Vector Algebra Question 243
Question: If the position vector of one end of the line segment AB be $ 2\mathbf{i}+3\mathbf{j}-\mathbf{k} $ and the position vector of its middle point be $ 3,(\mathbf{i}+\mathbf{j}+\mathbf{k}), $ then the position vector of the other end is
Options:
A) $ 4\mathbf{i}+3\mathbf{j}+5\mathbf{k} $
B) $ 4\mathbf{i}-3\mathbf{j}+7\mathbf{k} $
C) $ 4\mathbf{i}+3\mathbf{j}+7\mathbf{k} $
D) $ 4\mathbf{i}+3\mathbf{j}-7\mathbf{k} $
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Answer:
Correct Answer: C
Solution:
- $ \overrightarrow{OA}=2\mathbf{i}+3\mathbf{j}-\mathbf{k}, $ $ \overrightarrow{OP}=3(\mathbf{i}+\mathbf{j}+\mathbf{k}), $ $ \overrightarrow{OB}=? $ we have $ \overrightarrow{OP}=\frac{\overrightarrow{OA}+\overrightarrow{OB}}{2} $
$ \Rightarrow \overrightarrow{OB}=2\overrightarrow{OP}-\overrightarrow{OA} $ $ =4\mathbf{i}+3\mathbf{j}+7\mathbf{k} $ Trick : By inspection, middle point of $ 4\mathbf{i}+3\mathbf{j}+7\mathbf{k} $ and $ 2\mathbf{i}+3\mathbf{j}-\mathbf{k} $ is $ 3,(\mathbf{i}+\mathbf{j}+\mathbf{k}). $
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