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Find the Coordinates of the Tip of the Position Vector Which is Equivalent to → a B , Where the Coordinates of a and B Are (−1, 3) and (−2, 1) Respectively.

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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.

Answer in Brief
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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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Position Vector of a Point Dividing a Line Segment in a Given Ratio
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Chapter 22: Algebra of Vectors - Exercise 23.4 [Page 42]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 22 Algebra of Vectors
Exercise 23.4 | Q 5 | Page 42
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