A force \[\overrightarrow F = 6\widehat i + 2\widehat j - 3\widehat k\] acts on a particle and produces a displacement of \[\overrightarrow S = 2\widehat i - 3\widehat j + x\widehat k\]. If the work done is zero, then find the value of x.
A. -2
B. \[\dfrac{1}{2}\]
C. 6
D. 2

Hint:Before we start addressing the problem, we need to know about the displacement of a particle. When a force is applied, the object changes its position which is known as displacement. Since it is a vector quantity it has both direction and magnitude.

Formula Used:
Formula to find the work done is,
\[W = \overrightarrow F \cdot \overrightarrow S \]
Where, \[\overrightarrow F \] is force applied and \[\overrightarrow S \] is displacement.

Complete step by step solution:
If a force \[\overrightarrow F = 6\widehat i + 2\widehat j - 3\widehat k\] acts on a particle by producing a displacement \[\overrightarrow S = 2\widehat i - 3\widehat j + x\widehat k\]. Now if the work done by this force is zero, we need to find the value of x. Here the force is applied to the body and the body displaces, but still, the work done is zero means the force and displacement must be perpendicular to each other. Because,
\[W = \overrightarrow F \cdot \overrightarrow S \]
\[\Rightarrow W = F \cdot S\cos \theta \]

When the angle between the force and displacement is \[{90^0}\] then,
\[W = F \cdot S\cos {90^0}\]
\[\Rightarrow W = 0\]
Now,
\[W = \overrightarrow F \cdot \overrightarrow S \]
\[\Rightarrow \left( {6\widehat i + 2\widehat j - 3\widehat k} \right) \cdot \left( {2\widehat i - 3\widehat j + x\widehat k} \right) = 0\]
 \[\Rightarrow 12 - 6 - 3x = 0\]
\[\Rightarrow 6 - 3x = 0\]
\[\Rightarrow 3x = 6\]
\[ \Rightarrow x = 2\]
Therefore, the value of x is 2.

Hence, option D is the correct answer.

Note:We can say that the work done is zero because the Work done on an object depends on the amount of force F causing the work, and the displacement d experienced by the object. So, if no displacement occurs then the work done upon the object is zero. And also, as we said earlier, in work when the force and the displacement are perpendicular to each other we do not have to do any work.