If M and N are the mid-points of the sides BC | vedclass

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(C) Let $\vec{AB} = \vec{a}$ and $\vec{AD} = \vec{b}$. Since $ABCD$ is a parallelogram, $\vec{BC} = \vec{AD} = \vec{b}$ and $\vec{CD} = \vec{AB} = \ve

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$\frac{3}{2} AC$

Vedclass · Answer

(C) Let $\vec{AB} = \vec{a}$ and $\vec{AD} = \vec{b}$. Since $ABCD$ is a parallelogram, $\vec{BC} = \vec{AD} = \vec{b}$ and $\vec{CD} = \vec{AB} = \vec{a}$.Given that $M$ is the mid-point of $BC$, we have $\vec{BM} = \frac{1}{2} \vec{BC} = \frac{1}{2} \vec{b}$.Given that $N$ is the mid-point of $CD$, we have $\vec{DN} = \frac{1}{2} \vec{CD} = \frac{1}{2} \vec{a}$.In $	riangle ABM$, by triangle law of vector addition:$\vec{AM} = \vec{AB} + \vec{BM} = \vec{a} + \frac{1}{2} \vec{b}$.In $	riangle ADN$, by triangle law of vector addition:$\vec{AN} = \vec{AD} + \vec{DN} = \vec{b} + \frac{1}{2} \vec{a}$.Adding these two vectors:$\vec{AM} + \vec{AN} = (\vec{a} + \frac{1}{2} \vec{b}) + (\vec{b} + \frac{1}{2} \vec{a})$$= (1 + \frac{1}{2}) \vec{a} + (1 + \frac{1}{2}) \vec{b}$$= \frac{3}{2} \vec{a} + \frac{3}{2} \vec{b} = \frac{3}{2} (\vec{a} + \vec{b})$.Since $\vec{AC} = \vec{AB} + \vec{BC} = \vec{a} + \vec{b}$, we get:$\vec{AM} + \vec{AN} = \frac{3}{2} \vec{AC}$.

If $M$ and $N$ are the mid-points of the sides $BC$ and $CD$ respectively of a parallelogram $ABCD$, then $AM + AN$ equals

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