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

Question

Dr. Abubacker · Accepted Answer

(a)
Now, $\overrightarrow{\mathbf{AB}} + 2\overrightarrow{\mathbf{AD}} + \overrightarrow{\mathbf{BC}} - 2\overrightarrow{\mathbf{DC}}$
$= \overrightarrow{\mathbf{AC}} + 2\overrightarrow{\mathbf{AD}} - 2\overrightarrow{\mathbf{DC}}$
$= \overrightarrow{\mathbf{AC}} + 2\left( \overrightarrow{\mathbf{AC}} + \overrightarrow{\mathbf{CD}} \right) - 2\overrightarrow{\mathbf{DC}}$
$= 3\overrightarrow{\mathbf{AC}} - 4\overrightarrow{\mathbf{DC}}$
$= 3\left( 2\overrightarrow{\mathbf{QC}} \right) - 4\left( \frac{3}{2}\overrightarrow{\mathbf{PC}} \right)$
$= 6\overrightarrow{\mathbf{QC}} - 6\overrightarrow{\mathbf{PC}} = 6\left( \overrightarrow{\mathbf{QC}} + \overrightarrow{\mathbf{CP}} \right)$
$\Longrightarrow k\overrightarrow{\mathbf{PQ}} = 6\overrightarrow{\mathbf{QP}} = - 6\overrightarrow{\mathbf{PQ\ }}\$ (given)
$\Longrightarrow k = - 6$
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Q. In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio 1:2 $Q$ is the mid point of $AC$. If $\overrightarrow{\mathbf{AB}} + 2\ \ \overrightarrow{\mathbf{AD}} + \ \overrightarrow{\mathbf{BC}} - 2\ \overrightarrow{\mathbf{DC}} = k\ \overrightarrow{\mathbf{PQ}}$, then $k$ is equal to

Explanation

Reviewed by Dr. Abubacker

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