For unit vectors $u$ and $v$ the scalar product gives a measure of how much of $u$ is in the direction of $v$ (and symmetrically, how much of $v$ is in the direction of $u$ ). That means that if we take the scalar product of orthogonal vectors, the result must be $0.$ That is clear from the geometrical version, because the angle $\theta$ will be 90 degrees for orthogonal vectors, so $\cos(\theta)$ is $0.$ The result is less obvious from the algebraic version.
The scalar product of two vectors is a scalar quantity.

Algebraically in two dimensions:

• $(a,b).(c,d)~=~ac+bd$
In higher dimensions we again take the sum of the products of the individual components.

Geometrically in two dimensions:

• $(a,b).(c,d)~=~|a|.|b|.\cos(\theta)$
where $|a|$ is the magnitude of a vector $a,$ and $\theta$ is the angle between the two vectors.

In higher dimensions the formula is the same, and we find the angle between the vectors by considering the plane that contains both vectors.

AbsoluteValue
ArgandDiagram
MathematicsTaxonomy
UnitVector
VectorSubtraction
(none) DifferenceOfTwoSquares
EquationOfALine
LinearFunction
Matrices
MatrixTransformation
TypesOfNumber
MagnitudeOfAVector
Vectors
MatrixMultiplication
Quaternion

ScalarProduct
UnitVector
AbsoluteValue