The Binormal(C, t) command computes a Vector in the direction of the binormal vector to a curve in R^3.  Note that this Vector is not normalized by default, so it is a scalar multiple of the unit binormal vector to the curve C. Therefore, by default, the result is generally different from the output of $\mathrm{TNBFrame}\left(C\,t\,\mathrm{output}=\left['B'\right]\right)$.

If n is given as either normalized=true or normalized, then the resulting vector will be normalized before it is returned. As discussed above, the default value is false, so that the result is not normalized.

The curve can be specified as a free or position Vector or a Vector valued procedure. This determines the returned object type. However, it must have exactly three components, that is, the curve that the Vector or Vector valued procedure represents is in R^3.

If t is not specified, the function tries to determine a suitable variable name by using the components of C.  To do this, it checks all of the indeterminates of type name in the components of C and removes the ones which are determined to be constants.

If the resulting set has a single entry, that entry is the variable name.  If it has more than one entry, an error is raised.

If a coordinate system attribute is specified on C, C is interpreted in that coordinate system.  Otherwise, the curve is interpreted as a curve in the current default coordinate system.  If the two are not compatible, an error is raised.

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

$\mathrm{Binormal}\left(⟨\cos \left(t\right)\,\sin \left(t\right)\,t⟩\,t\right)$

$\left[\begin{array}{c}\frac{\sin \left(t\right)}{2}\\ -\frac{\cos \left(t\right)}{2}\\ \frac{1}{2}\end{array}\right]$

$\mathrm{Binormal}\left(⟨\exp \left(-t\right)\cos \left(t\right)\,\exp \left(-t\right)\sin \left(t\right)\,t⟩\right)$

$\left[\begin{array}{c}\frac{2\cos \left(t\right)}{2{\ⅇ}^{-t}+{\ⅇ}^t}\\ \frac{2\sin \left(t\right)}{2{\ⅇ}^{-t}+{\ⅇ}^t}\\ \frac{2}{2+{\ⅇ}^{2t}}\end{array}\right]$

$\mathrm{B1}≔\mathrm{Binormal}\left(t\mapsto ⟨t\,t^2\,t^3⟩\right)\:$

$\mathrm{B1}\left(t\right)$

$\left[\begin{array}{c}\frac{6t^2}{9t^4+4t^2+1}\\ -\frac{6t}{9t^4+4t^2+1}\\ \frac{2}{9t^4+4t^2+1}\end{array}\right]$

$\mathrm{B2}≔\mathrm{Binormal}\left(t\mapsto ⟨t\,t^2\,t^3⟩\,\mathrm{normalized}\right)\:$

$\mathrm{B2}\left(t\right)$

$\left[\begin{array}{c}\frac{3t^2}{\sqrt{\frac{9t^4+9t^2+1}{{\left(9t^4+4t^2+1\right)}^2}}\left(9t^4+4t^2+1\right)}\\ -\frac{3t}{\sqrt{\frac{9t^4+9t^2+1}{{\left(9t^4+4t^2+1\right)}^2}}\left(9t^4+4t^2+1\right)}\\ \frac{1}{\sqrt{\frac{9t^4+9t^2+1}{{\left(9t^4+4t^2+1\right)}^2}}\left(9t^4+4t^2+1\right)}\end{array}\right]$

$\mathrm{SetCoordinates}\left('\mathrm{cylindrical}'\right)$

$\mathrm{cylindrical}$

$\mathrm{Binormal}\left(⟨a\,t\,t⟩\right)\mathbf{assuming}a::\left(\mathrm{And}\left(\mathrm{positive}\,\mathrm{constant}\right)\right)$

$\left[\begin{array}{c}0\\ -\frac{a}{a^2+1}\\ \frac{a^2}{a^2+1}\end{array}\right]$