A particle P is moving with a constant speed of $6m/s$ in a direction $2\hat{i}-\hat{j}-2\hat{k}$. When $t=0,P$ is at a point whose position vector is $3\hat{i}+4\hat{j}-7\hat{k}$. Find the position vector of the particle $P$ after $4$ seconds.

The correct option is A $19\hat{i}-4\hat{j}-23\hat{k}$

We have

Unit vector in the direction of velocity$=\frac{2\hat{i}-\hat{j}-2\hat{k}}{\sqrt{4+1+4}}$

So velocity$=6\times \frac{2\hat{i}-\hat{j}-2\hat{k}}{3}=4\hat{i}-2\hat{j}-2\hat{j}$

Hence displacement in $4sec=(4\hat{i}-2\hat{j}-4\hat{k})\times 4=16\hat{i}-8\hat{j}-16\hat{k}$

Displacement hence $=$ final position $-$ Initial position

Final position$=(16\hat{i}-8\hat{j}-16\hat{k})+(3\hat{i}+4\hat{j}-7\hat{k})=(19\hat{i}-4\hat{j}-23\hat{k})$