The dot product

Geometry: Parametric curves and vectors

The dot product

Previously, we discussed vectors and how to add them together or multiply them with a scalar. The next step is to define the dot product, which can be seen as a way to multiply one vector with another.

Let $\blue{\vec{x}} =\blue{\cv{x_1\\x_2}}$ and $\green{\vec{y}}=\green{\cv{y_1\\y_2}}$ be two vectors. We define the dot product between $\blue{\vec{x}}$ and $\green{\vec{y}}$ to be $\dotprod{\blue{\vec{x}} }{ \green{\vec{y}}} = \blue{x_1}\cdot \green{y_1} + \blue{x_2}\cdot\green{y_2}$ Note that the dot product between two vectors always produces a number, not a vector.

Example

Let $\blue{\vec{x}} =\blue{\cv{1\\1}}$ and $\green{\vec{y}}=\green{\cv{5\\0}}$ . Then $\dotprod{\blue{\vec{x}}}{ \green{\vec{y}}} = \blue{1}\cdot\green{5} + \blue{1}\cdot\green{0} = 5$ .

Notations
In different literature, you may find other ways to denote the dot product between two vectors $\blue{\vec{x}}$ and $\green{\vec{y}}$ . Below we list some of the possibilities.
$\blue{\vec{x}} \cdot \green{\vec{y}}, \enspace \blue{\vec{x}} \bullet \green{\vec{y}}, \enspace (\blue{\vec{x}}, \green{\vec{y}}), \enspace \left<\blue{\vec{x}},\green{\vec{y}}\right>$ The dot product can also be referred to as the (standard) inner product or the scalar product.

Properties
Let $\blue{\vec{x}}$ , $\green{\vec{y}}$ and $\orange{\vec{z}}$ be vectors and let $c$ be a number. The following properties hold.

1. $\quad \dotprod{\blue{\vec{x}}}{\green{\vec{y}}} = \dotprod{\green{\vec{y}}}{\blue{\vec{x}}}$
2. $\quad \dotprod{\blue{\vec{x}}}{ (\green{\vec{y}} + \orange{\vec{z}})} = \dotprod{\blue{\vec{x}}}{\green{\vec{y}}} + \dotprod{\blue{\vec{x}}}{\orange{\vec{z}}}$
3. $\quad c \cdot (\dotprod{\blue{\vec{x}} }{ \green{\vec{y}}}) = \dotprod{(c\cdot \blue{\vec{x}}) }{ \green{\vec{y}}} = \dotprod{\blue{\vec{x}} }{ (c\cdot \green{\vec{y}})}$
4. $\quad \dotprod{\blue{\vec{x}} }{ \blue{\vec{x}}} = ||\blue{\vec{x}}||^2$ where $||\blue{\vec{x}}||$ is the length of $\blue{\vec{x}}$ .

The dot product gives rise to many interesting applications, for instance, it allows us to compute the angle between two vectors. To calculate this angle we need another formula.

Let $\blue{\vec{x}} =\blue{\cv{x_1\\x_2}}$ and $\green{\vec{y}}=\green{\cv{y_1\\y_2}}$ be vectors and let $\phi$ be the smallest angle between them. Then $\cos(\phi) = \frac{\dotprod{\blue{\vec{x}} }{ \green{\vec{y}}}}{||\blue{\vec{x}}|| \cdot ||\green{\vec{y}}||}$ where $||\blue{\vec{x}}||$ and $||\green{\vec{y}}||$ denote the lengths of vectors $\blue{\vec{x}}$ and $\green{\vec{y}}$ respectively.

Example
Let again $\blue{\vec{x}} =\blue{\cv{1\\1}}$ and $\green{\vec{y}}=\green{\cv{5\\0}}$ .

GeoGebra

$\cos(\phi)= \frac{\dotprod{\blue{\vec{x}} }{ \green{\vec{y}}}}{||\blue{\vec{x}}|| \cdot ||\green{\vec{y}}||} = \frac{5}{\sqrt{2} \cdot 5} = \frac{1}{\sqrt{2}},$ meaning that $\phi = \frac{\pi}{4}$ .

Smallest angle
Note that $\phi$ is the smallest angle between the two vectors, meaning that $0 \leq \phi \leq \pi$ .

In many applications, it is useful to work with vectors that are orthogonal.

Two vectors $\blue{\vec{x}}$ and $\green{\vec{y}}$ are orthogonal, or perpendicular, if the smallest angle $\phi$ between them is $\frac{\pi}{2}$ .

Since we know $\cos\left(\frac{\pi}{2}\right)=0$ , we have that $\blue{\vec{x}}$ and $\green{\vec{y}}$ are orthogonal if and only if $\dotprod{\blue{\vec{x}} }{ \green{\vec{y}}}=0$ .

Example
Let $\blue{\vec{x}} =\blue{\cv{1\\2}}$ and $\green{\vec{y}}=\green{\cv{2\\-1}}$ . We have $\dotprod{\blue{\vec{x}}}{ \green{\vec{y}}}=\blue{1}\cdot\green{2} + \blue{2} \cdot \green{-1} = 0$ and indeed $\phi=\frac{\pi}{2}$ .

GeoGebra

Normal vector
Let $\blue{\vec{x}}$ and $\vec{n}$ be non-zero vectors. If $\blue{\vec{x}}$ and $\vec{n}$ are orthogonal, or in other words: if $\dotprod{\blue{\vec{x}} }{ \vec{n}}=0$ , we say that $\vec{n}$ is a normal vector to $\blue{\vec{x}}$ .

If $\blue{\vec{x}} =\blue{\cv{x_1\\x_2}}$ , then the vector $\vec{n}=\cv{-x_2\\x_1}$ is normal to $\blue{\vec{x}}$ since $\dotprod{\blue{\vec{x}} }{ \vec{n}}=\blue{x_1} \cdot -x_2 + \blue{x_2} \cdot x_1=0$ . Note that a normal vector is not unique: for example the vector $\vec{m}=\cv{-3\cdot x_2\\3\cdot x_1}$ is also a normal vector to $\blue{\vec{x}}$ . In fact, all non-zero scalar multiples of $\vec{n}=\cv{-x_2\\x_1}$ are normal vectors to $\blue{\vec{x}}$ .

Let $\vec{x}=\cv{-2\\4}$ and $\vec{y}=\cv{-2\\2}$ . Find $\cos(\phi)$ , where $\phi$ is the smallest angle between $\vec{x}$ and $\vec{y}$ .

$\cos(\phi)={{3}\over{\sqrt{2}\cdot \sqrt{5}}}$

We want to use the formula $\cos(\phi)=\frac{\dotprod{\vec{x}}{ \vec{y}}}{||\vec{x}|| \cdot ||\vec{y}||}$ . First, we calculate the dot product between $\vec{x}$ and $\vec{y}$ : $\dotprod{\vec{x}}{\vec{y}} = -2 \cdot -2 + 4 \cdot 2 = 12$ Next, we find the lengths of vectors $\vec{x}$ and $\vec{y}$ : $||\vec{x}||= \sqrt{(-2)^2 + (4)^2} = 2\cdot \sqrt{5} \quad \text{and} \quad ||\vec{y}||=\sqrt{(-2)^2 + (2)^2}=2^{{{3}\over{2}}}$ Filling in these numbers gives $\cos(\phi)=\frac{\dotprod{\vec{x}}{ \vec{y}}}{||\vec{x}|| \cdot ||\vec{y}||}= \frac{12}{2\cdot \sqrt{5} \cdot 2^{{{3}\over{2}}}} = {{3}\over{\sqrt{2}\cdot \sqrt{5}}}$

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