If the position vector of one end of the line | vedclass

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(C) Let $\overrightarrow{OA} = 2\hat{i} + 3\hat{j} - \hat{k}$ be the position vector of one end $A$ and $\overrightarrow{OP} = 3\hat{i} + 3\hat{j} + 3

Vedclass · Accepted Answer

$4\hat{i} + 3\hat{j} + 7\hat{k}$

Vedclass · Answer

(C) Let $\overrightarrow{OA} = 2\hat{i} + 3\hat{j} - \hat{k}$ be the position vector of one end $A$ and $\overrightarrow{OP} = 3\hat{i} + 3\hat{j} + 3\hat{k}$ be the position vector of the midpoint $P$ of the line segment $AB$.Let $\overrightarrow{OB}$ be the position vector of the other end $B$.Since $P$ is the midpoint of $AB$,we have the relation:$\overrightarrow{OP} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2}$Multiplying by $2$,we get:$2\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{OB}$$\overrightarrow{OB} = 2\overrightarrow{OP} - \overrightarrow{OA}$Substituting the given vectors:$\overrightarrow{OB} = 2(3\hat{i} + 3\hat{j} + 3\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k})$$\overrightarrow{OB} = (6\hat{i} + 6\hat{j} + 6\hat{k}) - (2\hat{i} + 3\hat{j} - \hat{k})$$\overrightarrow{OB} = (6-2)\hat{i} + (6-3)\hat{j} + (6-(-1))\hat{k}$$\overrightarrow{OB} = 4\hat{i} + 3\hat{j} + 7\hat{k}$Thus,the position vector of the other end is $4\hat{i} + 3\hat{j} + 7\hat{k}$.

If the position vector of one end of the line segment $AB$ is $2\hat{i} + 3\hat{j} - \hat{k}$ and the position vector of its midpoint is $3\,(\hat{i} + \hat{j} + \hat{k}),$ then the position vector of the other end is

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