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Definition of Hyperbola

In a parallel...

Question

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

The position vector of point p is

$\frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} {\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}}$

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$\frac{| \to a | | \to c |}{3 | \to c | + 2 | \to a |} {\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}}$

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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 A
$\frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} {\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}}$

Since E divides BC in the ratio of 2 : 1,

Therefore, position vector of

$E = \frac{\to b + 2 \to c}{3} . . . . . . . . (i)$

But, $\to O A + \to A B = \to O B$

$\Rightarrow \to a + \to c = \to b . . . . . . . . (i i)$

$∴$ Position vector of $E = \frac{\to a + 3 \to c}{3},$ ...... from (i) and (ii)

The vector equation of the bisector of $∠ A O C i s \to r = λ {\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}} . . . . . (i i i)$

The vector equation of AE is $\to r = \to a + μ {\frac{\to a + 3 \to c}{3} - \to a}$

$o r \to r = \to a + μ {\frac{3 \to c - 2 \to a}{3}} . . . . . . . (i v)$

Lines (iii) and (iv) intersect at P.

$∴$ For point P, we must have $λ {\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}} = \to a + μ {\frac{3 \to c - 2 \to a}{3}}$

$\Rightarrow {\frac{λ}{| \to a |} - 1 + \frac{2 μ}{3}} \to a + {\frac{λ}{| \to c |} - μ} \to c = \to 0$

$\Rightarrow \frac{λ}{| \to a |} - 1 + \frac{2 μ}{3} = 0 a n d \frac{λ}{| \to c |} - μ = 0 [∵ \to a a n d \to c are non-collinear]$

$\Rightarrow λ = \frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} a n d μ = \frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |}$

Putting the value of $λ$ in (iii) or that of $μ$ in (iv), we obtain the position vector of P as

$_{1} = \frac{3 | \to a | | \to c |}{3 | \to c | + 2 | \to a |} {\frac{\to a}{| \to a |} + \frac{\to c}{| \to c |}}$

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