Vector Algebra Question 322
Question: If a, b, c are mutually perpendicular vectors of equal magnitudes, then the angle between the vectors a and $ \mathbf{a}+\mathbf{b}+\mathbf{c} $ is
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{6} $
C) $ {{\cos }^{-1}}\frac{1}{\sqrt{3}} $
D) $ \frac{\pi }{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Since $ \mathbf{a},\mathbf{b} $ and $ \mathbf{c} $ are mutually perpendicular, so $ \mathbf{a},.,\mathbf{b}=\mathbf{b},.,\mathbf{c}=\mathbf{c},.,\mathbf{a}=0 $ Angle between $ \mathbf{a} $ and $ \mathbf{a}+\mathbf{b}+\mathbf{c} $ is $ \cos \theta =\frac{\mathbf{a}.(\mathbf{a}+\mathbf{b}+\mathbf{c})}{|\mathbf{a}||\mathbf{a}+\mathbf{b}+\mathbf{c}|} $ …..(i) Now $ |\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=a $ $ |\mathbf{a}+\mathbf{b}+\mathbf{c}{{|}^{2}}={{\mathbf{a}}^{2}}+{{\mathbf{b}}^{2}}+{{\mathbf{c}}^{2}}+2,\mathbf{a},.,\mathbf{b}+2,\mathbf{b},.,\mathbf{c}+2,\mathbf{c},.,\mathbf{a} $ $ =a^{2}+a^{2}+a^{2}+0+0+0 $
Þ $ |\mathbf{a}+\mathbf{b}+\mathbf{c}{{|}^{2}}=3a^{2}\Rightarrow |\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{3}a $ Putting this value in (i), we get $ \theta ={{\cos }^{-1}}\frac{1}{\sqrt{3}}. $
BETA
