Vectors Question 207
Unit vector perpendicular to vector $ \vec{A},=3\hat{i}+,\hat{j} $ and $ \vec{B},=2\hat{i}-,\hat{j}-5\hat{k} $ is
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 The unit vector in normal direction is $ \hat{n} = \frac{\vec{A}\times ,\vec{B}}{|\vec{A}|,|\vec{B}|,\sin ,\theta } $
Here, $ \vec{A}\times ,\vec{B},=,| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & 1 & 0 \\ 2 & -1 & -5 \\ \end{matrix} | $
$ =-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{33}}{6} $ $ \ Therefore $ $ \hat{n}\pm ,\frac{-5\hat{i}+,15\hat{j}-5\hat{k}}{\sqrt{10}\cdot\sqrt{30}\cdot\frac{\sqrt{11}}{2\sqrt{3}}} $
$ =,\pm ,\frac{(\hat{i}-3\hat{j}+,\hat{k})}{\sqrt{11}} $Show Answer
Answer:
Solution:
BETA
