Non-Zero Vectors Orthogonal iff Perpendicular

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Theorem

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


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


Proof

Necessary Condition

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

\(\displaystyle 0\) \(=\) \(\displaystyle \mathbf u \cdot \mathbf v\) Definition of Orthogonal Vectors
\(\displaystyle \) \(=\) \(\displaystyle \left\Vert{\mathbf u}\right\Vert \left\Vert{\mathbf v}\right\Vert \cos \theta\) Cosine Formula for Dot Product

where $\left\Vert{\cdot}\right\Vert$ denotes the Euclidean norm.

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

Then $\cos \theta = 0$.

It follows from Zeroes of Sine and Cosine that $\theta = \dfrac \pi 2$, given that $\theta \in \left[{0\,.\,.\,\pi}\right]$ 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.

$\Box$


Sufficient Condition

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.

$\blacksquare$