A) \[\pm \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}}\]
B) \[\pm \frac{\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\]
C) \[\pm \frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}\]
D) \[\pm \,\hat{k}\]
Correct Answer: A
Solution :
[a] Let \[\vec{d}=x\hat{i}+y\hat{j}+z\hat{k}\] Where \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=1\] ?(i) \[\therefore \vec{a}\cdot \vec{d}=0\Rightarrow x-y=0\] or \[x=y\] ?(ii) \[[\vec{b}\,\vec{c}\,\vec{d}]=0\Rightarrow \left| \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ x & y & z \\ \end{matrix} \right|=0\] Or \[x+y+z=0\] Or \[2x+z=0\] [Using (ii)] Or z = - 2x ?(iii) From (i), (ii) and (iii), we have \[{{x}^{2}}+{{x}^{2}}+4{{x}^{2}}=1\] \[\Rightarrow x=\pm \frac{1}{\sqrt{6}}\] \[\therefore \vec{d}=\pm \left( \frac{1}{\sqrt{6}}\hat{i}+\frac{1}{\sqrt{6}}\hat{j}-\frac{2}{\sqrt{6}}\vec{k} \right)=\pm \left( \frac{\hat{i}+\hat{j}-2\hat{k}}{\sqrt{6}} \right)\]
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