Points, lines, and planes in space

A pair $(x,y)$ names a point in the plane. In space that pair is a flattening: infinitely many heights share the same $(x,y)$. The slope formula $y=mx+b$ from lines in the plane cannot write a line that leaves the $xy$-plane. The question on this page is how a third number names a point so that distance, a line, and a plane become equations you can check.

The frame here is Cartesian and right-handed: the $x$-, $y$-, and $z$-axes are pairwise perpendicular, matching the thumb, index finger, and middle finger of the right hand.

Three drops name a point

A point is an ordered triple $(x_1,y_1,z_1)$. Extra assumption for every formula below: those coordinates are read on perpendicular axes in that right-handed order. The three numbers are the signed lengths of the perpendicular drops from the point to the three axes. Drag $P$. The dashed box is those three drops. The far corner is $(x_1,y_1,z_1)$.

Drop the right angle between the axes and the next formula — a space diagonal built from $\Delta x$, $\Delta y$, and $\Delta z$ — is the wrong distance.

The space diagonal is the distance

Once two points have triples, the three edges $|x_1-x_2|$, $|y_1-y_2|$, and $|z_1-z_2|$ make a rectangular box because the axes are perpendicular. The distance $d$ is that space diagonal:

$d=\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2+(z_1-z_2{)}^2}.$

From $(1,1,1)$ to $(2,2,2)$ that is $\sqrt{3}$. When $z_1=z_2$ the third square vanishes and this is the plane distance. Drag $A$ or $B$. The dashed edges are the three coordinate differences; the solid segment is $d$.

A line is a time $t$

Among all named points, which ones lie on a line through two of them? Extra freeze: the two points are not the same. Then a point on the line is reached at a real number $t$:

$x=x_1+t(x_2-x_1),y=y_1+t(y_2-y_1),z=z_1+t(z_2-z_1).$

$t$ runs through all real numbers. Read $t$ as time: $t=0$ sits at the first point, $t=1$ at the second, and $t$ outside $[0,1]$ is still on the same line. So this writing does not need a slope: a line parallel to an axis is just one of the three differences equal to zero. Drag $A$ or $B$. $P$ is the point at time $t$.

One linear equation is a plane

In the plane, one linear equation $Ax+By+C=0$ is already a line. In space the same count — one linear constraint on three coordinates — cuts a plane:

$Ax+By+Cz=D$

for constants $A,B,C$ with $A^2+B^2+C^2\ne 0$. If all three coefficients vanish, the display is no longer an equation of a plane. That is why a line in space needed the extra parameter $t$: a single equation has already used up one degree of freedom, and a line still has one left. Drag $C$. $A$ and $B$ stay fixed. If $C$ lines up with $A$ and $B$, the three points no longer span a plane — that is the same non-degeneracy the coefficients are paying for.

The object is a named point $(x,y,z)$ on a right-handed perpendicular frame. Distance is the space diagonal of $\Delta x$, $\Delta y$, and $\Delta z$. A line through two distinct points is the motion at time $t\in \mathbb{R}$. A non-degenerate linear equation is a plane. The plane case — two coordinates, one equation already a line — is lines in the plane.