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Question: If the position vector of one end of the line segment AB be \(2\mathbf{i} + 3\mathbf{j} - \mathbf{k}...

If the position vector of one end of the line segment AB be 2i+3j−k2\mathbf{i} + 3\mathbf{j} - \mathbf{k} and the position vector of its middle point be 3(i+j+k),3(\mathbf{i} + \mathbf{j} + \mathbf{k}), then the position vector of the other end is

A

4i+3j+5k4\mathbf{i} + 3\mathbf{j} + 5\mathbf{k}

B

4i−3j+7k4\mathbf{i} - 3\mathbf{j} + 7\mathbf{k}

C

4i+3j+7k4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}

D

4i+3j−7k4\mathbf{i} + 3\mathbf{j} - 7\mathbf{k}

Answer

4i+3j+7k4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}

Explanation

Solution

OA→=2i+3j−k,\overset{\rightarrow}{OA} = 2\mathbf{i} + 3\mathbf{j} - \mathbf{k}, OP→=3(i+j+k),\overset{\rightarrow}{OP} = 3(\mathbf{i} + \mathbf{j} + \mathbf{k}), OB→=?\overset{\rightarrow}{OB} = ?

we have OP→=OA→+OB→2\overset{\rightarrow}{OP} = \frac{\overset{\rightarrow}{OA} + \overset{\rightarrow}{OB}}{2}

⇒OB→=2OP→−OA→\Rightarrow \overset{\rightarrow}{OB} = 2\overset{\rightarrow}{OP} - \overset{\rightarrow}{OA}

=4i+3j+7k= 4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k}

Trick : By inspection, middle point of 4i+3j+7k4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k} and 2i+3j−k2\mathbf{i} + 3\mathbf{j} - \mathbf{k} is 3(i+j+k).3(\mathbf{i} + \mathbf{j} + \mathbf{k}).