If $\to a$ and $\to b$ are position vectors of $A$ and $B$ respectively, then the position vector of a point $C$ in $A B$ produced such that $- - \to A C = 3 - - \to A B$ is

3 \to a - \to b

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3 \to b - \to a

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3 \to a - 2 \to b

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3 \to b - 2 \to a

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Solution

The correct option is D $3 \to b - 2 \to a$
$\frac{A C}{A B} = \frac{3}{1}$
$B C = A C - A B$
$= 3 - 1$
$= 2$
$\frac{A C}{B C} = \frac{3}{2}$
$∴ b^{- 1} = \frac{1 \times \to c + 2 \to a}{1 + 2}$
$∴ \frac{A B}{B C} = \frac{1}{2}$ $\Rightarrow \to b = \frac{\to c + 2 \to a}{3}$
$∴ \to c = 3 \to b - 2 \to a$ .

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