Vector Algebra Question 368
Question: If a, b and c are unit vectors such that $ \mathbf{a}+\mathbf{b}-\mathbf{c}=0, $ then the angle between a and b is
[Roorkee Qualifying 1998; MP PET 1999; UPSEAT 2000; RPET 2002]
Options:
A) $ \pi /6 $
B) $ \pi /3 $
C) $ \pi /2 $
D) $ 2\pi /3 $
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Answer:
Correct Answer: D
Solution:
- Given condition is $ \mathbf{a}+\mathbf{b}=\mathbf{c}. $ Using dot product, $ (\mathbf{a}+\mathbf{b}).(\mathbf{a}+\mathbf{b})=\mathbf{c}.\mathbf{c} $
$ \Rightarrow \mathbf{a}.\mathbf{a}+\mathbf{b}.\mathbf{b}+2\mathbf{a}.\mathbf{b}=\mathbf{c}.\mathbf{c} $
$ \Rightarrow ,|\mathbf{a}|.|\mathbf{a}|\cos 0{}^\circ +|\mathbf{b}|.|\mathbf{b}|\cos 0{}^\circ +2|\mathbf{a}|.|\mathbf{b}|\cos \alpha $ $ =,|\mathbf{c}|.|\mathbf{c}|\cos 0{}^\circ $ , $ (\because ,|\mathbf{a}|,=,|\mathbf{b}|,=,|\mathbf{c}|,=1) $
$ \Rightarrow 1+1+2\cos \alpha =1\Rightarrow \cos \alpha =-\frac{1}{2}\Rightarrow \alpha =\frac{2\pi }{3} $ .
BETA
