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The routine computes the distance between two given objects: points, lines, and planes, or any combination of these.

The command with(geom3d,distance) allows the use of the abbreviated form of this command.

$\mathrm{with}\left(\mathrm{geom3d}\right)\:$

$\mathrm{assume}\left(l\ne 0\right)$

$\mathrm{plane}\left(\mathrm{pp}\,lx+my+nz+p=0\,\left[x\,y\,z\right]\right)\:$

$\mathrm{point}\left(A\,\left[\mathrm{x1}\,\mathrm{y1}\,\mathrm{z1}\right]\right)\:$

$\mathrm{distance}\left(A\,\mathrm{pp}\right)$

$\frac{\left|\mathrm{x1}\mathrm{l~}+\mathrm{y1}m+\mathrm{z1}n+p\right|}{\sqrt{{\mathrm{l~}}^2+m^2+n^2}}$

$l≔'l'\:$

$\mathrm{point}\left(A\,6\,6\,-1\right),\mathrm{line}\left(l\,\left[2+t\,1+2t\,-3-t\right]\,t\right)\:$

$\mathrm{distance}\left(A\,l\right)$

$\frac{\sqrt{14}\sqrt{6}}{2}$

$\mathrm{point}\left(o\,t\,\frac{1}{2}t\,-2+\frac{t}{3}\right)\:$

$\mathrm{plane}\left(p\,2x+y-2z=5\,\left[x\,y\,z\right]\right)\:$

$d≔\mathrm{distance}\left(o\,p\right)\:$

$\mathrm{ans}≔\left[\mathrm{solve}\left(d=7\,t\right)\right]\:$

$$\begin{gathered} \mathbf{for}i\mathbf{to}\mathrm{nops}\left(\mathrm{ans}\right)\mathbf{do} \\ \mathrm{ans}≔\mathrm{subsop}\left(i=\mathrm{subs}\left(t=\mathrm{op}\left(i\,\mathrm{ans}\right)\,\mathrm{coordinates}\left(o\right)\right)\,\mathrm{ans}\right) \\ \mathbf{end}\mathbf{do}\: \end{gathered}$$

$\mathrm{ans}$

$\left[\left[12\,6\,2\right]\,\left[-\frac{120}{11}\,-\frac{60}{11}\,-\frac{62}{11}\right]\right]$