question_answer

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

Correct Answer: C

Solution :

[c] Let P. V of A, B and D be \[\vec{0},\,\vec{b}\] and \[\vec{d}\], respectively. Then P.V. of C, \[\vec{c}=\vec{b}+\vec{d}\]. Also P.V. of \[{{A}_{1}}=\vec{b}+\frac{{\vec{d}}}{2}\] and P.V. of \[{{B}_{1}}\] \[=\vec{d}+\frac{{\vec{b}}}{2}\Rightarrow \overrightarrow{A{{A}_{1}}}=\frac{3}{2}(\vec{b}+\vec{d})=\frac{3}{2}\overrightarrow{AC}\]