question_answer

If P is any point with in a triangle ABC, then \[\overrightarrow{PA}+\overrightarrow{CP}\] is equal to

A)  \[\overrightarrow{AC}+\overrightarrow{CB}\]     

B)  \[\overrightarrow{BC}+\overrightarrow{BA}\]

C)  \[\overrightarrow{CB}+\overrightarrow{AB}\]      

D)  \[\overrightarrow{CB}+\overrightarrow{BA}\]

Correct Answer: D

Solution :

\[\therefore \] \[\overrightarrow{CP}+\overrightarrow{PA}=\overrightarrow{CA}\] By triangles law, \[\overrightarrow{CA}+\overrightarrow{CB}+\overrightarrow{BA}\]  \[\therefore \] \[\overrightarrow{CP}+\overrightarrow{PA}=\overrightarrow{CB}+\overrightarrow{BA}\]