A) \[\pm \frac{\hat{j}-\hat{j}}{\sqrt{2}}\]
B) \[\pm \frac{\hat{j}+\hat{k}}{\sqrt{2}}\]
C) \[\pm \frac{\hat{j}-\hat{k}}{\sqrt{2}}\]
D) \[\pm \frac{\hat{i}+\hat{j}}{\sqrt{2}}\]
Correct Answer: D
Solution :
Let \[\vec{a}=-\hat{i}-\hat{j}+\hat{k},\,\vec{b}=-\hat{i}+\hat{j}+\hat{k}\]\[\underrightarrow{A}\] unit vector perpendicular to the plane a and b \[=\pm \frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}\] Now, \[\vec{a}\times \vec{b}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & -1 & 1 \\ -1 & 1 & 1 \\ \end{matrix} \right|\] \[=\,\hat{i}(-1-1)-\hat{j}\,(1+1)\] \[+\,\,\hat{k}\,\,(1-1)\] \[=-2\,\hat{i}-2\,\hat{j}\] \[|\vec{a}\times \vec{b}|=\sqrt{{{2}^{2}}+{{2}^{2}}}=2\sqrt{2}\] \[\therefore \] \[\frac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}=\mp \frac{2\,(\hat{i}+\hat{j})}{2\sqrt{2}}=\mp \frac{(\hat{i}+\hat{j})}{\sqrt{2}}\]
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