P is the point of intersection of the diagona | vedclass

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(B) Let $S$ be the origin. Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ respectively.In a parallelogra

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $S$ be the origin. Let the position vectors of $A, B, C,$ and $D$ be $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ respectively.In a parallelogram $ABCD$,the diagonals $AC$ and $BD$ bisect each other at $P$.Therefore,$P$ is the midpoint of $AC$ and also the midpoint of $BD$.Since $P$ is the midpoint of $AC$,the position vector of $P$ is $\vec{p} = \frac{\vec{a} + \vec{c}}{2}$,which implies $\vec{a} + \vec{c} = 2\vec{p}$.Since $P$ is the midpoint of $BD$,the position vector of $P$ is $\vec{p} = \frac{\vec{b} + \vec{d}}{2}$,which implies $\vec{b} + \vec{d} = 2\vec{p}$.Adding these two equations,we get:$(\vec{a} + \vec{c}) + (\vec{b} + \vec{d}) = 2\vec{p} + 2\vec{p} = 4\vec{p}$.Since $S$ is the origin,$\vec{a} = \vec{SA}, \vec{b} = \vec{SB}, \vec{c} = \vec{SC}, \vec{d} = \vec{SD},$ and $\vec{p} = \vec{SP}$.Thus,$\vec{SA} + \vec{SB} + \vec{SC} + \vec{SD} = 4\vec{SP}$.Comparing this with the given equation $\vec{SA} + \vec{SB} + \vec{SC} + \vec{SD} = \lambda \vec{SP}$,we get $\lambda = 4$.

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $S$ is any point in space and $\vec{SA} + \vec{SB} + \vec{SC} + \vec{SD} = \lambda \vec{SP}$,then $\lambda$ equals

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