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Question
If \[\vec{a,} \vec{b}\] are the position vectors of A, B respectively, find the position vector of a point C in AB produced such that AC = 3 AB and that a point D in BA produced such that BD = 2BA.
Solution
Let the position vectors of C and D are \[\vec{c}\] and \[\vec{d}\] respectively. We have,
\[AC = 3 AB . \]
\[ \Rightarrow AC = 3 (AC - BC) . \]
\[ \Rightarrow 2AC = 3BC .\]
\[\Rightarrow \frac{AC}{BC} = \frac{3}{2} .\]
So C divides AB in the ratio of \[3: 2\] externally.
\[\vec{c} = \frac{2 \vec{a} - 3 \vec{b}}{2 - 3} = 3 \vec{b} - 2 \vec{a} .\]
Position vector of point C is \[3 \vec{b} - 2 \vec{a}\] Moreover,
\[BD = 2 BA . \]
\[ \Rightarrow BD = 2(BD - AD) . \]
\[ \Rightarrow BD = 2AD .\]
\[\Rightarrow \frac{BD}{AD} = \frac{2}{1}\].
∴ \[\vec{d} = \frac{\vec{b} - 2 \vec{a}}{1 - 2} = 2 \vec{a} - \vec{b} .\]
Position vector of point D is \[2 \vec{a} - \vec{b}\]
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