• # question_answer If $\mathbf{u}=2\,\mathbf{i}+2\mathbf{j}-\mathbf{k}$and $\mathbf{v}=6\,\mathbf{i}-3\,\mathbf{j}+2\,\mathbf{k},$ then a unit vector perpendicular to both u and v is [MP PET 1987] A) $\mathbf{i}-10\mathbf{j}-18\mathbf{k}$ B) $\frac{1}{\sqrt{17}}\,\left( \frac{1}{5}\mathbf{i}-2\mathbf{j}-\frac{18}{5}\mathbf{k} \right)$ C) $\frac{1}{\sqrt{473}}\,(7\mathbf{i}-10\mathbf{j}-18\mathbf{k})$ D) None of these

Solution :

• $\mathbf{u}=2\mathbf{i}+2\mathbf{j}-\mathbf{k}$ and $\mathbf{v}=6\mathbf{i}-3\mathbf{j}+2\mathbf{k}$
• Let vector $\mathbf{c}={{c}_{1}}\mathbf{i}+{{c}_{2}}\mathbf{j}+{{c}_{3}}\mathbf{k}$ is perpendicular to both $\mathbf{u}$ and $\mathbf{v},$ then $\mathbf{c}\,.\,\mathbf{u}=0$
• $\Rightarrow 2{{c}_{1}}+2{{c}_{2}}-{{c}_{3}}=0$                           ...(i)
• and $\mathbf{c}\,.\,\mathbf{v}=0\Rightarrow 6{{c}_{1}}-3{{c}_{2}}+2{{c}_{3}}=0$ ...(ii)
• Solving equation (i) and (ii) by cross multiplication
• $\frac{{{c}_{1}}}{4-3}=\frac{{{c}_{2}}}{-6-4}=\frac{{{c}_{3}}}{-6-12}=\lambda$,    (say)
• $\Rightarrow \frac{{{c}_{1}}}{1}=\frac{{{c}_{2}}}{-10}=\frac{{{c}_{3}}}{-18}=\lambda$
• $\Rightarrow {{c}_{1}}=\lambda ,$ ${{c}_{2}}=-10\lambda$ and ${{c}_{3}}=-18\lambda$
• Thus $\mathbf{c}=\lambda (\mathbf{i}-10\mathbf{j}-18\mathbf{k})$
• $|\mathbf{c}|=\lambda \sqrt{1+100+324}=\lambda \sqrt{425}$
• Hence required unit vector is, $\frac{\mathbf{c}}{|\mathbf{c}|}$
• $=\frac{\lambda (\mathbf{i}-10\mathbf{j}-18\mathbf{k})}{\lambda \sqrt{425}}=\frac{1}{\sqrt{425}}(\mathbf{i}-10\mathbf{j}-18\mathbf{k})$ = $\frac{1}{5\sqrt{17}}(\mathbf{i}-10\mathbf{j}-18\mathbf{k})=\frac{1}{\sqrt{17}}\left( \frac{1}{5}\mathbf{i}-2\mathbf{j}-\frac{18}{5}\mathbf{k} \right)$                    Aliter : Required vector is $\frac{\mathbf{u}\times \mathbf{v}}{|\mathbf{u}\times \mathbf{v}|}=\frac{\mathbf{i}-10\mathbf{j}-18\mathbf{k}}{\sqrt{425}}.$

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