Kinematics Question 604
Question: Let two vectors $ \vec{A}=3\hat{i}+\hat{j}+2\hat{k} $
and $ \vec{B}=2\hat{i}-2\hat{j}+4\hat{k} $ . Consider the unit vector perpendicular to both A and b is
Options:
A) $ \frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} $
B)$ \frac{\hat{i}-\hat{j}-\hat{k}}{2\sqrt{3}} $
C) $ \frac{-\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} $
D)$ \frac{\hat{i}-\hat{j}-\hat{k}}{2\sqrt{3}} $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Angle between $ \vec{A} $ and $ \vec{B} $
it’s given by $ \cos \theta =\frac{\vec{A}\text{.}\vec{B}}{\text{AB}}=\frac{3}{\sqrt{21}} $
The unit vector perpendicular to $ \vec{A} $
and $ \vec{B} $ it’s given by $ \hat{n}=\frac{\vec{A}\times \vec{B}}{|\vec{A}\times \vec{B}|}=\frac{(3\hat{i}+\hat{j}+2\hat{k})\times (2\hat{i}-2\hat{j}+4\hat{k})}{|(3\hat{i}+\hat{j}+2\hat{k})\times (2\hat{i}-2\hat{j}+4\hat{k})|} $
$ =\frac{\hat{i}-\hat{j}-\hat{k}}{\sqrt{3}} $
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