Scalar (Dot) Product of Two Vectors

There are two distinct ways to "multiply" two vectors together, and they produce results of

entirely different character. The scalar product (also called the dot product) of

two vectors $\vec A$ and $\vec B$ is defined as

where $A$ and $B$ are the magnitudes of the two vectors and $\theta$ is the angle between

them when placed tail-to-tail. As the name says, the result is a pure scalar (a number,

with appropriate units), not a vector.

Geometric meaning. $B\cos\theta$ is the length of the projection ("shadow") of $\vec B$

onto the direction of $\vec A$. So $\vec A \cdot \vec B$ can be read as "the magnitude of

$\vec A$ times the component of $\vec B$ along $\vec A$" (or, equivalently, the magnitude of

$\vec B$ times the component of $\vec A$ along $\vec B$, since the dot product is symmetric).

Properties. The dot product is commutative, $\vec A \cdot \vec B = \vec B \cdot \vec A$,

and distributive over vector addition,

$\vec A \cdot (\vec B + \vec C) = \vec A\cdot\vec B + \vec A\cdot\vec C$. For the standard

unit vectors, $\hat i \cdot \hat i = \hat j \cdot \hat j = 1$ (a unit vector with itself, at

$\theta = 0^\circ$, gives $1 \times 1 \times \cos 0^\circ = 1$) and

$\hat i \cdot \hat j = 0$ (perpendicular unit vectors, $\theta = 90^\circ$, give

$\cos 90^\circ = 0$). Using these facts and the distributive property, the dot product of two

vectors given in component form works out to

a purely algebraic formula that needs no angle at all (its derivation is worked through step

by step as an exercise).

Special cases. When $\vec A$ and $\vec B$ are parallel ($\theta = 0^\circ$), the dot

product takes its largest possible value, $AB$. When they are perpendicular …