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**Hint:**Here, we will use the concept of the properties of the vectors. First, we will write the definition of the work. Then we will write the formula of the work and see how the scalar product of vectors related to work.

**Complete step by step solution:**

Firstly we will write the basic definition of the work.

We know that the work is defined as the amount of force required to displace an object through some distance and work done by this force also depends upon the angle between the direction of force and the direction of the displacement.

So, the work is the dot product of the force and the displacement.

\[ W = \vec F \cdot \vec r\]

\[ \Rightarrow W = Fr\cos \theta \]

Where F is the magnitude of force vector and r is the magnitude of the displacement and \[\theta \] is the angle between them.

Hence, this is how the scalar product of vectors is related to the work.

**Note:**

Here we have to keep in mind that while expanding the dot product of vectors is related to cos function, not the sine function. The sine function is related to the cross product of the vectors. Also, we have to remember that the dot product of the perpendicular vectors is always zero as the angle between them is \[{90^0}\] and dot product is related to cos function. Also, the cross product of the parallel vectors is always zero as the angle between the parallel vectors is \[{0^0}{\rm{or18}}{{\rm{0}}^0}\] and cross product is related to the sin function.

Dot product of two vectors is \[a.b = \left| a \right|\left| b \right|\cos \theta \]

Cross product of two vectors is \[a \times b = \left| a \right|\left| b \right|\sin \theta \]

Where \[\theta \] is the angle between the vectors.

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