If $A$ is a vector in the context of the Physics[Vectors] package, Norm returns its Euclidean norm, that is, $\sqrt{A·A}$ unless $A$ is not commutative or the option conjugate = true is passed, or the value of Physics:-Setup(normusesconjugate) is set to true - in all these cases it returns $\sqrt{{\left(A·\stackrel{\&conjugate0;}{A}\right)}^{}}$. To expand the norm of a sum of unprojected vectors in terms of their dot product, use expand. In the case of a scalar, Norm returns the absolute value $\left|A\right|$. Regarding how a vector is identified in the context of the Physics[Vectors] package see Identify.

The %Norm is the inert form of Norm, that is, it represents the same mathematical operation while holding the operation unperformed. To activate the operation use value.

$\mathrm{with}\left(\mathrm{Physics}\left[\mathrm{Vectors}\right]\right)$

$\left[\mathrm{\&x}\,\mathrm{`+`}\,\mathrm{`.`}\,\mathrm{Assume}\,\mathrm{ChangeBasis}\,\mathrm{ChangeCoordinates}\,\mathrm{CompactDisplay}\,\mathrm{Component}\,\mathrm{Curl}\,\mathrm{DirectionalDiff}\,\mathrm{Divergence}\,\mathrm{Gradient}\,\mathrm{Identify}\,\mathrm{Laplacian}\,\nabla \,\mathrm{Norm}\,\mathrm{ParametrizeCurve}\,\mathrm{ParametrizeSurface}\,\mathrm{ParametrizeVolume}\,\mathrm{Setup}\,\mathrm{Simplify}\,\mathrm{`^`}\,\mathrm{diff}\,\mathrm{int}\right]$

$\mathrm{Setup}\left(\mathrm{mathematicalnotation}=\mathrm{true}\right)$

$\left[\mathrm{mathematicalnotation}=\mathrm{true}\right]$

$a\mathrm{\_i}+b\mathrm{\_j}+c\mathrm{\_k}$

$a\stackrel{\wedge }{i}+b\stackrel{\wedge }{j}+c\stackrel{\wedge }{k}$

$\mathrm{Norm}\left(\right)$

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

$\mathrm{Norm}\left(\,\mathrm{conjugate}\right)$

$\sqrt{a\stackrel{\&conjugate0;}{a}+b\stackrel{\&conjugate0;}{b}+c\stackrel{\&conjugate0;}{c}}$

$\mathrm{convert}\left(\,\mathrm{abs}\right)$

$\sqrt{{\left|a\right|}^2+{\left|b\right|}^2+{\left|c\right|}^2}$

$\mathrm{Norm}\left(\mathrm{A\_}\right)$

$‖\overrightarrow{A}‖$

$\mathrm{Norm}\left(A\right)$

$\left|A\right|$