What does a pair of orthonormal vectors in 2-D Euclidean space look like? Let u = (x1, y1) and v = (x2, y2). Consider the restrictions on x1, x2, y1, y2 required to make u and v form an orthonormal pair. From the orthogonality restriction, u • v = 0. From the unit length restriction on u, ||u|| = 1. From the unit length restriction on v, ||v|| = 1. Expanding these terms gives 3 equations: x 1 x 2 + y 1 y 2 = 0 {\displaystyle x_{1}x_{2}+y_{1}y_{2}=0\quad } x 1 2 + y 1 2 = 1 {\displaystyle {\sqrt {{x_{1}}^{2}+{y_{1}}^{2}}}=1} x 2 2 + y 2 2 = 1 {\displaystyle {\sqrt {{x_{2}}^{2}+{y_{2}}^{2}}}=1} Converting from Cartesian to polar coordinates, and considering Equation ( 2 ) {\displaystyle (2)} and Equation ( 3 ) {\displaystyle (3)} immediately gives the result r1 = r2 = 1. In other words, requiring the vectors be of unit length restricts the vectors to lie on the unit circle. After substitution, Equation ( 1 ) {\displaystyle (1)} becomes cos ⁡ θ 1 cos ⁡ θ 2 + sin ⁡ θ 1 sin ⁡ θ 2 = 0 {\displaystyle \cos \theta _{1}\cos \theta _{2}+\sin \theta _{1}\sin \theta _{2}=0} . Rearranging gives tan ⁡ θ 1 = − cot ⁡ θ 2 {\displaystyle \tan \theta _{1}=-\cot \theta _{2}} . Using a trigonometric identity to convert the cotangent term gives tan ⁡ ( θ 1 ) = tan ⁡ ( θ 2 + π 2 ) {\displaystyle \tan(\theta _{1})=\tan \left(\theta _{2}+{\tfrac {\pi }{2}}\right)} ⇒ θ 1 = θ 2 + π 2 {\displaystyle \Rightarrow \theta _{1}=\theta _{2}+{\tfrac {\pi }{2}}} It is clear that in the plane, orthonormal vectors are simply radii of the unit circle whose difference in angles equals 90°.