Q. If $P$ is any point with in a triangle $ABC,$ then $\overrightarrow{\mathbf{PA}} + \overrightarrow{\mathbf{CP}}$ is equal to
A) $\overrightarrow{\mathbf{AC}} + \overrightarrow{\mathbf{CB}}$
B) $\overrightarrow{\mathbf{BC}} + \overrightarrow{\mathbf{BA}}$
C) $\overrightarrow{\mathbf{CB}} + \overrightarrow{\mathbf{AB}}$
D) $\overrightarrow{\mathbf{CB}} + \overrightarrow{\mathbf{BA}}$
Explanation
(d)
$\therefore\ \overrightarrow{\mathbf{CP}} + \overrightarrow{\mathbf{PA}} + \overrightarrow{\mathbf{BA}}$
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By triangle law,
$\overrightarrow{\mathbf{CA}} = \overrightarrow{\mathbf{CB}} + \overrightarrow{\mathbf{BA}}$
$\therefore\ \overrightarrow{\mathbf{CP}} + \overrightarrow{\mathbf{PA}} = \overrightarrow{\mathbf{CB}} + \overrightarrow{\mathbf{BA}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.