Vector Algebra Question 158
Question: The points $ O,,A,,B,,C,,D $ are such that $ \overrightarrow{OA}=\mathbf{a}, $ $ \overrightarrow{OB}=\mathbf{b},, $ $ \overrightarrow{OC}=2\mathbf{a}+3\mathbf{b} $ and $ \overrightarrow{OD}=\mathbf{a}-2\mathbf{b}. $ If $ |\mathbf{a}|,=3,|\mathbf{b}|, $ then the angle between $ \overrightarrow{BD} $ and $ \overrightarrow{AC} $ is
Options:
A) $ \frac{\pi }{3} $
B) $ \frac{\pi }{4} $
C) $ \frac{\pi }{6} $
D) None of these
Show Answer
Answer:
Correct Answer: D
Solution:
- We have $ \overrightarrow{BD}=\overrightarrow{OD}-\overrightarrow{OB}=\mathbf{a}-2\mathbf{b}-\mathbf{b}=\mathbf{a}-3\mathbf{b} $ and $ \overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=2\mathbf{a}+3\mathbf{b}-\mathbf{a}=\mathbf{a}+3\mathbf{b}. $
Let $ \theta $ be the angle between $ \overrightarrow{BD} $ and $ \overrightarrow{AC}. $
Then $ \cos \theta =\frac{\overrightarrow{BD},.,\overrightarrow{AC}}{|\overrightarrow{BD}|,|\overrightarrow{AC}|}=\frac{|\mathbf{a}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|,|\overrightarrow{AC}|} $
$ =\frac{9|\mathbf{b}{{|}^{2}}-9|\mathbf{b}{{|}^{2}}}{|\overrightarrow{BD}|,|\overrightarrow{AC}|} $ ,
$ (\because ,| ,\mathbf{a}, |=3|\mathbf{b}|) $
$ \Rightarrow \cos \theta =0{}^\circ \Rightarrow \theta =\frac{\pi }{2}. $
BETA
