Vector Algebra Question 59
Question: What is the interior acute angle of the parallelogram whose sides are represented by the vectors $ \frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k} $ and $ \frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k} $ ?
Options:
A) $ 60{}^\circ $
B) $ 45{}^\circ $
C) $ 30{}^\circ $
D) $ 15{}^\circ $
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Answer:
Correct Answer: A
Solution:
- [a] Let $ a=\frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k} $ and $ b=\frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k} $
$ \therefore ,\cos \theta =\frac{a.b}{|a||b|} $
$ =\frac{( \frac{1}{\sqrt{2}}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\hat{k} ).( \frac{1}{\sqrt{2}}\hat{i}-\frac{1}{\sqrt{2}}\hat{j}+\hat{k} )}{\sqrt{\frac{1}{2}+\frac{1}{2}+1}\sqrt{\frac{1}{2}+\frac{1}{2}+1}} $
$ =\frac{1}{2}[ \frac{1}{2}-\frac{1}{2}+1 ]=\frac{1}{2}=\cos 60{}^\circ $
$ \therefore \theta =60{}^\circ $
BETA
