Question

$ABCD$ is a parallelogram and $A_1$ 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

The correct option is 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 $B_1=\overrightarrow{d}+\frac{\overrightarrow{b}}{2}$
$\Rightarrow \overrightarrow{A{A}_1}+\overrightarrow{A{B}_1}=\frac{3}{2}(\overrightarrow{b}+\overrightarrow{d})\Rightarrow \lambda \overrightarrow{AC}=\frac{3}{2}\overrightarrow{AC}\Rightarrow \lambda =\frac{3}{2}$