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

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(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.Using the triangle law of vector additio

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-$6$

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(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.Using the triangle law of vector addition,$\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$.So,the expression becomes: $\overrightarrow{AC}+2\overrightarrow{AD}-2\overrightarrow{DC}$.Since $\overrightarrow{AD}+\overrightarrow{DC} = \overrightarrow{AC}$,we have $\overrightarrow{AD} = \overrightarrow{AC}-\overrightarrow{DC}$.Substituting this: $\overrightarrow{AC}+2(\overrightarrow{AC}-\overrightarrow{DC})-2\overrightarrow{DC} = 3\overrightarrow{AC}-4\overrightarrow{DC}$.Given $Q$ is the midpoint of $AC$,$\overrightarrow{AC} = 2\overrightarrow{QC}$.Given $P$ divides $DC$ in ratio $1:2$,$\overrightarrow{DP} = \frac{1}{3}\overrightarrow{DC}$ and $\overrightarrow{PC} = \frac{2}{3}\overrightarrow{DC}$,so $\overrightarrow{DC} = \frac{3}{2}\overrightarrow{PC}$.Substituting these: $3(2\overrightarrow{QC})-4(\frac{3}{2}\overrightarrow{PC}) = 6\overrightarrow{QC}-6\overrightarrow{PC} = 6(\overrightarrow{QC}+\overrightarrow{CP}) = 6\overrightarrow{QP}$.Since $\overrightarrow{QP} = -\overrightarrow{PQ}$,we have $6\overrightarrow{QP} = -6\overrightarrow{PQ}$.Thus,$k\overrightarrow{PQ} = -6\overrightarrow{PQ}$,which implies $k = -6$.

In a quadrilateral $ABCD$,the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$,then $k$ is equal to

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