Editing Euclidean space (section)

===Orthogonality===
{{main|Perpendicular|Orthogonality}}
Two nonzero vectors {{mvar|u}} and {{mvar|v}} of <math>\overrightarrow E</math> (the associated vector space of a Euclidean space {{mvar|E}}) are ''perpendicular'' or ''orthogonal'' if their inner product is zero:

<math display="block"> u \cdot v =0</math>

Two linear subspaces of <math>\overrightarrow E</math> are orthogonal if every nonzero vector of the first one is perpendicular to every nonzero vector of the second one. This implies that the intersection of the linear subspaces is reduced to the zero vector.

Two lines, and more generally two Euclidean subspaces (A line can be considered as one Euclidean subspace.) are orthogonal if their directions (the associated vector spaces of the Euclidean subspaces) are orthogonal. Two orthogonal lines that intersect are said ''perpendicular''.

Two segments {{math|''AB''}} and {{math|''AC''}} that share a common endpoint {{math|''A''}} are ''perpendicular'' or ''form a [[right angle]]'' if the vectors <math>\overrightarrow {AB}\vphantom{\frac){}}</math> and <math>\overrightarrow {AC}\vphantom{\frac){}}</math> are orthogonal.

If {{math|''AB''}} and {{math|''AC''}} form a right angle, one has

<math display="block">|BC|^2 = |AB|^2 + |AC|^2.</math>

This is the [[Pythagorean theorem]]. Its proof is easy in this context, as, expressing this in terms of the inner product, one has, using bilinearity and symmetry of the inner product:

<math display="block">\begin{align}
|BC|^2 &= \overrightarrow {BC}\cdot \overrightarrow {BC} \vphantom{\frac({}}\\[2mu]
&=\Bigl(\overrightarrow {BA}+\overrightarrow {AC}\Bigr) \cdot \Bigl(\overrightarrow {BA}+\overrightarrow {AC}\Bigr)\\[4mu]
&=\overrightarrow {BA}\cdot \overrightarrow {BA}+ \overrightarrow {AC}\cdot \overrightarrow {AC} -2 \overrightarrow {AB}\cdot \overrightarrow {AC}\\[6mu]
&=\overrightarrow {AB}\cdot \overrightarrow {AB} + \overrightarrow {AC}\cdot\overrightarrow {AC}\\[6mu]
&=|AB|^2 + |AC|^2.
\end{align}</math>

Here, <math>\overrightarrow {AB}\cdot \overrightarrow {AC} = 0 \vphantom{\frac({}}</math> is used since these two vectors are orthogonal.