Q. In $\mathrm{\Delta}ABC,\ if\ 2\ \overrightarrow{\mathbf{AC}} = 3\overrightarrow{\mathbf{CB}},\$then$\ 2\overrightarrow{\mathbf{OA}} + 3\overrightarrow{\mathbf{OB}}$ equals
A) $5\ \overrightarrow{\mathbf{OC}}$
B) $- \overrightarrow{\mathbf{OC}}$
C) $\overrightarrow{\mathbf{OC}}$
D) $4\ \overrightarrow{\mathbf{OC}}$
Explanation
(a)
$2\overrightarrow{\mathbf{OA}} + 3\overrightarrow{\mathbf{OB}} = 2\left( \overrightarrow{\mathbf{OC}} + \overrightarrow{\mathbf{CA}} \right) + 3\left( \overrightarrow{\mathbf{OC}} + \overrightarrow{\mathbf{CB}} \right)$
$= 5\overrightarrow{\mathbf{OC}} + 2\overrightarrow{\mathbf{CA}} + 3\overrightarrow{\mathbf{CB}}$
$= 5\overrightarrow{\mathbf{OC}}\ \ \ \ \left\lbrack \because 2\overrightarrow{\mathbf{CA}} = - 3\overrightarrow{\mathbf{CB}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.