Two constant force $\vec{F}_1=2 \hat{i}-5 \hat{j}+6 \hat{k}$ and $\vec{F}_2=-\hat{i}+2 \hat{j}-\hat{k}$ act on a particle. If particle is displaced from a point A with position vector $4 \hat{i}-3 \hat{j}-\hat{k}$ to the point B with position vector $6 \hat{i}+\hat{j}-3 \hat{k}$ Then work done in the process is equal to:

-15 units

Total force, $\vec{F}=\vec{F}_1+\vec{F}_2=\hat{i}-3 \hat{j}+5 \hat{k}$

Displacement, $\vec{d}=6 \hat{i}+\hat{j}-3 \hat{k}-(4 \hat{i}-3 \hat{j}-2 \hat{k})$

$=2 \hat{i}+4 \hat{j}-\hat{k}$

Work done $=\vec{F} .\vec{d}$

= 2 – 12 – 5 = -15 units

Hence (3) is correct answer.