The Norm(v, p) command computes the p-norm of the Vector or vector field v. If p is omitted, it defaults to 2.

If v is a RootedVector, its norm is computed 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.

If v is a vector field, the result is a procedure which at any point (Vector) w evaluates to the p-norm of the value of v at w.

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

If $0\le p<1$ , then the value computed by this command defines a metric, but not a norm.  See LinearAlgebra[Norm] for details.

Note: If the Norm command is applied to a vector field, vf, using the context menu, the result will be Norm(Vector(vf)) rather than Norm(vf), as this results in a more readable expression.

$\mathrm{with}\left(\mathrm{VectorCalculus}\right)\&colon;$

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

$\mathrm{Norm}\left(⟨3\&comma;4⟩\right)$

$5$

$n≔\mathrm{Norm}\left(\mathrm{VectorField}\left(⟨xy\&comma;\frac{x}{y}⟩\right)\&comma;3\right)\&colon;$

$n\left(⟨2\&comma;3⟩\right)$

$\frac{2{730}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{3}$

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

$\mathrm{Norm}\left(⟨2\&comma;0\&comma;3⟩\&comma;0\right)$

$2$

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

$\mathrm{Norm}\left(\mathrm{PositionVector}\left(\left[1\&comma;\frac{\mathrm{\pi }}{2}\&comma;\frac{\mathrm{\pi }}{3}\right]\right)\&comma;\mathrm{Euclidean}\right)$

$1$

$\mathrm{Norm}\left(\mathrm{PositionVector}\left(\left[2\&comma;\frac{\mathrm{\pi }}{3}\&comma;\frac{\mathrm{\pi }}{4}\right]\right)\&comma;\mathrm{\infty }\right)$

$\frac{\sqrt{6}}{2}$

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

$\mathrm{Norm}\left(\mathrm{RootedVector}\left(\mathrm{root}=\left[1\&comma;2\right]\&comma;\left[a\&comma;b\right]\right)\right)$

$\sqrt{a^2+b^2}$