Vector Dot Product Euclidean Inner Product

Dot Product (Euclidean Inner Product)

Definition: The Dot Product (for vectors in 2-space and 3-space), also known as the Euclidean Inner Product (for vectors in n-space) of two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ denoted $\vec{u} \cdot \vec{v}$ is a method of multiplying two vectors resulting in a numerical quantity defined by $\vec{u} \cdot \vec{v} = \mid \vec{u} \mid \mid \vec{v} \mid \cos \theta$ or $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n$ (component form of the dot product).

Note that we have two forms of the dot product. We will now go on to prove the component form of the dot product, that is $\vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n$ in Euclidean 3-space is equivalent to our original definition so $\vec{u} \cdot \vec{v} = \mid \vec{u} \mid \mid \vec{v} \mid \cos \theta = u_1v_1 + u_2v_2 +... + u_nv_n$.

Theorem 1: If $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $\mid \vec{u} \mid \mid \vec{v} \mid \cos \theta = u_1v_1 + u_2v_2 + ... + u_nv_n$.

• Proof in 3-Space: Let $\vec{u} = (u_1, u_2, u_3), \vec{v} = (v_1, v_2, v_3) \in \mathbb{R}^3$, and position $\vec{u}$ and $\vec{v}$ so that their initial points are at the origin. Now construct a vector that connects these vectors together. This vector will be $\vec{v} - \vec{u}$ as illustrated:

• Recall the formula for the law of cosines ($a^2 = b^2 + c^2 - 2bc \cos(\theta_a)$) that relates the angle of two connected sides to the lengths of the three sides of the triangle. In this case, the length of the three sides of our triangle will be $\| \vec{u} \|$, $\| \vec{v} \|$ and $\| \vec{v} - \vec{u} \|$. Substituting what we know into the law of cosines we get that:

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\begin{align} a^2 = b^2 + c^2 - 2bc \cos(\theta_a) \\ \| \vec{v} - \vec{u} \|^2 = \| \vec{u} \|^2 + \| \vec{v} \|^2 - 2 \| \vec{u} \| \| \vec{v} \| \cos (\theta_{\| \vec{u} \|}) \\ 2\| \vec{u} \| \| \vec{v} \| \cos(\theta) = \| \vec{u} \|^2 + \| \vec{v} \|^2 - \| \vec{v} - \vec{u} \|^2 \\ \| \vec{u} \| \| \vec{v} \| \cos(\theta) = \frac{1}{2} ( \| \vec{u} \|^2 + \| \vec{v} \|^2 - \| \vec{v} - \vec{u} \|^2 ) \end{align}

• We defined the dot product of two vectors to be $\vec{u} \cdot \vec{v} = \mid \vec{u} \mid \mid \vec{v} \mid \cos \theta$, so we can make that substitution from our equation.

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\begin{align} \vec{u} \cdot \vec{v} = \frac{1}{2} ( \| \vec{u} \|^2 + \| \vec{v} \|^2 - \| \vec{v} - \vec{u} \|^2 ) \\ \vec{u} \cdot \vec{v} = \frac{1}{2} \left (\left( \sqrt{u_{1}^2 + u_{2}^2 + u_{3}^2} \right)^2 + \left( \sqrt{v_{1}^2 + v_{2}^2 + v_{3}^2} \right)^2 - \left(\sqrt{(v_{1} - u_{1})^2 + (v_{2} - u_{2})^2 + (v_{3} - u_{3})^2}\right)^2 \right) \\ \quad \vec{u} \cdot \vec{v} = \frac{1}{2} ( [u_{1}^2 + u_{2}^2 + u_{3}^2] + [v_{1}^2 + v_{2}^2 + v_{3}^2] - [(v_{1}^2 - 2v_{1}u_{1} + u_{1}^2) + (v_{2}^2 - 2v_{2}u_{2} + u_{2}^2 ) + (v_{3}^2 - 2v_{3}u_{3} + u_{3}^2)]) \\ \vec{u} \cdot \vec{v} = \frac{1}{2} ( u_{1}^2 + u_{2}^2 + u_{3}^2 + v_{1}^2 + v_{2}^2 + v_{3}^2 - v_{1}^2 + 2v_{1}u_{1} - u_{1}^2 - v_{2}^2 + 2v_{2}u_{2} - u_{2}^2 - v_{3}^2 + 2v_{3}u_{3} - u_{3}^2) \\ \vec{u} \cdot \vec{v} = \frac{1}{2} (2v_{1}u_{1} + 2v_{2}u_{2}+ 2v_{3}u_{3}) \\ \vec{u} \cdot \vec{v} = v_{1}u_{1} + v_{2}u_{2} + v_{3}u_{3} \\ \vec{u} \cdot \vec{v} = u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} \\ \blacksquare \end{align}

Angles Between Vectors

The original formula for the dot product is helpful in finding the angle between two vectors $\vec{u}$ and $\vec{v}$. For example, if we take the formula for the dot product and rearrange it to isolate cosine, we get that:

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\begin{align} cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{ \mid \mid \vec{u} \mid \mid \mid \mid \vec{v} \mid \mid } \end{align}

Acute, Obtuse, and Perpendicular Vector Angles

Recall the following three classifications of angles:

Name of Angle	Range of Angle	Sign of Cosine
Acute Angles	$0 < \theta < \frac{pi}{2}$	$\cos \theta > 0$
Perpendicular (Right) Angles	$\theta = \frac{pi}{2}$	$\cos \theta = 0$
Obtuse Angles	$\frac{\pi}{2} < \theta < \pi$	$\cos \theta < 0$

We also note that the norm of any vector is always positive, hence, we have now found an important property of the vector dot product. If we know the sign of the dot product between two vectors, then we also know if the angle between them is acute or obtuse since the value of cosine affects the entire sign of the dot product. Furthermore, if $\vec{u} \cdot \vec{v} = 0$, then these two vectors are perpendicular to each other.

