In a quadrilateral ABCD,the point P divides D | vedclass

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(D) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Since $P$ divides $DC$ in the ratio $1:3$ i

Vedclass · Accepted Answer

-$4$

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(D) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Since $P$ divides $DC$ in the ratio $1:3$ internally,the position vector of $P$ is $\vec{p} = \frac{3\vec{d} + 1\vec{c}}{1+3} = \frac{3\vec{d} + \vec{c}}{4}$.Since $Q$ is the mid-point of $AC$,the position vector of $Q$ is $\vec{q} = \frac{\vec{a} + \vec{c}}{2}$.Now,$\vec{PQ} = \vec{q} - \vec{p} = \frac{\vec{a} + \vec{c}}{2} - \frac{3\vec{d} + \vec{c}}{4} = \frac{2\vec{a} + 2\vec{c} - 3\vec{d} - \vec{c}}{4} = \frac{2\vec{a} + \vec{c} - 3\vec{d}}{4}$.Given expression: $\vec{AB} + \vec{AD} + \vec{BC} - 2\vec{DC} = (\vec{b} - \vec{a}) + (\vec{d} - \vec{a}) + (\vec{c} - \vec{b}) - 2(\vec{c} - \vec{d})$.Simplifying this: $\vec{b} - \vec{a} + \vec{d} - \vec{a} + \vec{c} - \vec{b} - 2\vec{c} + 2\vec{d} = -2\vec{a} - \vec{c} + 3\vec{d} = -(2\vec{a} + \vec{c} - 3\vec{d})$.Comparing with $\lambda \vec{PQ}$:$-(2\vec{a} + \vec{c} - 3\vec{d}) = \lambda \left( \frac{2\vec{a} + \vec{c} - 3\vec{d}}{4} \right)$.Thus,$\lambda = -4$.

In a quadrilateral $ABCD$,the point $P$ divides $DC$ in the ratio $1:3$ internally and $Q$ is the mid-point of $AC$. If $\vec{AB} + \vec{AD} + \vec{BC} - 2\vec{DC} = \lambda \vec{PQ}$,then the value of $\lambda$ is

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