Non-Zero Vectors are Orthogonal iff Perpendicular

Theorem

Let $\mathbf u$, $\mathbf v$ be non-zero vectors in the real Euclidean space $\R^n$.

Then $\mathbf u$ and $\mathbf v$ are orthogonal if and only if they are perpendicular.

When $\theta$ denotes the angle between $\mathbf u$ and $\mathbf v$ measured in radians, we have:

$\ds 0$

$=$

$\ds \mathbf u \cdot \mathbf v$

$\ds $

$=$

$\ds \norm {\mathbf u} \norm {\mathbf v} \cos \theta$

where $\norm {\,\cdot\,}$ denotes the Euclidean norm.

As $\mathbf u, \mathbf v$ are non-zero vectors, it follows from the norm axiom of positive definiteness that $\norm {\mathbf u} \ne 0$, and $\norm {\mathbf v} \ne 0$.

Then $\cos \theta = 0$.

It follows from Zeroes of Sine and Cosine that $\theta = \dfrac \pi 2$, given that $\theta \in \closedint 0 \pi$ by the convention of the definition of angle between vectors.

Then $\mathbf u$ and $\mathbf v$ are perpendicular by the measurement of a right angle.

$\Box$

If $\mathbf u$ and $\mathbf v$ are perpendicular, then the angle between them measures $\dfrac \pi 2$ in radians.

Then $\cos \theta = 0$, and the calculations above show that $\mathbf u \cdot \mathbf v = 0$.

Hence, $\mathbf u$ and $\mathbf v$ are orthogonal.

$\blacksquare$