Kinematics Question 441
Question: Unit vector perpendicular to vector $ \vec{A}=3\hat{i}+\hat{j} $ and $ \vec{B}=2\hat{i}-\hat{j}-5\hat{k} $ both it’s
Options:
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 [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}= \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & 0 \\ 2 & -1 & -5 \\ \end{vmatrix} $ $ =-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}} $Show Answer
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