Vector Algebra Question 370
Question: The angle between the vector $ 2i+3j+k $ and $ 2i-j-k $ is
[MNR 1990; UPSEAT 2000]
Options:
A) $ \pi /2 $
B) $ \pi /4 $
C) $ \pi /3 $
D) 0
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Answer:
Correct Answer: A
Solution:
- Let $ \mathbf{a}=2\mathbf{i}+3\mathbf{j}+\mathbf{k} $ and $ \mathbf{b}=2\mathbf{i}-\mathbf{j}+\mathbf{k} $ Since $ \cos ,\theta =\frac{\mathbf{a},.\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} $ $ =\frac{(2\mathbf{i}+3\mathbf{j}+\mathbf{k}),.,(2\mathbf{i}-\mathbf{j}-\mathbf{k}),}{\sqrt{{{(2)}^{2}}+{{(3)}^{2}}+{{(1)}^{2}}}\sqrt{{{(2)}^{2}}+{{(-1)}^{2}}+{{(-1)}^{2}}}} $ $ =\frac{4-3-1}{\sqrt{(4+9+1)}\sqrt{(4+1+1)}}=0 $
$ \therefore \theta =\frac{\pi }{2} $ .
BETA
