Gram–Schmidt in Motion

Steps

1. Normalize \(\VA1\) Start with \(\VA1\) and divide it by its length. The result, \(\UA1 = \VA1/\lVert\VA1\rVert\), is a unit vector pointing the same way.
2. Draw \(\VA2\) Bring in the second basis vector \(\VA2\). It is not orthogonal to \(\UA1\).
3. Project \(\VA2\) onto \(\UA1\) Drop a perpendicular from the tip of \(\VA2\) to the line through \(\UA1\). That gives \(\VPAR2 = (\UA1\cdot\VA2)\,\UA1\). What is left, \(\VPERP2 = \VA2 - \VPAR2\), is orthogonal to \(\UA1\).
4. Normalize \(\VPERP2\) Slide \(\VPERP2\) back to the origin and divide by its length: \(\UA2 = \VPERP2/\lVert\VPERP2\rVert\). Now \(\UA1\) and \(\UA2\) are orthonormal.
5. Reveal the plane \(\PV\) Tilt the camera. Everything so far lies in the plane \(\PV = \operatorname{span}(\UA1, \UA2)\), which sits inside three-dimensional space.
6. Draw \(\VA3\) Bring in a third vector \(\VA3\), which points out of the plane \(\PV\).
7. Project \(\VA3\) onto \(\PV\) Drop a perpendicular from the tip of \(\VA3\) to the plane. That gives \(\VPAR3 = (\UA1\cdot\VA3)\,\UA1 + (\UA2\cdot\VA3)\,\UA2\). The rest, \(\VPERP3 = \VA3 - \VPAR3\), is perpendicular to \(\PV\).
8. Normalize \(\VPERP3\) Slide \(\VPERP3\) back to the origin and divide by its length: \(\UA3 = \VPERP3/\lVert\VPERP3\rVert\).
9. An orthonormal basis \(\UA1, \UA2, \UA3\) each have length 1, and each is perpendicular to the other two: \(\vec u_i\cdot\vec u_j\) is 1 when \(i = j\) and 0 otherwise.