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Question
Find the coordinates of the tip of the position vector which is equivalent to \[\vec{A} B\], where the coordinates of A and B are (−1, 3) and (−2, 1) respectively.
Solution
Let O be the origin. Let \[P\left( x, y \right)\] be the required point. Then, \[\overrightarrow{P}\] is the tip of the position vector\[\overrightarrow {OP}\]of the point P.
We have,
\[\vec{OP} = x \stackrel\frown{i} + y\stackrel\frown{j}.\]
and, \[\vec{AB} =\] Position vector of B - Position vector of A
\[= \left( - 2 \stackrel\frown{i} + \stackrel\frown{j} \right) - \left( - \stackrel\frown{i} + 3 \stackrel\frown{j} \right)\]
\[ = - 2 \stackrel\frown{i} + \stackrel\frown{j} + \stackrel\frown{i} - 3 \stackrel\frown{j}\]
\[ = - \stackrel\frown{i} - 2 \stackrel\frown{j}\]
Given that \[\vec{OP} = \vec{AB}\]
So,
\[x \stackrel\frown{i} + y \stackrel\frown{j} = - \stackrel\frown{i} - 2 \stackrel\frown{j} \Leftrightarrow x = - 1 , y = - 2\]
Hence, coordinated of the required point is \[\left( - 1 . - 2 \right)\]
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