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If → a Be the Position Vector Whose Tip is (5, −3), Find the Coordinates of a Point B Such that −−→ a B = → a , the Coordinates of a Being (4, −1). - Mathematics

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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).

Sum
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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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Position Vector of a Point Dividing a Line Segment in a Given Ratio
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Chapter 23: Algebra of Vectors - Exercise 23.4 [Page 43]

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RD Sharma Mathematics [English] Class 12
Chapter 23 Algebra of Vectors
Exercise 23.4 | Q 8 | Page 43

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