Class content > Kinematics > Vectors > Multiplying vectors
Prerequisites
The dot product is useful when only one part of a vector plays a physical role -- the part that is in the same direction as another vector. One example is the work. That's basically how a force changes the magnitude (rather than the direction) of an object's velocity. We can get that by taking the part of the force in the direction of the velocity (dot producting the force with the displacement). We can figure out the math of this product either geometrically or algebraically.
Geometrical approach
The dot product is defined to give the product of two vectors projected on one another. That is, if we want to take the dot product of two vectors, A and B , we can take the part of A that is in the direction of B and multiply it by the length of B; or we can take the part of B that is in the direction of A and multiply it by the length of A.
These are shown in the figure below.
The first gives us A|| x B, the second gives us A x B||. Since A|| = A cos(θ) and since B|| = B cos(θ), our dot product in either case is
where θ is the angle between the two vectors.
Algebraic approach
Instead of using geometry to get the dot product, we can use algebra. Since we want the dot product to be the parts of two vectors that are in the same direction, we define the dot product of the basic vectors, i-hat, j-hat (and k-hat if we are in 3D rather than 2D) as being 1 if they are the same and 0 if they are different. So we define the dot product between two vectors by the rules:
Since dot products are products of vectors -- not numbers -- any numbers multiplying them don't do anything special. So we can work this all out using the distributive rule.
Well that's a royal mess! But it kind of makes sense. It's all possible products of the components of A with those of B multiplied by the dot products of the appropriate directions. The result, using our rules for the dot products of the basic i-hat and j-hat vectors now collapses everything! All the terms with both an i-hat and a j-hat drop out and the other terms just become 1. The result is rather simple and elegant:
You can imagine what the result would be if we were in 3D -- you just keep the z-terms as well! Showing that this result is the same as what we got by our geometrical argument is an exercise in trigonometric identities so we won't go through it here, but you're welcome to try!
Joe Redish 11/6/11
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