Find the Work Done by Force \[\vec{F}=\left( y\hat{i}+x\hat{j} \right)N\] in taking a particle from A (2, 4) to B (5, 7) in a straight line.

Hint: In the question it is said that a particle is moved from one point to another. The force applied to move the particle is given. To find the work done in moving the particle we first convert the given Cartesian coordinates of the points into its vector form. Then we find the displacement of the particle and substitute it in the equation to get the work done.

Formula used:
Work done, $W=\vec{F}.\vec{S}$

Complete step by step answer:
In the question it is said that a particle is taken from the point A (2, 4) to the point B (5, 7) on a straight line.
The force exerted to move the particle is given to us.
$\vec{F}=\left( y\hat{i}+x\hat{j} \right)N$
Since the particle is moved from point A to point B, we can write the position vectors of these two points.
$\vec{a}=2\hat{i}+4\hat{j}$
$\vec{b}=5\hat{i}+7\hat{j}$
Now we can find the distance from point A to B.
$\overrightarrow{AB}=\vec{b}-\vec{a}$
$\begin{align}
  & \overrightarrow{AB}=\left( 5\hat{i}+7\hat{j} \right)-\left( 2\hat{i}+4\hat{j} \right) \\
 & \overrightarrow{AB}=5\hat{i}+7\hat{j}-2\hat{i}-4\hat{j} \\
 & \overrightarrow{AB}=3\hat{i}+3\hat{j} \\
 & \overrightarrow{AB}=3\left( \hat{i}+\hat{j} \right) \\
\end{align}$
This is the displacement (S) of the particle due to the applied force.
In the question we are asked to find the work done in moving the particle from point A to point B.
We know that work done is the product of the force in the direction of the displacement and the magnitude of this displacement, i.e.
$W=\vec{F}.\vec{S}$, where ‘W’ is the work done, ‘F’ is the force exerted and ‘S’ is the displacement.
We know the displacement and the force exerted in the given situation.
Therefore the work done,
$\begin{align}
  & W=\left( y\hat{i}+x\hat{j} \right)N.3\hat{i}+3\hat{j} \\
 & W=\left( Ny\hat{i}+Nx\hat{j} \right).3\hat{i}+3\hat{j} \\
 & W=3Ny+3Nx\text{ units} \\
\end{align}$
Hence the work done in moving the given particle is $3Ny+3Nx\text{ units}$.

Note:
A physical quantity with both direction and magnitude is known as a vector quantity.
We know that, $W=\vec{F}.\vec{S}$
Here we can see that work is not a vector quantity even though force and displacement is a vector quantity. This is because the dot product of two vectors will always result in a scalar quantity.
Hence work is a scalar quantity.
A physical quantity which has only magnitude and has no direction is a scalar quantity.
Therefore work has only magnitude and has no direction.