Question

$ABCD$ a parallelogram, $A_1$ and $B_1$ are the midpoints of sides $BC$ and $CD$, respectively. If $\overrightarrow{A{A}_1}+\overrightarrow{A{B}_1}=\lambda \overrightarrow{AC}$, then $\lambda$ is equal to

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

Answer: (C)

$\frac{3}{2}$

Let P.V. of $A,B$ and $D$ be $\overrightarrow{0},\overrightarrow{b},$ and $\overrightarrow{d}$, respectively.
Then P.V. of $C,\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d}$
Also, P.V of $A_1=\overrightarrow{b}+\frac{\overrightarrow{d}}{2}$
and P.V. of $B_1=\overrightarrow{d}+\frac{\overrightarrow{b}}{2}$
$\Rightarrow \overrightarrow{A{A}_1}+\overrightarrow{A{B}_1}=\frac{3}{2}(\overrightarrow{b}+\overrightarrow{d})=\frac{3}{2}\overrightarrow{AC}$