Kinematics Question 292
Question: The area of the parallelogram whose sides are represented by the vectors $ \hat{j}+3\hat{k} $ and $ \hat{i}+2\hat{j}-\hat{k} $ it’s
Options:
A) $ \frac{\hat{i}+10\hat{j}-18\hat{k}}{5\sqrt{17}} $
B) $ \frac{\hat{i}-10\hat{j}+18\hat{k}}{5\sqrt{17}} $
C) $ \frac{\hat{i}-10\hat{j}-18\hat{k}}{5\sqrt{17}} $
D) $ \frac{\hat{i}+10\hat{j}+18\hat{k}}{5\sqrt{17}} $
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Answer:
Correct Answer: C
Solution:
$ \vec{A}=2\hat{i}+2\hat{j}-\hat{k} $ and $ \vec{B}=6\hat{i}-3\hat{j}+2\hat{k} $
$ \vec{C}=\vec{A}\times \vec{B}=( 2\hat{i}+2\hat{j}-\hat{k} )\times ( 6\hat{i}-3\hat{j}+2\hat{k} ) $
$ = \begin{vmatrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 2 & 2 & -1 \\ 6 & -3 & 2 \\ \end{vmatrix} $
$ =\hat{i}-10\hat{j}-18\hat{k} $ Unit vector perpendicular to both $ \vec{A} $ and $ \vec{B} $
$ =\frac{\hat{i}-10\hat{j}-18\hat{k}}{\sqrt{1^{2}+10^{2}+18^{2}}} $
$ =\frac{\hat{i}-10\hat{j}-18\hat{k}}{5\sqrt{17}} $
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