Orthogonal vectors

In the plane, \(u\cdot v=\|u\|\|v\|\cos\theta\) vanishes with positive norms only when \(\cos\theta=0\), that is, at \(90^\circ\). In \(\mathbb{R}^n\) that computation is taken as the definition, so no picture is needed. The zero vector is orthogonal to everything, since \(0\cdot v=0\) for any \(v\); it is the only vector orthogonal to itself, because \(u\cdot u=\|u\|^2\).

The most used consequence is the Pythagorean theorem: if \(u\perp v\), then \(\|u+v\|^2=\|u\|^2+\|v\|^2\), because the cross term \(2\,u\cdot v\) disappears. For example, \((1,2)\) and \((-2,1)\) are orthogonal since \(-2+2=0\), and in the plane a vector orthogonal to \((a,b)\) is always obtained by rotating: \((-b,a)\).

The vectors orthogonal to a fixed nonzero \(u\) form a subspace: the equation \(u\cdot x=0\) is linear and homogeneous. In \(\mathbb{R}^3\) it is a plane through the origin; in \(\mathbb{R}^2\), a line. A common mistake is to think orthogonal and linearly independent mean the same: \((1,0)\) and \((1,1)\) are independent but not orthogonal. What is true is that nonzero, mutually orthogonal vectors are always independent.

This notion is the step before projecting one vector onto another: the remainder of a projection is precisely an orthogonal vector. Later it will be used to build bases of mutually perpendicular vectors, where coordinates are found with a single dot product.