Question

A particle is projected with initial velocity $\overrightarrow{v}=3\hat{i}+4\hat{j}+12\hat{k}$ & $\overrightarrow{a}$ = $8\hat{i}+6\hat{j}-4\hat{k}$ then path of particle is

• A. parabolic
• B. curved but not parabolic
• C. straight line
• D. not defined as vector in 3D

Answer 1

The correct option is A parabolic

$\overrightarrow{a}.\overrightarrow{v}=(8\hat{i}+6\hat{j}-4\hat{k}).(3\hat{i}+4\hat{j}+12\hat{k})$

=24+24-48=0

Hence $\overrightarrow{a}$ is $\perp$ to $\overrightarrow{v}$ and not parallel. So path will not be straight line. At any other angle between constant acceleration and velocity only parabolic path is possible.