In the last section we covered vector addition. Adding two vectors will always yield a vector. With multiplication, it is perform two different operation. One of those operations yields a scalar. One yields a vector. In this section we will dive into the scalar product, otherwise known as the dot product. It is called the dot product because the symbol designating the scalar product is a dot. Written out, it looks something like this:
When not being denoted with an arrow, vectors are also commonly denoted in a bold font. Above, scalar a is equal to the scalar product between vectors B and C. The process of performing the dot product operation can be broken into 5 easy steps:
1) Identify two vectors- in this step you will need to simply locate the two vectors you’ll be “dotting”.
2) Identify number of dimensions- i.e. [2,0] and [4,3] are both two dimensional, because there are two numbers, or components, inside the brackets, [2,0,5] and [4,3,-9] are 3-dimensional vectors.
3) Multiply like components from each dimension together- Example: [bx,by,bz] and [cx,cy,cz], we now perform bx*cx, by*cy, and bz*cz.
4) Add Together All Component Multiples- For example, with our three dimensional vector: bx*cx+by*cy+bz*cz.
5) Box Your Answer, You’re Done!
So this is a step-by-step skeleton to follow any time you need to perform a dot product operation, of which there will be no short supply during your 1st year of physics. Next, we will roll through one example just to understand exactly what’s going on with a little vector subtraction and then a little bit of dot product.
Question: A strapping young man exerted a force of [10N,10N,10N] moving a box from position (1,2,3) to position (4,5,6) on a 3-D cartesian axis, where all positions are defined in meters. What is the total work done?
Answer: Work is defined as follows:W = F ⋅ d where W is the work (scalar), F is the force vector, and d is the displacement vector. In the problem, the displacement is not exactly given to us. For that, we need to know the following relationship: d = xf - xi where xf is the final position and xi is the initial position. First, since we know the initial and final displacements, let’s calculate the displacement vector: d = (4,5,6) - (1,2,3) d = [4-1,5-2,6-3] d = [3,3,3] So, our displacement vector is [3,3,3]. Notice in the vector subtraction that each component of the position was subtracted individually, and all three directions combined then give the final displacement vector. With both vectors in hand, we can now calculate the work: