If force \[\overrightarrow F = \left( {4\widehat i + 5\widehat j} \right)\] and displacement \[\overrightarrow S = \left( {3\widehat i + 6\widehat k} \right)\] then find the work done.
A. \[4 \times 3\]
B. \[5 \times 6\]
C. \[6 \times 3\]
D. \[4 \times 6\]

Hint:We start proceeding with this problem by understanding the term displacement and work done. 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. Work is nothing but a force needed to move an object from one place to another. Work done can be positive, negative, or zero depending on the angle between the force applied and the displacement.

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

Complete step by step solution:
If a force of \[\overrightarrow F = \left( {4\widehat i + 5\widehat j} \right)\]acts on a particle and produces a displacement of\[\overrightarrow S = \left( {3\widehat i + 6\widehat k} \right)\] then we need to find the work done. Now by using the formula for work done we have,
\[W = \overrightarrow F \cdot \overrightarrow S \]
Here, \[\overrightarrow F \]is the force applied on a particle and \[\overrightarrow S \] is the displacement of a particle due to the force applied.

Here, in the force vector, the unit vector k is not given and in the displacement vector the unit vector j is not given. So, we can replace \[0\widehat k\] in force vector and \[0\widehat j\] in displacement vector, then,
\[\overrightarrow F = \left( {4\widehat i + 5\widehat j + 0\widehat k} \right)\] and \[\overrightarrow S = \left( {3\widehat i + 0\widehat j + 6\widehat k} \right)\]

Now, substitute the values of force, displacement then we get,
\[W = \left( {4\widehat i + 5\widehat j + 0\widehat k} \right) \cdot \left( {3\widehat i + 0\widehat j + 6\widehat k} \right)\]
\[\Rightarrow W = 12 + 0 + 0\]
\[\therefore W = \left( {4 \times 3} \right)J\]
Therefore, the work done is \[\left( {4 \times 3} \right)J\].

Hence, option A is the correct answer.

Note: Remember to check if the coefficient of the unit vectors of force and displacement are properly given. Suppose in the force vector, the unit vector k is not given and in the displacement vector, the unit vector j is not given. So, in order to get the missing unit vectors, we can assume j and k values as \[0\widehat j\] and \[0\widehat k\] in their respective vectors. write as zero for those coefficients that are not given.