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In a quadrilateral $A B C D$, the point $P$ divides $D C$ in the ratio $1: 2$ and $Q$ is the mid point of $A C$. If $\overrightarrow{\mathbf{A B}}+2 \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{B C}}-2 \overrightarrow{\mathbf{D C}}=k \overrightarrow{\mathbf{P Q}}$, then $k$ is equal to
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Verified Answer
The correct answer is:
-6
Now, $\overrightarrow{\mathbf{A B}}+2 \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{B C}}-2 \overrightarrow{\mathbf{D C}}$
$=\overrightarrow{\mathbf{A C}}+2 \overrightarrow{\mathbf{A D}}-2 \overrightarrow{\mathbf{D C}}$
$=\overrightarrow{\mathbf{A C}}+2(\overrightarrow{\mathbf{A C}}+\overrightarrow{\mathbf{C D}})-2 \overrightarrow{\mathbf{D C}}$
$\begin{aligned}
& =3 \overrightarrow{\mathbf{A C}}-4 \overrightarrow{\mathbf{D C}} \\
& =3(2 \overrightarrow{\mathbf{Q C}})-4\left(\frac{3}{2} \overrightarrow{\mathbf{P C}}\right) \\
& =6 \overrightarrow{\mathbf{Q C}}-6 \overrightarrow{\mathbf{P C}}=6(\overrightarrow{\mathbf{Q C}}+\overrightarrow{\mathbf{C P}})
\end{aligned}$
$\Rightarrow \quad k \overrightarrow{\mathbf{P Q}}=6 \overrightarrow{\mathbf{Q P}}=-6 \overrightarrow{\mathbf{P Q}} \quad$ (given)
$\Rightarrow \quad k=-6$
$=\overrightarrow{\mathbf{A C}}+2 \overrightarrow{\mathbf{A D}}-2 \overrightarrow{\mathbf{D C}}$
$=\overrightarrow{\mathbf{A C}}+2(\overrightarrow{\mathbf{A C}}+\overrightarrow{\mathbf{C D}})-2 \overrightarrow{\mathbf{D C}}$
$\begin{aligned}
& =3 \overrightarrow{\mathbf{A C}}-4 \overrightarrow{\mathbf{D C}} \\
& =3(2 \overrightarrow{\mathbf{Q C}})-4\left(\frac{3}{2} \overrightarrow{\mathbf{P C}}\right) \\
& =6 \overrightarrow{\mathbf{Q C}}-6 \overrightarrow{\mathbf{P C}}=6(\overrightarrow{\mathbf{Q C}}+\overrightarrow{\mathbf{C P}})
\end{aligned}$
$\Rightarrow \quad k \overrightarrow{\mathbf{P Q}}=6 \overrightarrow{\mathbf{Q P}}=-6 \overrightarrow{\mathbf{P Q}} \quad$ (given)
$\Rightarrow \quad k=-6$
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