ABCD is a parallelogram $A_1$ and $B_1$ are the midpoints of side BC and CD respectively. If $\vec{A A}_1+\vec{AB}_1=\lambda \vec{A C}$, then $\lambda$ is equal to:
Answer & explanation
Correct answer: option 3
Let P.V. of A, B, D be $\vec{0}, \vec{b}$ and $\vec{d}$ respectively.
Then P.V. of C = $\vec{b}+\vec{d}$
Also, P.V. of $A_1=\vec{b}+\frac{\vec{d}}{2}$
and, P.V. of $B_1=\vec{d}+\frac{b}{2}$
$\Rightarrow \vec{AA_1}+\vec{AB_1}=\frac{3}{2}\left(\vec{b}+\vec{d}\right)=\frac{3}{2} \vec{AC}$
Hence (3) is correct answer.