A) \[\pm \,\frac{3j-2\hat{k}}{\sqrt{11}}\]
B) \[\pm \,\frac{(\hat{i}-3\hat{j}+\,\hat{k})}{\sqrt{11}}\]
C) \[\pm \,\frac{-\hat{j}+2\,\hat{k}}{\sqrt{13}}\]
D) \[\pm \,\frac{\hat{i}+3\hat{j}-\,\hat{k}}{\sqrt{13}}\]
Correct Answer: B
Solution :
| [b] The unit vector in normal direction is \[\hat{n}\pm \,\frac{\vec{A}\times \,\vec{B}}{|\vec{A}|\,|\vec{B}|\,\sin \,\theta }\] |
| Here, \[\vec{A}\times \,\vec{B}\,=\,\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & 0 \\ 2 & -1 & -5 \\ \end{matrix} \right|\] |
| \[=-5\hat{i}+15\hat{j}-5\hat{k}\] |
| \[|\vec{A}|=\,\sqrt{{{3}^{2}}+{{1}^{2}}}\,=\sqrt{10}\] |
| \[|\vec{B}|=\,\sqrt{{{(2)}^{2}}\,+\,{{(-1)}^{2}}+\,{{(-5)}^{2}}}=\sqrt{30}\] |
| \[\cos \,\theta \,=\frac{\vec{A}\,.\vec{B}}{|\vec{A}|\,|\vec{B}|}=\frac{1}{2\sqrt{3}}\] |
| \[\therefore \] \[\sin \,\theta =\,\sqrt{1-\,{{\cos }^{2}}\theta }\,=\frac{\sqrt{11}}{2\sqrt{3}}\] |
| \[\therefore \] \[\hat{n}\,\,\pm \,\frac{-5\hat{i}+\,15\hat{j}-5\hat{k}}{\sqrt{10}\,.\,\sqrt{30}\,.\,\frac{\sqrt{11}}{2\sqrt{3}}}\] |
| \[=\,\pm \,\frac{(\hat{i}-3\hat{j}+\,\hat{k})}{\sqrt{11}}\] |
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