If O is any point and OA + OB + OC + OD = xOE | vedclass

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(A) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Let $P, Q, R, S$ be the mid-points of sides

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the position vectors of vertices $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively.Let $P, Q, R, S$ be the mid-points of sides $AB, BC, CD, DA$ respectively.Then $P = \frac{\vec{a}+\vec{b}}{2}$,$Q = \frac{\vec{b}+\vec{c}}{2}$,$R = \frac{\vec{c}+\vec{d}}{2}$,$S = \frac{\vec{d}+\vec{a}}{2}$.The mid-point of $PR$ is $\frac{P+R}{2} = \frac{\frac{\vec{a}+\vec{b}}{2} + \frac{\vec{c}+\vec{d}}{2}}{2} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.The mid-point of $SQ$ is $\frac{S+Q}{2} = \frac{\frac{\vec{d}+\vec{a}}{2} + \frac{\vec{b}+\vec{c}}{2}}{2} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.Since the mid-points of $PR$ and $SQ$ coincide,let this point be $E$.Thus,the position vector of $E$ is $\vec{e} = \frac{\vec{a}+\vec{b}+\vec{c}+\vec{d}}{4}$.This implies $4\vec{e} = \vec{a}+\vec{b}+\vec{c}+\vec{d}$.For any point $O$ with position vector $\vec{o}$,we have $\vec{OA} = \vec{a}-\vec{o}$,$\vec{OB} = \vec{b}-\vec{o}$,$\vec{OC} = \vec{c}-\vec{o}$,$\vec{OD} = \vec{d}-\vec{o}$,and $\vec{OE} = \vec{e}-\vec{o}$.Then $\vec{OA} + \vec{OB} + \vec{OC} + \vec{OD} = (\vec{a}+\vec{b}+\vec{c}+\vec{d}) - 4\vec{o} = 4\vec{e} - 4\vec{o} = 4(\vec{e}-\vec{o}) = 4\vec{OE}$.Comparing this with $x\vec{OE}$,we get $x = 4$.

If $O$ is any point and $OA + OB + OC + OD = xOE$,where $ABCD$ is a quadrilateral and $E$ is the point of intersection of the lines joining the mid-points of opposite sides,then find the value of $x$.

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