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.

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