Dot product

← Back: Simple arithmetic operations

Overview: Vector algebra

Next: Cross product →

There are exercises as selftest for this article.

You can use MATLAB for the content of this article.

The dot product of two vectors results in a scalar value and is defined as

$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot b\vec{\mathbf{e}}_{b} = a b \cos \alpha$

where $\alpha$ describes the angle between the two vectors which ranges from $0$ to $\pi$ (see figure). The dot product is denoted with a simple point between the vectors or without any sign.

Regarding the right side of the above equation, the following correlation can be noted: If you project the vector $\vec{\mathbf{b}}$ on the vector $\vec{\mathbf{a}}$, you get the distance $b\cos\alpha$. As a consequence the result of the dot product can be seen as the area of a rectangle with the side legths $a$ and $b\cos\alpha$. The projection can also be done contrariwise (projection of vector $\vec{\mathbf{a}}$ on vector $\vec{\mathbf{b}}$). So that you get the distance $a\cos\alpha$. The multiplication of this term with $b$ leads to a rectangle with equivalent area but different aspect ratio (see figure).

Another possibility to compute the dot product is to multiply the corresponding components and sum them up:

$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a b\cos\alpha = a_x b_x + a_y b_y + a_z b_z$

In general the dot product of n-dimensional vectors is computed as follows:

$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a b \cos\alpha = \sum_{i=1}^{n} a_i b_i$

which is nothing else than the matrix product of the transpose of the first vector with the second vector denoted in matrix algebra:

$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = \vec{\mathbf{a}}^T \vec{\mathbf{b}} = \left[\begin{array}{ccc} a_1 & \dots & a_n \end{array}\right] \left[\begin{array}{c} b_1 \\ \vdots \\ b_n \end{array}\right] = \sum_{i=1}^{n} a_i b_i$

On the basis of the described relations it appears, that the commutative law holds:

$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = \vec{\mathbf{b}} \cdot \vec{\mathbf{a}}$

Furthermore the following special cases can be considered, that often lead to simplifications in technical context:

$\begin{align} \vec{\mathbf{a}} \cdot \vec{\mathbf{b}} &= ab& &\text{if}& \vec{\mathbf{a}} & \upuparrows \vec{\mathbf{b}}& &(\text{because} \cos(0) = 1)\\ \vec{\mathbf{a}} \cdot \vec{\mathbf{b}} &= 0& &\text{if}& \vec{\mathbf{a}} &\perp \vec{\mathbf{b}}& &(\text{because} \cos(\frac{\pi}{2}) = 0)\\ \vec{\mathbf{a}} \cdot \vec{\mathbf{b}} &= -ab& &\text{if}& \vec{\mathbf{a}} &\downarrow\uparrow \vec{\mathbf{b}}& &(\text{because} \cos(\pi) = -1) \end{align}$

Multimedial educational material

http://demonstrations.wolfram.com/DotProduct/ Applet: Dot product of two vectors (free CDF-Player required)

Literature

• Kurt Meyberg und Peter Vachenauer, Höhere Mathematik 1: Differential- und Integralrechnung. Vektor- und Matrizenrechnung, 6. Edition (Springer Berlin Heidelberg, 2001)
• Manfred Albach, Grundlagen der Elektrotechnik 1: Erfahrungssätze, Bauelemente, Gleichstromschaltungen, 3. Edition (Pearson Studium, 2011)