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Question
If \[\vec{a}\] be the position vector whose tip is (5, −3), find the coordinates of a point B such that \[\overrightarrow{AB} =\] \[\vec{a}\], the coordinates of A being (4, −1).
Solution
Let O be the origin and let \[P\left( 5, - 3 \right)\] be the tip of the position vector \[\vec{a}\]. Then, \[\vec{a} = \overrightarrow {OP} = 5 \hat{i}-3 \hat{j}.\] Let the coordinate of B be \[\left( x, y \right)\] and A has coordinates \[\left( 4, - 1 \right)\]
Therefore,
\[\overrightarrow{AB}\] = Position vector of B - Position vector of A
\[= \left( x \hat{i} + y \hat{j} \right) - \left( 4 \hat{i} - \hat{j} \right)\]
\[ = \left( x - 4 \right) \hat{i} + \left( y + 1 \right) \bar{j}\]
Now,
\[\overrightarrow {AB} = \vec{a} \]
\[ \Rightarrow \left( x - 4 \right) \hat{i} + \left( y + 1 \right) \hat{j} = 5 \hat{i} - 3 \hat{j} \]
\[ \Rightarrow x - 4 = 5\text{ and }y + 1 = - 3\]
\[ \Rightarrow x = 9\text{ and }y = - 4\]
Hence, the coordinates of B are \[\left( 9, - 4 \right)\].
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