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I cant figure out how to solve this vector question, all i know is that i need to use the dot product?

An object is being acted upon by a force of 15N acting in the direction of the vector (6,2,-2). Consequentially the object is being moved from A(-12,0,4) to B(2,-4,6) and then to C(8,4,-2). Calculate the work done, if distances are measured in meters.

I will denote the inner product of x and y by

Now, by definition, W done in moving the object from A to B is defined as the path integral from A to B of where dl is the differential vector in the direction of path. But recall that work is path independent.

Moreover, in this case, force is constant (denote the force vector by F), so we can simplify things further by eliminating the integral, and this just becomes the inner product. It ends up being the following:

= Work done to move object from A to B
= Work done to move object from B to C

So total work done to move object from A to C is just
+ =
(note I have used linearity of the inner product here, although another way to see this is the path independence of work. It didn't matter if it stopped at B or if it stopped at a point 300 miles to the left of B. As counterintuitive as this sounds, it's the way line and path integrals work, so the only thing that mattered was that it started at A and ended at C)

Now, normalize the direction vector of the force:
First find the length of the vector given:
|(6, 2, -2)| = Sqrt(44) = 2*Sqrt(11)

So the unit vector (direction vector) of the force is (call this unit vector n):
n = 1/(2*Sqrt(11)) * (6, 2, -2)

And (C-A) = (20, 4, -6)

So what you have is
W = 15* = 15*(120 + 8 + 12) / (2*Sqrt(11)) = 10500/Sqrt(11) = 3165.87 J

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