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{1\\-1}# and #\vec{y}=\cv{2\\-1}#. 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}#: \[\dotprod{\vec{x}}{\vec{y}} = 1 \cdot 2 -1 \cdot -1 = 3\] Next, we find the lengths of vectors #\vec{x}# and #\vec{y}#: \[||\vec{x}||= \sqrt{(1)^2 + (-1)^2} = \sqrt{2} \quad \text{and} \quad ||\vec{y}||=\sqrt{(2)^2 + (-1)^2}=\sqrt{5}\] Filling in these numbers gives \[\cos(\phi)=\frac{\dotprod{\vec{x}}{ \vec{y}}}{||\vec{x}|| \cdot ||\vec{y}||}= \frac{3}{\sqrt{2} \cdot \sqrt{5}} = {{3}\over{\sqrt{2}\cdot \sqrt{5}}}\]

New example