The Normalize(v, p) command normalizes the Vector or vector field v with respect to the p-norm.  If p is omitted, it defaults to 2.

If v is a RootedVector, it is normalized in the corresponding vector space, that is, relative to the root point of that vector space.  See VectorCalculus details for more information about rooted vectors.

The 2-norm can also be specified using the value Euclidean for the parameter p.

If the p-norm of the input Vector v is 0, then v is returned.

$\mathrm{with}\left(\mathrm{VectorCalculus}\right)\:$

$\mathrm{SetCoordinates}\left(\mathrm{cartesian}\left[x,y\right]\right)\:$

$\mathrm{Normalize}\left(⟨3\,4⟩\right)$

$\left(\frac{3}{5}\right)e_x\+\left(\frac{4}{5}\right)e_y$

$n≔\mathrm{Normalize}\left(\mathrm{VectorField}\left(⟨xy\,\frac{x}{y}⟩\right)\,3\right)$

$n≔\left(\frac{xy\left|\frac{y}{x}\right|}{{\left({\left|y\right|}^6+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right){\stackrel{\_}{e}}_x\+\left(\frac{x\left|\frac{y}{x}\right|}{y{\left({\left|y\right|}^6+1\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}\right){\stackrel{\_}{e}}_y$

$\mathrm{simplify}\left(\mathrm{evalVF}\left(n\,⟨2\,3⟩\right)\right)$

$\left[\begin{array}{c}\frac{9{730}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{730}\\ \frac{{730}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{730}\end{array}\right]$

$\mathrm{SetCoordinates}\left(\mathrm{cartesian}\left[x,y,z\right]\right)\:$

$\mathrm{Normalize}\left(⟨2\,0\,3⟩\,0\right)$

$\left(1\right)e_x\+\left(0\right)e_y\+\left(\frac{3}{2}\right)e_z$

$\mathrm{SetCoordinates}\left(\mathrm{spherical}\left[r,\mathrm{\phi },\mathrm{\theta }\right]\right)\:$

$\mathrm{Normalize}\left(\mathrm{PositionVector}\left(\left[1\,\frac{\mathrm{\pi }}{2}\,\frac{\mathrm{\pi }}{3}\right]\right)\,\mathrm{Euclidean}\right)$

$\left[\begin{array}{c}\frac{1}{2}\\ \frac{\sqrt{3}}{2}\\ 0\end{array}\right]$

$v≔\mathrm{Normalize}\left(\mathrm{PositionVector}\left(\left[2\,\frac{\mathrm{\pi }}{3}\,\frac{\mathrm{\pi }}{4}\right]\right)\,\mathrm{\infty }\right)$

$v≔\left[\begin{array}{c}1\\ 1\\ \frac{\sqrt{6}}{3}\end{array}\right]$

$\mathrm{Norm}\left(v\,\mathrm{\infty }\right)$

$1$

$\mathrm{SetCoordinates}\left(\mathrm{polar}\left[r,\mathrm{\theta }\right]\right)\:$

$\mathrm{Normalize}\left(\mathrm{RootedVector}\left(\mathrm{root}=\left[1\,2\right]\,\left[a\,b\right]\right)\right)$

$\left[\begin{array}{c}\frac{a}{\sqrt{a^2+b^2}}\\ \frac{b}{\sqrt{a^2+b^2}}\end{array}\right]$