Q. The initial and final position vectors for a particle are respectively $( - 3.0\ m)\widehat{\mathbf{i}} +$(2.0 m)$\widehat{\mathbf{j}}$+(8.0 m)$\widehat{\mathbf{k}}$ and$\ (9.0\ m)\widehat{\mathbf{i}} + (2.0\ m)\widehat{\mathbf{j}} + ( - 8.0\ m)\widehat{\mathbf{k}}.$The displacement of the particle is

A) $(6.0\ m)\widehat{\mathbf{i}} + (4.0\ m)\widehat{\mathbf{j}} + (16.0\ m)\widehat{\mathbf{k}}$
B) $(6.0\ m)\widehat{\mathbf{i}}$
C) $(12.0\ m)\widehat{\mathbf{i}} - (16.0\ m)\widehat{\mathbf{j}}$
D) $(12.0\ m)\widehat{\mathbf{i}}$
Explanation

(c) $\mathbf{r}_{\mathbf{i}} = ( - 3.0\ m)\widehat{\mathbf{i}} + (2.0\ m)\widehat{\mathbf{j}} + (8.0\ m)\widehat{\mathbf{k}}$
$r_{f} = (9.0\ m)\widehat{\mathbf{i}}\mathbf{+}(2.0\ m)\widehat{\mathbf{j}}\mathbf{+}( - 8.0\ m)\widehat{\mathbf{k}}$
$\therefore$Displacement $= \mathbf{r}_{f} - \mathbf{r}_{i}$
$= \left\lbrack (9.0\ m)\widehat{\mathbf{i}} + (2.0\ m)\widehat{\mathbf{j}} + ( - 8.0\ m)\widehat{\mathbf{k}} \right\rbrack$
$\ \ \ \ \ \ \ \ \ \ \ \ \ - \lbrack( - 3.0m)\widehat{\mathbf{i}} + (2.0m)\widehat{\mathbf{j}} + (8.0m)\widehat{\mathbf{k}}\mathbf{\rbrack}$
$= \left\lbrack (12.0m)\widehat{\mathbf{i}} - (16.0m)\widehat{\mathbf{j}} \right\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.