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Section Formula

Let A and ...

Question

Let $A$ and $B$ be points with position vectors $¯ ¯ ¯ a$ and $¯ ¯ b$ with respect to origin $O$ . If the point $C$ on $O A$ is such that $2 ¯ ¯¯¯¯¯¯ ¯ A C = ¯ ¯¯¯¯¯¯ ¯ C O$ , $¯ ¯¯¯¯¯¯¯ ¯ C D$ is parallel to $¯ ¯¯¯¯¯¯ ¯ O B$ and $| ¯ ¯¯¯¯¯¯¯ ¯ C D | = 3 | ¯ ¯¯¯¯¯¯ ¯ O B |$ then $A D$ is

A

¯ ¯ b - \frac{a}{9}

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B

3 ¯ ¯ b - \frac{a}{3}

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C

¯ ¯ b - \frac{a}{3}

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D

¯ ¯ b + \frac{a}{3}

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Solution

The correct option is B $3 ¯ ¯ b - \frac{a}{3}$
Given that $2 - - \to A C = - - \to C O$
$2 (\to c - \to a) = - \to c$
$3 \to c = 2 \to a$
$\to c = \frac{2 \to a}{3}$
Also given $- - \to C D | | - - \to O B$
$\Rightarrow \to a - \to c = k \to b$
$\Rightarrow ∣ ∣ \to d - \to c ∣ ∣ = 3 ∣ ∣ \to b ∣ ∣$
$\Rightarrow k ∣ ∣ \to b ∣ ∣ = 3 ∣ ∣ \to b ∣ ∣$
$\Rightarrow k = 3$
and $\to d = \to c + 3 \to b$
$\Rightarrow - - \to A D = \to d - \to a$ $= 3 \to b + \to c - \to a$
$\Rightarrow - - \to A D = 3 \to b - \frac{\to a}{3}$

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