gamma = \mp \frac{2}{\sqrt{6}}$

$\therefore$ $\widehat{\mathbf{d}} = \pm \frac{1}{\sqrt{6}}(\mathbf{i} + \mathbf{j} - 2\mathbf{k})$