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Let A,B,C be vertices of a triangle ABC ...

Let A,B,C be vertices of a triangle ABC in which B is taken as origin of reference and position vectors of A and C are `veca and vecc` respectively. A line AR parallel to BC is drawn from A PR (P is the mid point of AB) meets AC and Q and area of triangle ACR is 2 times area of triangle ABC Position vector of R in terms `veca and vecc` is (A) `veca+2vecc` (B) `veca+3vecc` (C) `veca+vecc` (D) `veca+4vecc`

A

`vec(a)+2vec( c )`

B

`vec(a)+3vec( c )`

C

`vec(a)+vec( c )`

D

`vec(a)+4vec( c )`

Text Solution

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The correct Answer is:
A
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Knowledge Check

  • The triangle is formed by joining the mid-points of the sides AB,BC and CA of triangle ABC and the area of triangle PQR is 6cm^2 , then the area of triangle ABC is :

    A
    `36cm^2`
    B
    `12cm^2`
    C
    `18cm^2`
    D
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  • In a triangle ABC, it is known that AB=AC. Suppose D is the mid-point of AC and BD=BC=2. Then the area of the triangle ABC is-

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    ` 1/2 |veca xx vecb +vecb xx vecc +vecc xx veca|`
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