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…

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
  1. -6
  2. -4
  3. 6
  4. 4

Solution

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) $\Rightarrow \quad k=-6$

Asked in: AP EAMCET 2009

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