Question

$ABCD$ is a parallelogram. If $L$ and $M$ are the middle points of $BC$ and $CD$ respectively, then find :
(i) $AL$ and $AM$ in terms of $AB$ and $AD$

Answer 1

Take $A$ as the orgin. Let $\overrightarrow{b},\overrightarrow{d}$ be postion vectos of $B$ and $D$ respectively such that

$\overrightarrow{AB}=\overrightarrow{b},\overrightarrow{AD}=\overrightarrow{d}$

$\overrightarrow{AL}=\overrightarrow{AB}+\overrightarrow{BL}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{BC}=\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AD}=\overrightarrow{b}+\overrightarrow{\frac{d}{2}}\overrightarrow{b}+\overrightarrow{\frac{d}{2}}$

$\therefore$ position vector of $L$ is $\overrightarrow{b}+\overrightarrow{\frac{d}{2}}$

Again, $\overrightarrow{AM}=\overrightarrow{AD}+\overrightarrow{DM}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{DC}=\overrightarrow{AD}+\frac{1}{2}\overrightarrow{AB}=\overrightarrow{d}+\frac{1}{2}\overrightarrow{b}\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{b}+\overrightarrow{d}⟶\left(i\right)\overrightarrow{AL}+\overrightarrow{AM}=(\overrightarrow{b}+\frac{\overrightarrow{d}}{2})+(\overrightarrow{d}+\frac{\overrightarrow{b}}{2})=\frac{3}{2}\overrightarrow{b}+\frac{3}{2}\overrightarrow{d}=\frac{3}{2}(\overrightarrow{b}+\overrightarrow{d})[\therefore of(i\left)\right]=\frac{3}{2}(\overrightarrow{AB}+\overrightarrow{AD})$