The Dot Product of Vectors

This page is intended to be a part of the Calculus hub. More in depth math on vectors and matrices can be found on the Linear Algebra hub.

The Dot Product of Vectors

We have just introduced vectors on the Introduction to Vectors page. We will now look at some more advanced material regarding vectors such as an important operation known as the dot product.

Definition: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$ where $\vec{u} = (u_1, u_2, ..., u_n)$ and $\vec{v} = (v_1, v_2, ..., v_n)$. Then the Dot Product between $\vec{u}$ and $\vec{v}$ is the sum of the products of corresponding components, that is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n$.

We should note that the dot product produces a number, NOT a vector. This is important to note as we will soon look at another operation known as the Cross Product by which we obtain a vector instead of a number. Do not confuse these two!

We will look at some theorems regarding the dot products of vectors. Their proofs are rather elementary, so the reader need not worry about them.

Theorem 1: Let $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$ and let $k$ be a scalar. Then the following properties hold:
a) $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$ (Commutativity Property of the Dot Product).
b) $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$ (Distributivity Property of the Dot Product.
c) $(k \vec{u} ) \cdot \vec{v} = k (\vec{u} \cdot \vec{v})$.
d) $\vec{u} \cdot \vec{u} = \| \vec{u} \| ^2$.
e) $\vec{u} \cdot \vec{0} = 0$.
  • Proof of a): Let $\vec{u}, \vec{v} \in \mathbb{R}^n$. Then we have that:
\begin{align} \quad \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n = v_1u_1 + v_2u_2 + ... + v_nu_n = \vec{v} \cdot \vec{u} \end{align}
  • Proof of b): Let $\vec{u}, \vec{v} \in \mathbb{R}^n$. Then we have that:
\begin{align} \quad \quad \vec{u} \cdot (\vec{v} + \vec{w}) = u_1(v_1 + w_1) + u_2(v_2 + w_2) + ... + u_n(v_n + w_n) = u_1v_1 + u_1w_1 + u_2v_2 + u_2w_2 + ... + u_nv_n + u_nw_n = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} \end{align}
  • Proof of c) Let $\vec{u}, \vec{v} \in \mathbb{R}^n$ and $k \in \mathbb{R}$. Then we have that:
\begin{align} \quad \quad (k \vec{u}) \cdot \vec{v} = (ku_1, ku_2, ..., ku_n) \cdot (v_1, v_2, ..., v_n) = ku_1v_1 + ku_2v_2 + ... + ku_nv_n = k(u_1v_1 + u_2v_2 + ... + u_nv_n) = k ( \vec{u} \cdot \vec{v} ) \end{align}
  • Proof of d): Let $\vec{u} \in \mathbb{R}^n$. Then we have that:
\begin{align} \quad \quad \vec{u} \cdot \vec{u} = u_1^2 + u_2^2 + ... + u_n^2 = \left(\sqrt{u_1^2 + u_2^2 + ... + u_n^2} \right)^2 = \| \vec{u} \| ^2 \end{align}
  • Proof of e) Let $\vec{u} \in \mathbb{R}^n$. Then we have that:
\begin{align} \quad \vec{u} \cdot \vec{0} = (u_1)(0) + (u_2)(0) + ... + (u_n)(0) = 0 \quad \blacksquare \end{align}

We will now look at a much more useful theorem regarding an alternate form of the dot product that says that $\vec{u} \cdot \vec{v}$ is equal to the product of the magnitudes of $\vec{u}$ and $\vec{v}$ multiplied by the cosine of the angle $\theta$ between these two vectors. We will look at theorem for $\mathbb{R}^2$, and the theorem for $\mathbb{R}^3$ is analogous. For $\mathbb{R}^n$, the angle $\theta$ between $\vec{u}$ and $\vec{v}$ is defined as the rearrangement of the dot product formula described below.

Theorem 2: If $\vec{u}, \vec{v} \in \mathbb{R}^2$ and $0 ≤ \theta ≤ \pi$ is the angle between $\vec{u}$ and $\vec{v}$ then $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta$.

We will not prove theorem 2, but the reader is advised to look up a geometric proof.

Corollary 1: If $\vec{u}, \vec{v} \in \mathbb{R}^n$ are nonzero vectors and if $\theta$ is the angle between $\vec{u}$ and $\vec{v}$ then $\cos \theta = \frac{\vec{u} \cdot \vec{v}}{\| \vec{u} \| \| \vec{v} \|}$.

We note that the formula in corollary one does not have division by zero since if $\vec{u}$ and $\vec{v}$ are nonzero vectors, then both $\| \vec{u} \|, \| \vec{v} \| > 0$.

Corollary 2: If $\vec{u}, \vec{v} \in \mathbb{R}^n$, then $\vec{u} \cdot \vec{v} = 0$ if and only if $\vec{u}$ is perpendicular/orthogonal to $\vec{u}$, that is $\vec{u} \perp \vec{v}$.
  • Proof: $\Rightarrow$ Suppose that $\vec{u} \cdot \vec{v} = 0$. Assume that neither $\vec{u}$ nor $\vec{v}$ are the zero vector since the zero vector is already perpendicular to every other vector. Then $\| \vec{u} \| \| \vec{v} \| \cos \theta = 0$, and so $\cos \theta = 0$. But this implies that $\theta = \frac{\pi}{2}$, and so the angle between $\vec{u}$ and $\vec{v}$ is $\frac{\pi}{2}$ so $\vec{u} \perp \vec{v}$.
  • $\Leftarrow$ Suppose that $\vec{u} \perp \vec{v}$. Then the angle between these vectors is $\theta = \frac{\pi}{2}$, and so $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \frac{\pi}{2} = 0$. $\blacksquare$
Corollary 3: If $\vec{u}, \vec{v} \in \mathbb{R}^n$ and if $\theta$ is the angle between $\vec{u}$ and $\vec{v}$ then:
If $\theta$ is an acute angle, that is $0 < \theta < \frac{\pi}{2}$ then $\vec{u} \cdot \vec{v} > 0$.
If $\theta$ is an obtuse angle, that is $\frac{\pi}{2} < \theta < \pi$ then $\vec{u} \cdot \vec{v} < 0$.
  • Proof: Suppose that $0 < \theta < \frac{\pi}{2}$. Then $\cos \theta > 0$ and therefore $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta > 0$.
  • Now suppose that $\frac{\pi}{2} < \theta < \pi$. Then $\cos \theta < 0$ and therefore $\vec{u} \cdot \vec{v} = \| \vec{u} \| \| \vec{v} \| \cos \theta < 0$. $\blacksquare$
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License