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\begin{align} \mathrm{if} \quad \vec{u} \cdot \vec{v} > 0, \quad 0 < \theta < \frac{\pi}{2} \\ \mathrm{if} \quad \vec{u} \cdot \vec{v} < 0, \quad \frac{\pi}{2} < \theta < \pi \\ \mathrm{if} \quad \vec{u} \cdot \vec{v} = 0, \quad \theta = \frac{\pi}{2} \end{align}

Properties of the Dot Product

Theorem 2 (Commutativity of the Dot Product): For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$, their dot product is commutative, that is $\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$.

• Proof: We will use the component form of the dot product to prove this. Let $\vec{u}, \vec{v} \in \mathbb{R}^n$. The proof will follow straight forward:

(5)

\begin{align} \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + ... + u_nv_n \\ \vec{u} \cdot \vec{v} = v_1u_1 + v_2u_2 + ... + v_nu_n \\ \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u} \\ \blacksquare \end{align}

Theorem 3 (Distributivity of the Dot Product): For any three vectors $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$, their dot product is distributive, that is $\vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w}$.

• Proof: Let $\vec{u}, \vec{v}, \vec{w} \in \mathbb{R}^n$.

(6)

\begin{align} \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}) \\ \vec{u} \cdot (\vec{v} + \vec{w}) = u_{1}v_{1} + u_{1}w_{1} + u_{2}v_{2} + u_{2}w_{2} + ... + u_{n}v_{n} + u_{n}w_{n}\\ \vec{u} \cdot (\vec{v} + \vec{w}) = u_{1}v_{1} + u_{2}v_{2} + ... + u_{n}v_{n} + u_{1}w_{1} + u_{2}w_{2} + ... + u_{n}w_{n} \\ \vec{u} \cdot (\vec{v} + \vec{w}) = \vec{u} \cdot \vec{v} + \vec{u} \cdot \vec{w} \\ \blacksquare \end{align}

Theorem 4: For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ and some scalar $k$, it follows that $k(\vec{u} \cdot \vec{v}) = (k\vec{u}) \cdot \vec{v}) = \vec{u} \cdot (k \vec{v})$.

• Proof: Let $\vec{u}, \vec{v} \in \mathbb{R}^n$ and let $k$ be a scalar.

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\begin{align} k(\vec{u} \cdot \vec{v}) = k(u_{1}v_{1} + u_{2}v_{2} + ... + u_{n}v_{n} \\ k(\vec{u} \cdot \vec{v}) = (ku_{1}v_{1} + ku_{2}v_{2} + ... + ku_{n}v_{n}) \\ k(\vec{u} \cdot \vec{v}) = u_{1}(kv_{1}) + u_{2}(kv_{2}) + ... + u_{n}(kv_{n}) \\ k(\vec{u} \cdot \vec{v}) = \vec{u} \cdot (k\vec{v}) \\ \blacksquare \end{align}

We will omit the rest of the proof as it is practically identical to the above.

Theorem 5: Given any vector $\vec{u} \in \mathbb{R}^n$, it follows that $\vec{u} \cdot \vec{u} ≥ 0$.

• Proof: We first note that $\vec{u} \cdot \vec{u} = u_1^2 + u_2^2 + ... + u_n^2$. All of the elements in this sum are nonnegative so the dot product is not negative either. We note that if $\vec{u} = \vec{0}$, then $\vec{u} \cdot \vec{u} = 0$. $\blacksquare$

Alternate Form of the Euclidean Inner Product

We will now look at yet another form of the dot product which can be useful in certain circumstances.

Theorem 6: For any two vectors $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $\vec{u} \cdot \vec{v} = \frac{1}{4} \mid \mid \vec{u} + \vec{v} \mid \mid^2 - \frac{1}{4} \mid \mid \vec{u} - \vec{v} \mid \mid^2$.

• Proof: We will prove our claim by showing that the righthand side is equal to the lefthand side.

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\begin{align} \quad \frac{1}{4} \mid \mid \vec{u} + \vec{v} \mid \mid ^2 - \frac{1}{4} \mid \mid \vec{u} - \vec{v} \mid \mid ^2 \\ \quad = \frac{1}{4} (\vec{u} + \vec{v}) \cdot (\vec{u} + \vec{v}) - \frac{1}{4}(\vec{u} - \vec{v}) \cdot (\vec{u} - \vec{v}) \\ \quad = \frac{1}{4} (\mid \mid \vec{u} \mid \mid ^2 + 2(\vec{u} \cdot \vec{v}) + \mid \mid \vec{v}\mid \mid ^2) - \frac{1}{4}(\mid \mid \vec{u} \mid \mid ^2 - 2(\vec{u} \cdot \vec{v}) + \mid \mid \vec{v}\mid \mid ^2) \\ \quad = \frac{1}{4} \mid \mid \vec{u} \mid \mid ^2 + \frac{1}{2}\vec{u} \cdot \vec{v} + \frac{1}{4}\mid \mid \vec{v}\mid \mid ^2 - \frac{1}{4}\mid \mid \vec{u} \mid \mid ^2 + \frac{1}{2}\vec{u} \cdot \vec{v} - \frac{1}{4}\mid \mid \vec{v}\mid \mid ^2 \\ \quad = \frac{1}{2} \vec{u} \cdot \vec{v} + \frac{1}{2}\vec{u} \cdot \vec{v} \\ \quad = \vec{u} \cdot \vec{v} \\ \blacksquare \end{align}