Normalize - Maple Help
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VectorCalculus

 Normalize
 normalize a Vector or vector field

 Calling Sequence Normalize(v, p)

Parameters

 v - Vector(algebraic); Vector to be normalized p - (optional) nonnegative, infinity, or identical(Euclidean); the norm to use

Description

 • 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.

Examples

 > $\mathrm{with}\left(\mathrm{VectorCalculus}\right):$
 > $\mathrm{SetCoordinates}\left(\mathrm{cartesian}\left[x,y\right]\right):$
 > $\mathrm{Normalize}\left(⟨3,4⟩\right)$
 > $n≔\mathrm{Normalize}\left(\mathrm{VectorField}\left(⟨xy,\frac{x}{y}⟩\right),3\right)$
 > $\mathrm{simplify}\left(\mathrm{evalVF}\left(n,⟨2,3⟩\right)\right)$
 $\left[\begin{array}{c}\frac{{9}{}{{730}}^{{2}}{{3}}}}{{730}}\\ \frac{{{730}}^{{2}}{{3}}}}{{730}}\end{array}\right]$ (1)
 > $\mathrm{SetCoordinates}\left(\mathrm{cartesian}\left[x,y,z\right]\right):$
 > $\mathrm{Normalize}\left(⟨2,0,3⟩,0\right)$
 > $\mathrm{SetCoordinates}\left(\mathrm{spherical}\left[r,\mathrm{\phi },\mathrm{\theta }\right]\right):$
 > $\mathrm{Normalize}\left(\mathrm{PositionVector}\left(\left[1,\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{3}\right]\right),\mathrm{Euclidean}\right)$
 > $v≔\mathrm{Normalize}\left(\mathrm{PositionVector}\left(\left[2,\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{4}\right]\right),\mathrm{\infty }\right)$
 > $\mathrm{Norm}\left(v,\mathrm{\infty }\right)$
 ${1}$ (2)
 > $\mathrm{SetCoordinates}\left(\mathrm{polar}\left[r,\mathrm{\theta }\right]\right):$
 > $\mathrm{Normalize}\left(\mathrm{RootedVector}\left(\mathrm{root}=\left[1,2\right],\left[a,b\right]\right)\right)$
 $\left[\begin{array}{c}\frac{{a}}{\sqrt{{{a}}^{{2}}{+}{{b}}^{{2}}}}\\ \frac{{b}}{\sqrt{{{a}}^{{2}}{+}{{b}}^{{2}}}}\end{array}\right]$ (3)