# Notes on How to Solve for Properties and Proofs of the Dot Product for Calculus

*Leonard Kelley holds a bachelor's in physics with a minor in mathematics. He loves the academic world and strives to constantly explore it.*

## 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

c^{2} = a^{2} + b^{2} – 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.

Notice that

||**a-b**||^{2} = ([(a_{1} – b_{1})^{2} + (a_{2} – b_{2})^{2 }+ (a_{3} – b_{3})^{2}]^{0.5})^{2}

= (a_{1} – b_{1})^{2} + (a_{2} – b_{2})^{2 }+ (a_{3} – b_{3})^{2}

= (a_{1} – b_{1})(a_{1} – b_{1}) + (a_{2} – b_{2})(a_{2} – b_{2})+ (a_{3} – b_{3})(a_{3} – b_{3})

= a_{1}^{2} – 2a_{1}b_{1} + b_{1}^{2} + a_{2}^{2} – 2a_{2}b_{2} + b_{2}^{2} + a_{3}^{2} – 2a_{3}b_{3} + b_{3}^{2}

Putting this back in for ||**a-b**||^{2} means that

a_{1}^{2} – 2a_{1}b_{1} + b_{1}^{2} + a_{2}^{2} – 2a_{2}b_{2} + b_{2}^{2} + a_{3}^{2} – 2a_{3}b_{3} + b_{3}^{2} = ||**a**||^{2} + ||**b**||^{2} – 2 ||**a**||||**b**||Cos Θ

But notice that

||**a**||^{2} + ||**b**||^{2} = [(a_{1}^{2} + a_{2}^{2} + a_{3}^{2})^{0.5}]^{2 }+ [(b_{1}^{2}+ b_{2}^{2} + b_{3}^{2})^{0.5}]^{2}

= a_{1}^{2} + a_{2}^{2} + a_{3}^{2 }+ b_{1}^{2}+ b_{2}^{2} + b_{3}^{2}

So

a_{1}^{2} – 2a_{1}b_{1} + b_{1}^{2} + a_{2}^{2} – 2a_{2}b_{2} + b_{2}^{2} + a_{3}^{2} – 2a_{3}b_{3} + b_{3}^{2} = a_{1}^{2} + a_{2}^{2} + a_{3}^{2 }+ b_{1}^{2}+ b_{2}^{2} + b_{3}^{2} – 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

– 2a_{1}b_{1} – 2a_{2}b_{2} – 2a_{3}b_{3} = – 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

a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3 }=||**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 = **a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}

And

**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.

## Additive Properties

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 = **a_{1}b_{1} + a_{2}b_{2 }+ a_{3}b_{3}

But because the components are real numbers and the order that I multiply real numbers does not matter,

a_{1}b_{1} + a_{2}b_{2 }+ a_{3}b_{3 }= b_{1}a_{1 }+ b_{2}a_{2} + b_{3}a_{3}

= **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 ∙ <(**b_{1} + c_{1}), (b_{2} + c_{2}), (b_{3} + c_{3})>

= a_{1}**(**b_{1} + c_{1}) + a_{2}(b_{2} + c_{2}) + a_{3}(b_{3} + c_{3})

But since I am multiplying a real number across a sum and the multiplication can be spread out,

a_{1}**(**b_{1} + c_{1}) + a_{2}(b_{2} + c_{2}) + a_{3}(b_{3} + c_{3}) = a_{1}b_{1} + a_{1}c_{1} + a_{2}b_{2} + a_{2}c_{2} + a_{3}b_{3} + a_{3}c_{3}

And because the order that I add real numbers doesn’t matter,

a_{1}b_{1} + a_{1}c_{1} + a_{2}b_{2} + a_{2}c_{2} + a_{3}b_{3} + a_{3}c_{3 }= a_{1}b_{1 }+ a_{2}b_{2} + a_{3}b_{3 }+ a_{1}c_{1} + a_{2}c_{2} + a_{3}c_{3}

_{= }**a ∙ b + a ∙ c**

Yes, the dot product can be distributed (781).

## Multiplicative Properties

What happens if I try to distribute a scalar across the dot product? Does c(**a ∙ b) = **c**a ∙ **c**b**? It doesn’t. Let’s see why.

c(**a ∙ b) = **c(a_{1}b_{1} + a_{2}b_{2 }+ a_{3}b_{3})

= ca_{1}b_{1} + ca_{2}b_{2 }+ ca_{3}b_{3}

= (ca_{1})b_{1} + (ca_{2})b_{2 }+ (ca_{3})b_{3 }OR = a_{1}(cb_{1}) + a_{2}(cb_{2})+ a_{3}(cb_{3})

Because of the associative property. Therefore,

(ca_{1})b_{1} + (ca_{2})b_{2 }+ (ca_{3})b_{3} = (c**a**) **∙ b**

And

a_{1}(cb_{1}) + a_{2}(cb_{2})+ a_{3}(cb_{3}) = **a ∙ **(c**b**)

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) = **(c**a**) **∙ b = a ∙ **(c**b**) (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 = **a_{1}0 + a_{2}0 + a_{3}0

= 0 + 0 + 0 = 0 (781).

What does the dot product of a vector and itself equal?

**a ∙ a **= a_{1}a_{1} + a_{2}a_{2} + a_{3}a_{3}

= a_{1}^{2} + a_{2}^{2 }+ a_{3}^{2}

Which is the length of the vector squared, so

a_{1}^{2} + a_{2}^{2 }+ a_{3}^{2} = ||**a**||^{2}

Therefore, **a ∙ a = **||**a**||^{2}

## Orthogonal Vectors

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

Therefore,

**a ∙ b =** 0

So if the dot product of two vectors equals zero, then you know that they are orthogonal (783).

I hope you enjoyed these proofs, and check out my webpage for even more useful tips!

## Works Cited

Larson, Ron, Robert Hostetler, and Bruce H. Edwards. __Calculus: Early Transcendental Functions.__ Maidenhead: McGraw-Hill Education, 2007. Print. 781-3.

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**© 2014 Leonard Kelley**