Dot Product

The dot product is a miracle composed of two definitions. The first definition, the geometric definition, is that

The algebraic definition of the dot product is that

This is by no means obvious, and indeed you should spend several minutes (or more) thinking about how, why, and if this could be true. Try applying both definitions to pairs of vectors with easily calculatable angles like (1,0,0) and (0,0,1), or between two arbitrary vectors in the plane, or between a vector and itself.

Before we prove that the two definitions are the same, let's look at some of the properties of the dot product. First and foremost, both definitions agree that the dot product of two vectors is a scalar real number. It is not a vector.

Second, both definitions agree that the dot product is commutative (that v · w = w · v) and that it absorbs constants: ( c v) · w = c ( v · w). Also, both definitions imply that 0 · w = 0.

Finally, both definitions (with a little bit of help from the Pythagorean theorem) give v · v = | v |² = v₁² + v₂² + v₃² . This is nice, and it further implies that v · v >= 0 and that v · v = 0 if and only if v = v = 0.

Okay, so our two definitions of the dot product agree on lots of things. What is even more important is their different strengths:

• The algrebraic definition of the dot product is great for calculation. Given the cartesian coordinates of two vectors, you can calculate their dot product lickety-split. It is also very helpful in proving algebraic identities like ( v₁ + v₂ ) · v = v₁ · v + v₂ · v. Try proving this with the geometric definition!

• The geometric definition of the dot product is great for, well, geometry. For example, if two vectors are orthogonal (perpendicular) than their dot product is 0 because the cosine of 90 (or 270) degrees is 0.

  Another example is finding the projection of a vector onto another vector. By trigonometry, the length of the projection of the vector w onto the vector v is

  |proj_v w| = |w| cos theta = v · w / | v |

  If you want the projection of the vector w onto the vector v as a vector, then just multiply the above magnitude by v normalized:

  proj_v w = ( v · w ) v / v · v These expressions get used all the time, so either remember them or remember how to derive them. Here's one application: the work done by exerting a force F over a displacement d is equal to the product of the magnitude of the displacement and the component of the force in the direction of the displacement: Work = |d| proj_{d F = d · F}

• Of course, the dot product is at its most powerful when you combine the strengths of its two definitions, when you use the fact that |v| |w| cos theta = v₁ w₁ + v₂ w₂ + v₃ w₃

  For example, here is a formula for the angle between two vectors that is great if you have a calculator with an inverse cosine button:

  theta = cos^-1 ( v · w )/( |v| |w| )

Exercises:

1. Why can't you prove that the dot product is associative?
2. Calculate the dot product of (1,2,3) and (4,5,6).
3. Calculate the dot product of two unit vectors separated by an angle of 60 degrees.
4. What is i · j? i · i?
5. Find a vector orthogonal to (1,2,3).
6. Prove that projection is additive; i.e., that proj_v ( w₁ + w₂ ) = ( proj_v w₁ ) + ( proj_v w₂ ).
7. Calculate the angle between (0,4,-6) and (3,0,-2).
8. Calculate the projection of (1,2,3) onto (4,4,4).
9. Calculate the work done against the force (0,0,-30) by moving an object from (1,1,1) to (-10,3,7).
10. Prove the Cauchy-Bunyakovskii-Schwarz inequality: v · w <= | v | | w |.
11. Prove the triangle inequality: | v + w | <= | v | + | w |. [Hint: square both sides and apply Cauchy-Bernard-Schwartz.] Where does the name triangle inequality come from?