Forces 3î + 2ĵ + 5k̂ and 2î + ĵ - 3k̂ are acting on a particle and displace it from the point 2î - ĵ - 3k̂ to the point 4î - 3ĵ + 7k̂. The work done by the force is:

Concept:If two points A and B have position vectors \(\rm \vec A\) and \(\rm \vec B\) respectively, then the vector \(\rm \vec {AB}=\vec B-\vec A\). For two vectors \(\rm \vec A\) and \(\rm \vec B\) at an ANGLE θ to each other:Dot Product is defined as: \(\rm \vec A.\vec B=|\vec A||\vec B|\cos \theta\).Resultant Vector is EQUAL \(\rm \vec A + \vec B\). Work: The work (W) done by a force (\(\rm \vec F\)) in moving (displacing) an object along a vector \(\rm \vec D\) is given by: W = \(\rm \vec F.\vec D=|\vec F||\vec D|\cos \theta\). Calculation:Let's say that the forces ACTING on the particle are \(\rm \vec F_1\) = 3î + 2ĵ + 5k̂ and \(\rm \vec F_2\) = 2î + ĵ - 3k̂.∴ The resulting force acting on the particle will be \(\rm \vec F=\vec F_1+\vec F_2\).⇒ \(\rm \vec F\) = (3î + 2ĵ + 5k̂) + (2î + ĵ - 3k̂)⇒ \(\rm \vec F\) = 5î + 3ĵ + 2k̂.Since the particle is moved from point 2î - ĵ - 3k̂ to the point 4î - 3ĵ + 7k̂, the displacement vector \(\rm \vec D\) will be:\(\rm \vec D\) = (4î - 3ĵ + 7k̂) - (2î - ĵ - 3k̂)⇒ ​\(\rm \vec D\) = 2î - 2ĵ + 10k̂.And finally, the work done W will be:W = \(\rm \vec F.\vec D\) = (5î + 3ĵ + 2k̂).(2î - 2ĵ + 10k̂)⇒ W = (5)(2) + (3)(-2) + (2)(10)⇒ W = 10 - 6 + 20 = 24 units.