OABC is a parallelogram, with O as origin and position vectors of A,C are respectively 3i¯+4j¯,4i¯+3j¯.Point E is on side BC, dividing it in the ratio 2:1. The angular bisector of ∠AOC meets AE at P then position vector of P is

|OA¯|=|OC¯|=5⇒ Rhombus Angular Bisector=OA¯+OB¯=7i¯+7j¯ OP¯=λ(i¯+j¯)E devides BC in the ratio's 2:1 ,then E5,133Then use A(3,4),P(λ,λ)and E5,133 are collinear. ⇒λ=215

OABC is a parallelogram, with O as origin and position vectors of A,C are respectively 3i¯+4j¯,4i¯+3j¯.Point E is on side BC, dividing it in the ratio 2:1. The angular bisector of ∠AOC meets AE at P then position vector of P is

|OA¯|=|OC¯|=5⇒ Rhombus Angular Bisector=OA¯+OB¯=7i¯+7j¯ OP¯=λ(i¯+j¯)E devides BC in the ratio's 2:1 ,then E5,133Then use A(3,4),P(λ,λ)and E5,133 are collinear. ⇒λ=215

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OABC is a parallelogram, with O as origin and position vectors of A,C are respectively $3 \bar{i} + 4 \bar{j}, 4 \bar{i} + 3 \bar{j}$ .Point E is on side BC, dividing it in the ratio 2:1. The angular bisector of $∠ AOC$ meets AE at P then position vector of P is

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$\frac{2}{3} (i + j)$

$\frac{13}{3} (i + j)$

i+j

$\frac{21}{5} (i + j)$

answer is D.

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Detailed Solution

$| \bar{OA} | = | \bar{OC} | = 5 \Rightarrow Rhombus$
Angular Bisector $= \bar{OA} + \bar{OB} = 7 \bar{i} + 7 \bar{j}$
$\bar{OP} = λ (\bar{i} + \bar{j})$
E devides BC in the ratio's 2:1 ,then $E (5, \frac{13}{3})$
Then use $A (3, 4), P (λ, λ)$ and $E (5, \frac{13}{3})$ are collinear. $\Rightarrow λ = \frac{21}{5}$