• # question_answer Let $\mathbf{a}=\mathbf{i}-\mathbf{j},\,\,\mathbf{b}=\mathbf{j}-\mathbf{k},\,\,\mathbf{c}=\mathbf{k}-\mathbf{i}.$ If $\mathbf{\hat{d}}$ is a unit vector such that $\mathbf{a}\,.\,\mathbf{\hat{d}}=0=[\mathbf{b}\,\,\mathbf{c}\,\,\mathbf{\hat{d}}],$ then $\mathbf{\hat{d}}$ is equal to [IIT 1995] A) $\pm \frac{\mathbf{i}+\mathbf{j}-\mathbf{k}}{\sqrt{3}}$ B) $\pm \frac{\mathbf{i}+\mathbf{j}+\mathbf{k}}{\sqrt{3}}$ C) $\pm \frac{\mathbf{i}+\mathbf{j}-2\mathbf{k}}{\sqrt{6}}$ D) $\pm \,\,\mathbf{k}$

Solution :

• Let $\mathbf{a}=\mathbf{i}-\mathbf{j},$ $\mathbf{b}=\mathbf{j}-\mathbf{k}$ and $\mathbf{c}=\mathbf{k}-\mathbf{i}$
• Let $\mathbf{\hat{d}}={{a}_{1}}\mathbf{i}+{{a}_{2}}\mathbf{j}+{{a}_{3}}\mathbf{k},$ $|\mathbf{\hat{d}}|\,=\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}}=1$     $\Rightarrow a_{1}^{2}+a_{2}^{2}+a_{3}^{2}=1$                                ......(i)
• $\mathbf{a}\,.\mathbf{\hat{d}}=0\Rightarrow {{a}_{1}}-{{a}_{2}}=0$    ..(ii)
• $[\mathbf{b}\,\mathbf{c}\,\mathbf{\hat{d}}]=0\Rightarrow \mathbf{b}\,.\,(\mathbf{c}\times \mathbf{\hat{d}})=0$
• $\Rightarrow \left| \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\ \end{matrix} \right|=-1(-{{a}_{3}}-{{a}_{1}})-1(-{{a}_{2}})$
• $\therefore \,{{a}_{1}}+{{a}_{2}}+{{a}_{3}}=0\Rightarrow {{a}_{1}}-{{a}_{2}}+0{{a}_{3}}=0$,    {from (ii)}
• $\therefore \,\frac{{{a}_{1}}}{0+1}=\frac{{{a}_{2}}}{1-0}=\frac{{{a}_{3}}}{-1-1}\Rightarrow \frac{{{a}_{1}}}{1}=\frac{{{a}_{2}}}{1}=\frac{{{a}_{3}}}{-2}=\lambda$, (say)
• $\Rightarrow {{a}_{1}}=\lambda ,$ ${{a}_{2}}=\lambda ,$ ${{a}_{3}}=-2\lambda$
• $\therefore \,{{\lambda }^{2}}+{{\lambda }^{2}}+4{{\lambda }^{2}}=1$ ,    {from (i)}
• $\Rightarrow 6{{\lambda }^{2}}=1\Rightarrow \lambda =\pm \frac{1}{\sqrt{6}};$ $\therefore \,\mathbf{\hat{d}}=\pm \frac{\mathbf{i}+\mathbf{j}-2\mathbf{k}}{\sqrt{6}}$.

You need to login to perform this action.
You will be redirected in 3 sec