If P and Q are the midpoints of the sides BC | vedclass

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(D) In parallelogram $ABCD$,let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$. Then $\overrightarrow{BC} = \overrightarrow{AD} =

Vedclass · Accepted Answer

$\frac{3}{2}\overrightarrow{AC}$

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(D) In parallelogram $ABCD$,let $\overrightarrow{AB} = \vec{a}$ and $\overrightarrow{AD} = \vec{b}$. Then $\overrightarrow{BC} = \overrightarrow{AD} = \vec{b}$ and $\overrightarrow{DC} = \overrightarrow{AB} = \vec{a}$.Since $P$ is the midpoint of $BC$,$\overrightarrow{BP} = \frac{1}{2}\overrightarrow{BC} = \frac{1}{2}\vec{b}$.Thus,$\overrightarrow{AP} = \overrightarrow{AB} + \overrightarrow{BP} = \vec{a} + \frac{1}{2}\vec{b}$.Since $Q$ is the midpoint of $CD$,$\overrightarrow{DQ} = \frac{1}{2}\overrightarrow{DC} = \frac{1}{2}\vec{a}$.Thus,$\overrightarrow{AQ} = \overrightarrow{AD} + \overrightarrow{DQ} = \vec{b} + \frac{1}{2}\vec{a}$.Adding these two vectors:$\overrightarrow{AP} + \overrightarrow{AQ} = (\vec{a} + \frac{1}{2}\vec{b}) + (\vec{b} + \frac{1}{2}\vec{a}) = \frac{3}{2}\vec{a} + \frac{3}{2}\vec{b} = \frac{3}{2}(\vec{a} + \vec{b})$.Since $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b}$,we have:$\overrightarrow{AP} + \overrightarrow{AQ} = \frac{3}{2}\overrightarrow{AC}$.

If $P$ and $Q$ are the midpoints of the sides $BC$ and $CD$ of the parallelogram $ABCD$,then $\overrightarrow{AP} + \overrightarrow{AQ} = $

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