Vector Algebra Question 259
Question: Let A and B be points with position vectors a and b with respect to the origin O. If the point C on OA is such that $ 2AC=CO,CD $ is parallel to OB and $ |\overrightarrow{CD}|=3|\overrightarrow{OB}|, $ then $ \overrightarrow{AD} $ is equal to
Options:
A) $ 3\mathbf{b}-\frac{\mathbf{a}}{2} $
B) $ 3\mathbf{b}+\frac{\mathbf{a}}{2} $
C) $ 3\mathbf{b}-\frac{\mathbf{a}}{3} $
D) $ 3\mathbf{b}+\frac{\mathbf{a}}{3} $
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Answer:
Correct Answer: C
Solution:
- Since $ \overrightarrow{OA}=\mathbf{a}, $ $ \overrightarrow{OB}=\mathbf{b} $ and $ 2AC=CO $ By section formula $ \overrightarrow{OC}=\frac{2}{3}\mathbf{a}. $ Therefore, $ |\overrightarrow{CD}|=3|\overrightarrow{OB}|,\Rightarrow \overrightarrow{CD}=3\mathbf{b} $
$ \Rightarrow \overrightarrow{OD}=\overrightarrow{OC}+\overrightarrow{CD}=\frac{2}{3}\mathbf{a}+3\mathbf{b} $ Hence, $ \overrightarrow{AD}=\overrightarrow{OD}-\overrightarrow{OA}=\frac{2}{3}\mathbf{a}+3\mathbf{b}-\mathbf{a}=3\mathbf{b}-\frac{1}{3}\mathbf{a}. $
BETA
