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Acceleration in 2D

If a and bare...

Question

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

A

3 \to a - \to b

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B

3 \to b - \to a

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C

3 \to a - 2 \to b

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D

3 \to b - 2 \to a

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Solution

The correct option is D $3 \to b - 2 \to a$
Let position vector of $C$ is $\to c$
Given $- - \to A C = 3 - - \to A B$
$\Rightarrow \to c - \to a = 3 (\to b - \to a)$
$\Rightarrow \to c = 3 \to b - 2 \to a$

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Acceleration in 2D

Standard XII Physics

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