Vector Algebra Question 218
Question: In a trapezium, the vector $ \overrightarrow{BC}=\lambda \overrightarrow{AD}. $ We will then find that $ \mathbf{p}=\overrightarrow{AC}+\overrightarrow{BD} $ is collinear with $ \overrightarrow{AD}, $ If $ \mathbf{p}=\mu \overrightarrow{AD}, $ then
Options:
A) $ \mu =\lambda +1 $
B) $ \lambda =\mu +1 $
C) $ \lambda +\mu =1 $
D) $ \mu =2+\lambda $
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Answer:
Correct Answer: A
Solution:
- We have, $ \mathbf{p}=\overrightarrow{AC}+\overrightarrow{BD}=\overrightarrow{AC}+\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{AC}+\lambda \overrightarrow{AD}+\overrightarrow{CD} $ $ =\lambda ,\overrightarrow{AD}+(\overrightarrow{AC}+\overrightarrow{CD})=\lambda ,\overrightarrow{AD}+\overrightarrow{AD}=(\lambda +1)\overrightarrow{AD}. $ Therefore $ \mathbf{p}=\mu \overrightarrow{AD}\Rightarrow \mu =\lambda +1. $
BETA
