In a parallelogram $O A B C,$ vectors $\to a, \to b, \to c$ are, respectively, the position vectors of vertices $A, B, C$ with reference to $O$ as origin. A point $E$ is taken on the side $B C$ which divides it in the ratio of $2 : 1$ . Also, the line segment $A E$ intersects the line bisecting the angle $∠$ AOC internally at point $P$ . If $C P$ when extended meets $A B$ in point $F,$ then the position vector of point $P$ is

\frac{| \to a | | \to c |}{3 | \to c | + 2 | \to a |} (\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |})

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\frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} (\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |})

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\frac{2 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} (\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |})

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None of these

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Solution

The correct option is C $\frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} (\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |})$
Let the position vector of $A$ and $C$ be $\to a$ and $\to c$ respectively. Therefore,
Position vector of $B = \to b = \to a + \to c$ ........................................................... $(i)$
Also Position vector of $E = \frac{\to b + 2 \to c}{3} = \frac{\to a + 3 \to c}{3}$ ............................................................ $(i i)$
Now point $P$ lies on angle bisector of $∠ A O C .$ Thus,
Position vector of point $P = λ (\frac{\to a}{| \to a |} + \frac{\to b}{| \to b |})$ ............................................................................ $(i i i)$
Also let $P$ divides $E A$ in ration $μ : 1$ . Therefore,
Position vector of $P$
$= \frac{μ \to a + \frac{\to a + 3 \to c}{3}}{μ + 1} = \frac{(3 μ + 1) \to a + 3 \to c}{3 (μ + 1)}$ ................................................................................ $(i v)$
Comparing $(i i i)$ and $(i v),$ we get
$λ (\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}) = \frac{(3 μ + 1) \to a + 3 \to c}{3 (μ + 1)}$
$\Rightarrow \frac{λ}{| \to a |} = \frac{3 μ + 1}{3 (μ + 1)}$ and $\frac{λ}{| \to c |} = \frac{1}{μ + 1}$
$\Rightarrow \frac{3 | \to c | - | \to a |}{3 | \to a |} = μ$
$\Rightarrow \frac{λ}{| \to c |} = \frac{1}{\frac{3 | \to c | - \to a}{3 | \to a |} + 1}$
$\Rightarrow λ = \frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |}$
Hence, position vector of $P$ is $\frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to c |} (\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |})$
Let $F$ divides $A B$ in ratio $t : 1,$ then position vector of $F$ is $\frac{t \to b + \to a}{t + 1}$

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In a parallelogram OABC, vectors $\to a, \to b, \to c$ are respectively the position vectors of vertices A, B, C with reference to O is origin. A point $\to E$ is taken on the side BC which divides it in the ratio of 2 : 1. Also, the line segment AE intersects the line bisecting the angel O internally in point P. If CP, when extended meets AB in point F. Then