two point line form in 3d

Question

the two-point form for a line in 2d is $$y-y_1 = \left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1);$$ what is it for 3d lines/planes?

Answer 1

For lines, you need two equations, so it is just duplicated: $y-y_1 = \frac{y_2-y_1}{x_2-x_1} (x-x_1)$ and $z-z_1 = \frac{z_2-z_1}{x_2-x_1} (x-x_1)$

For planes, you need three points.
Three approaches are shown in Wikipedia under "Define a plane through three points"

Answer 2

For a line, one can express it in parametric form. To give you an idea on how to derive it, note that the expression

$$(1-t)x_1+tx_2$$

gives $x_1$ for $t=0$ and $x_2$ for $t=1$. You can do this for the $y$ and $z$ components as well to arrive at a parametric equation of a line.

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Another way to go about representing a line is to represent it in symmetric form (from which a parametric form can also be obtained by equating to a parameter each of the components):

$$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$$

where

$$\begin{align*}a&=\frac{x_2-x_1}{\rho}\\b&=\frac{y_2-y_1}{\rho}\\c&=\frac{z_2-z_1}{\rho}\\\rho&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\end{align*}$$

and $a,b,c$ are so-called direction cosines, cosines of the angles the line makes with the positive axes.

This notation is sometimes (ab)used even when any of the direction cosines are 0; this just means that the line is parallel to an axis. From the symmetric form, taking any two of the three implied equations determines two planes whose intersection is the line in question.