Q. If $D,E\ and\ F$ are respectively the mid points of $AB,\ AC\ and\ BC\ in\ \mathrm{\Delta}ABC,\ then\$ $\overrightarrow{\mathbf{BE}} + \overrightarrow{\mathbf{AF}}$ is equal to

A) $\overrightarrow{\mathbf{DC}}$
B) $\frac{1}{2}\overrightarrow{\mathbf{BF}}$
C) $2\ \overrightarrow{\mathbf{BF}}$
D) $\frac{3}{2}\overrightarrow{\mathbf{BF}}$
Explanation

(a)
Let $\overrightarrow{\mathbf{OA}} = \overrightarrow{a},\ \overrightarrow{\mathbf{OB}} = \overrightarrow{\mathbf{b}}\$and$\ \overrightarrow{\mathbf{OC}} = \overrightarrow{\mathbf{c}}\ ,\$then
$\overrightarrow{\mathbf{OD}} = \frac{\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}}}{2},\ \overrightarrow{\mathbf{OE}} = \frac{\overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{c}}}{2},\ \overrightarrow{\mathbf{OF}} = \frac{\overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}}}{2}$
$Now,\ \overrightarrow{\mathbf{AF}} = \frac{1}{2}\left( \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right) - \overrightarrow{\mathbf{a\ }},\ \overrightarrow{\mathbf{BE}} = \frac{1}{2}\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{c}} \right) - \overrightarrow{\mathbf{b\ }}$
$and\ \overrightarrow{\mathbf{CD}} = \frac{1}{2}\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right) - \overrightarrow{\mathbf{c\ }}$
$\therefore\ \overrightarrow{\mathbf{AF}} + \overrightarrow{\mathbf{BE}} = \frac{1}{2}\left( \overrightarrow{\mathbf{b}} + \overrightarrow{\mathbf{c}} \right) - \overrightarrow{\mathbf{a\ }} + \frac{1}{2}\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{c}} \right) - \overrightarrow{\mathbf{b\ }}$
$= \overrightarrow{\mathbf{c}} - \frac{1}{2}\left( \overrightarrow{\mathbf{a}} + \overrightarrow{\mathbf{b}} \right) = \overrightarrow{\mathbf{DC}}\ \ \$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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