Vector Algebra Question 156
Question: The vector $ \mathbf{a}+\mathbf{b} $ bisects the angle between the vectors a and b, if
Options:
A) $ |\mathbf{a}|,=,|\mathbf{b}| $
B) $ |\mathbf{a}|,=,|\mathbf{b}| $ or angle between a and b is zero
C) $ |\mathbf{a}|=m,|\mathbf{b}| $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Since the angle between $ \mathbf{a}+\mathbf{b} $ and $ \mathbf{a} $ and the angle between $ \mathbf{a}+\mathbf{b} $ and $ \mathbf{b} $ are the same, so we have
$ \frac{(\mathbf{a}+\mathbf{b}),.,\mathbf{a}}{|\mathbf{a}+\mathbf{b}|,|\mathbf{a}|}=\frac{(\mathbf{a}+\mathbf{b}),.,\mathbf{b}}{|\mathbf{a}+\mathbf{b}|,|\mathbf{b}|} $
$ \Rightarrow \frac{|\mathbf{a}{{|}^{2}}}{|\mathbf{a}+\mathbf{b}|,|\mathbf{a}|}+\frac{\mathbf{b},.,\mathbf{a}}{|\mathbf{a}+\mathbf{b}|,|\mathbf{a}|}=\frac{\mathbf{a},.,\mathbf{b}}{|\mathbf{a}+\mathbf{b}|,|\mathbf{b}|}+\frac{|\mathbf{b}{{|}^{2}}}{|\mathbf{a}+\mathbf{b}|,|\mathbf{b}|} $
$ \Rightarrow \frac{|\mathbf{a}|-|\mathbf{b}|}{|\mathbf{a}+\mathbf{b}|}( 1-\frac{\mathbf{a},.,\mathbf{b}}{|\mathbf{a}|,|\mathbf{b}|} )=0 $
Hence $ |\mathbf{a}|,=,|\mathbf{b}| $ or angle between $ \mathbf{a} $ and $ \mathbf{b} $ is 0.
BETA
