Vector Pythagorean Theorem

# The Pythagorean Theorem in Vector Geometry

Recall that for a right triangle with side lengths $a$, $b$, and $c$ where $c$ is the hypotenuse, then $a^2 + b^2 = c^2$. We will prove a similar result with regards to vectors. Suppose that we have two vectors $\vec{u}$ and $\vec{v}$ that are perpendicular to each other, that is $\vec{u} \perp \vec{v}$. The vector $\vec{u} + \vec{v}$ will be the hypotenuse of our triangle:

We will go on to prove an analogue to the Pythagorean Theorem for vectors.

Theorem 1 (The Pythagorean Theorem for Vectors): If $\vec{u}, \vec{v} \in \mathbb{R}^n$ then $\| \vec{u} + \vec{v} \| ^2 = \| \vec{u} \| ^2 + \| \vec{v} \| ^2$. |

**Proof:**The prove is done algebraically while noting that $\vec{u} \cdot \vec{v} = 0$ since $\vec{u} \perp \vec{v}$.

\begin{align} \| \vec{u} + \vec{v} \|^2 = (\vec{u} + \vec{v})(\vec{u} + \vec{v}) \\ \| \vec{u} + \vec{v} \|^2 = \|\vec{u}\| ^2 + 2(\vec{u} \cdot \vec{v}) + \|\vec{v}\|^2 \\ \| \vec{u} + \vec{v} \|^2 = \| \vec{u} \|^2 + \| \vec{v} \| ^2 \\ \blacksquare \end{align}