Question

ABCD is a parallelogram. If L, M be the middle points of BC and CD, express $\overrightarrow{AL}$ and $\overrightarrow{AM}$ in terms of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ also show that $\overrightarrow{AL}+\overrightarrow{AM}=\left(\frac{3}{2}\right)\overrightarrow{AC}$.

Answer 1

Take A as the origin let $\overrightarrow{b}$ and $\overrightarrow{d}$

be position vectors of B and D respectively

such that

$\begin{array}{c}\overrightarrow{AB}=\overrightarrow{b}A\overrightarrow{D}=\overrightarrow{d}\\ \overrightarrow{AL}=A\overrightarrow{B}+B\overrightarrow{L}\\ =\overrightarrow{AB}+\frac{1}{2}B\overrightarrow{C}\\ =A\overrightarrow{B}+\frac{1}{2}A\overrightarrow{D}\\ =\overrightarrow{b}+\frac{\overrightarrow{d}}{2}\\ positionvectorsofLis\overrightarrow{b}+\frac{\overrightarrow{d}}{2}\\ AgainA\overrightarrow{M}=A\overrightarrow{D}+D\overrightarrow{M}\\ =A\overrightarrow{D}+\frac{1}{2}D\overrightarrow{C}\\ =A\overrightarrow{D}+\frac{1}{2}A\overrightarrow{B}\\ =\overrightarrow{d}+\frac{1\overrightarrow{b}}{2}\\ A\overrightarrow{C}=A\overrightarrow{B}+B\overrightarrow{C}=A\overrightarrow{B}+A\overrightarrow{D}=\overrightarrow{b}+\overrightarrow{d}\\ A\overrightarrow{L}+A\overrightarrow{M}=\left(\overrightarrow{b}+\frac{\overrightarrow{d}}{2}\right)+\left(\overrightarrow{d}+\frac{\overrightarrow{b}}{2}\right)\\ =\frac{3}{2}\overrightarrow{b}+\frac{3}{2}\overrightarrow{d}\\ =\frac{3}{2}\left(\overrightarrow{b}+\overrightarrow{d}\right)=\frac{3}{2}A\overrightarrow{C}\end{array}$