# If the force $\left( {3\hat i - 2\hat j + \hat k} \right)N$ , produces a displacement of $\left( {2\hat i - 4\hat j + c\hat k} \right)m$. If the work done is $16J$ then, find the value of $c$.

A) $ - 1$

B) $ - 2$

C) $1$

D) $2$

Answer

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**Hint:**We can define the work done as the product of force applied on a body and displacement produced by it. Now, put the values of force and displacement and evaluate the dot product of force and displacement. Also assume the initial position of the particle to be at origin.

**Complete step by step answer:**

Let the force applied on a body be $\vec F$, displacement produced by the body be $\vec s$ and the work done by the body be $\vec W$.

According to the question, it is given that –

$\implies \vec F = \left( {3\hat i - 2\hat j + \hat k} \right)N$

$\implies \vec s = \left( {2\hat i - 4\hat j + c\hat k} \right)m$ and

$\implies \vec W = 16J$

Work done is the transfer of energy for the displacement of an object using the application of force. Work has only magnitude and no direction therefore, it is the scalar quantity. The S.I unit of work is Joule. It transfers energy from one place to another.

Now, we know that the work done can be defined as the product of force applied on a body and displacement done by the body. So, in the vectors, it can be said that work done is the dot product of force applied on a body and displacement done by the body.

Mathematically, the above statement can be represented as –

$\vec W = \vec F.\vec s$

Putting the values of work done, force and displacement in the above equation –

$

\implies 16 = \left( {3\hat i - 2\hat j + \hat k} \right).\left( {2\hat i - 4\hat j + c\hat k} \right) \\

\implies 16 = 6 + 8 + c \\

$

Now, doing transposition method, we get –

$

\implies c = 16 - 14 \\

\implies c = 2 \\

$

Hence, the value of $c$ is $2$.

Therefore, the vector of displacement can be expressed as –

$\vec s = \left( {2\hat i - 4\hat j + 2\hat k} \right)m$

Hence, the correct option for this question is (D).

**Note:**The dot product of any orthogonal vector with itself is always one and the dot product of any orthogonal vector with any other orthogonal vector is always zero.

$

\hat i.\hat i = 1 \\

\hat j.\hat j = 1 \\

\hat k.\hat k = 1 \\

\hat i.\hat j = 0 \\

\hat i.\hat k = 0 \\

\hat j.\hat k = 0 \\

$

Let there be any two vectors such as –

$

\vec A = a\hat i + b\hat j + c\hat k \\

\vec B = x\hat i + y\hat j + z\hat k \\

$

Then, $\vec A.\vec B = ax + by + cz$

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