Question

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

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

Answer 1

The correct option is C

$\frac{3}{2}$

$\text{Let the position vectors of A, B, C, D be}\overrightarrow{0},\overrightarrow{b},\overrightarrow{c}and\overrightarrow{d}$
$\text{Then position vector of C,}\overrightarrow{c}=\overrightarrow{b}+\overrightarrow{d}$
$\text{Also, position vector of}{A}_1=\overrightarrow{b}+\frac{\overrightarrow{d}}{2}$

$\text{and position vector of}{B}_1=\overrightarrow{d}+\frac{\overrightarrow{b}}{2}\Rightarrow {\overrightarrow{AA}}_1+{\overrightarrow{AB}}_1=\frac{3}{2}(\overrightarrow{a}+\overrightarrow{d})=\frac{3}{2}\overrightarrow{AC}$