A) \[\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\]
B) \[\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}\]
C) \[\pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\]
D) \[\pm \hat{k}\]
E) \[\hat{i}+\hat{j}\]
Correct Answer: B
Solution :
Let\[\overrightarrow{d}={{d}_{1}}\hat{i}+{{d}_{2}}\hat{j}+{{d}_{3}}\hat{k}\] \[\overrightarrow{a}.\overrightarrow{b}={{d}_{1}}-{{d}_{2}}=0\Rightarrow {{d}_{1}}={{d}_{2}}\] ...(i) Also,\[\overrightarrow{d}\]is a unit vector \[\Rightarrow \] \[d_{1}^{2}+d_{2}^{2}+d_{3}^{2}=1\] ...(ii) \[[\overrightarrow{b}\overrightarrow{c}\overrightarrow{d}]=0\Rightarrow \left| \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ {{d}_{1}} & {{d}_{2}} & {{d}_{3}} \\ \end{matrix} \right|=0\] \[\Rightarrow \]\[-1(-{{d}_{3}}-{{d}_{1}})-1(-{{d}_{2}})=0\] \[\Rightarrow \]\[{{d}_{1}}+{{d}_{2}}+{{d}_{3}}=0\Rightarrow 2{{d}_{1}}+{{d}_{3}}=0\] \[\Rightarrow \] \[{{d}_{3}}=-2{{d}_{1}}\] ...(iii) Using Eqs. (iii) and (i) in Eq. (ii) we get \[d_{1}^{2}+d_{1}^{2}+4d_{1}^{2}=1\] \[\Rightarrow \] \[6d_{1}^{2}=1\Rightarrow {{d}_{1}}=\pm \frac{1}{\sqrt{6}}\] \[\Rightarrow \] \[{{d}_{2}}=\pm \frac{1}{\sqrt{6}}\] and \[{{d}_{3}}=\mp \frac{2}{\sqrt{6}}\] \[\therefore \]Required vector is\[\pm \frac{1}{\sqrt{6}}(\hat{i}+\hat{j}-2\hat{k})\]
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