How to Solve for Properties and Proofs of the Dot Product for Calculus
The Definition of the Dot Product
The dot product is an incredible tool of higher-level math, though many may not know how we formally arrive at its results but just know its applications. While it is good to know how tu use the properties, it is often insightful to know where they came from. To see this at work, observe two vectors and the difference between them. To find the hypotenuse, or the difference between vectors, we need to take advantage of the Law of Cosines, or that
c2 = a2 + b2 – 2abCos Θ.
In our case, we need to use the lengths of the vectors for our sides of the triangle, or ||a||, ||b||, and ||a-b||. So
||a-b||2 = ||a||2 + ||b||2 – 2 ||a||||b||Cos Θ.
||a-b||2 = ([(a1 – b1)2 + (a2 – b2)2 + (a3 – b3)2]0.5)2
= (a1 – b1)2 + (a2 – b2)2 + (a3 – b3)2
= (a1 – b1)(a1 – b1) + (a2 – b2)(a2 – b2)+ (a3 – b3)(a3 – b3)
= a12 – 2a1b1 + b12 + a22 – 2a2b2 + b22 + a32 – 2a3b3 + b32
Putting this back in for ||a-b||2 means that
a12 – 2a1b1 + b12 + a22 – 2a2b2 + b22 + a32 – 2a3b3 + b32 = ||a||2 + ||b||2 – 2 ||a||||b||Cos Θ
But notice that
||a||2 + ||b||2 = [(a12 + a22 + a32)0.5]2 + [(b12+ b22 + b32)0.5]2
= a12 + a22 + a32 + b12+ b22 + b32
a12 – 2a1b1 + b12 + a22 – 2a2b2 + b22 + a32 – 2a3b3 + b32 = a12 + a22 + a32 + b12+ b22 + b32 – 2 ||a||||b||Cos Θ
Wow, quite a lot there. But notice how we can cancel out all the squared terms from both sides! When we finish with this, we arrive at
– 2a1b1 – 2a2b2 – 2a3b3 = – 2 ||a||||b||Cos Θ
We can simplify the -2 out of the left hand side and then divide both sides by it, giving us
a1b1 + a2b2 + a3b3 =||a||||b||Cos Θ
So what is all that stuff on the left? We define that as the dot product and is known as a ∙ b. So we now know that
a ∙ b = a1b1 + a2b2 + a3b3
a ∙ b = ||a||||b||Cos Θ
This also gives us another way to find Cos Θ, for by moving terms to the other side,
Cos Θ = (a ∙ b) / (||a||||b||)
We have a new way to find the angle between vectors (Larson 782). Now let’s see what else we can use the dot product for.
It is important to note that the final result of the dot product is a number without direction, or a scalar. If you were to end up with a vector as your final answer then you know something is wrong and it is best to look over your work. It is also worth knowing if any of the previous properties of vectors apply to the dot product.
Does the commutative property apply? That is, does a ∙ b = b ∙ a? Well,
a ∙ b = a1b1 + a2b2 + a3b3
But because the components are real numbers and the order that I multiply real numbers does not matter,
a1b1 + a2b2 + a3b3 = b1a1 + b2a2 + b3a3
= b ∙ a
Yes, the dot product is commutative (781)
How about the distributive property? Will a ∙ (b + c) = a ∙ b + a ∙ c?
a ∙ (b + c) = a ∙ <(b1 + c1), (b2 + c2), (b3 + c3)>
= a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3)
But since I am multiplying a real number across a sum and the multiplication can be spread out,
a1(b1 + c1) + a2(b2 + c2) + a3(b3 + c3) = a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3
And because the order that I add real numbers doesn’t matter,
a1b1 + a1c1 + a2b2 + a2c2 + a3b3 + a3c3 = a1b1 + a2b2 + a3b3 + a1c1 + a2c2 + a3c3
= a ∙ b + a ∙ c
Yes, the dot product can be distributed (781).
What happens if I try to distribute a scalar across the dot product? Does c(a ∙ b) = ca ∙ cb? It doesn’t. Let’s see why.
c(a ∙ b) = c(a1b1 + a2b2 + a3b3)
= ca1b1 + ca2b2 + ca3b3
= (ca1)b1 + (ca2)b2 + (ca3)b3 OR = a1(cb1) + a2(cb2)+ a3(cb3)
Because of the associative property. Therefore,
(ca1)b1 + (ca2)b2 + (ca3)b3 = (ca) ∙ b
a1(cb1) + a2(cb2)+ a3(cb3) = a ∙ (cb)
Indeed, the dot product is not a direct summation but the sum of products, so you cannot distribute as we normally would. The dot product c(a ∙ b) = (ca) ∙ b = a ∙ (cb) (781).
The dot product of any vector and 0 is equal to 0. Note that it equals the number 0 and not the vector. To see this,
a ∙ 0 = a10 + a20 + a30
= 0 + 0 + 0 = 0 (781).
What does the dot product of a vector and itself equal?
a ∙ a = a1a1 + a2a2 + a3a3
= a12 + a22 + a32
Which is the length of the vector squared, so
a12 + a22 + a32 = ||a||2
Therefore, a ∙ a = ||a||2
Another important feature of the dot product tells us if two vectors are orthogonal, or perpendicular. Remember back to the unit circle for a moment. When we were at a 90 degree angle, the cosine (or x-component) was equal to zero. Therefore, if two vectors are orthogonal,
Cos 90 = (a ∙ b) / (||a||||b||) = 0
a ∙ b = 0
So if the dot product of two vectors equals zero, then you know that they are orthogonal (783).
Larson, Ron, Robert Hostetler, and Bruce H. Edwards. Calculus: Early Transcendental Functions. Maidenhead: McGraw-Hill Education, 2007. Print. 781-3.
- Deriving the Quadratic Formula
While often a useful tool in our math class, how many of us know why it works? Here is that reason.
- Solving Terminating Fractions and Repeating Decimals...
Repeating decimals and terminating fractions can seen foreign to us and hard to utilize. But for certain ones, they have a relation with 9 that makes them easy to solve.
© 2014 Leonard Kelley
More by this Author
Vectors have components that may overlap other vectors. We are interested in these projections for various reasons, and their proofs offer further insight into vector calculus.
These proofs go beyond the standard properties of the dot and cross product and explore new applications and surprising results.
While they are powerful engines of destruction, the beams they shoot into space are a separate breed of chaos which deserves a closer look.
No comments yet.