Three-dimensional geometry

Axes, planes, octants. The $x$-, $y$- and $z$-axes are mutually perpendicular lines through a common origin $O$. Each pair of axes spans a coordinate plane: the $XY$-plane is every point with $z=0$, the $YZ$-plane is $x=0$, and the $ZX$-plane is $y=0$. These three planes slice all of space into eight octants, distinguished purely by the signs of $x$, $y$, $z$ — the familiar positive quadrant of 2-D work becomes the octant where all three coordinates are positive. A point $P$ is written $P(x,y,z)$, where $x$, $y$, $z$ are its signed perpendicular distances from the $YZ$-, $ZX$-, and $XY$-planes respectively. Compare this with the plane: there, a point needed only its distances from two axes; here, a third perpendicular measurement — height above or below the $XY$-plane — is unavoidable, because the room a fly buzzes around genuinely has three independent directions, not two. The axes themselves are simply the three edges of a corner of a room meeting at the origin, each at right angles to the other two, and the coordinate planes are the three walls-plus-floor meeting at that corner. Just as the sign of $x$ and $y$ told you which quadrant a 2-D point sat in, the signs of $x$, $y$, $z$ together tell you which of the eight octants a 3-D point sits in. There is no separate name to memorise for each octant beyond the sign pattern itself — for instance $(2,-3,5)$ has $x>0,y<0,z>0$, so it sits in the octant where the first and third coordinates are positive and the second is negative, and no two octants share the same sign pattern.

Distance formula. For $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$,

where: $PQ$ is the straight-line distance, and each squared term is the change in one coordinate. Setting $P$ at the origin gives the special case $OP=\sqrt{x^2+y^2+z^2}$.

Section formula. If $R(x,y,z)$ divides segment $PQ$ internally in the ratio $m:n$, then

where: $m:n$ is read as "$m$ parts near $Q$, $n$ parts near $P$" — $m$ multiplies $Q$'s coordinates and $n$ multiplies $P$'s. For external division, flip the sign of the $n$ terms (equivalently use ratio $m:(-n)$). Setting $m=n=1$ gives the mid-point, $M=\left(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2},\dfrac{z_1+z_2}{2}\right)$, and applying the formula to a triangle's median (ratio $2:1$) gives its centroid, $\left(\dfrac{x_1+x_2+x_3}{3},\dfrac{y_1+y_2+y_3}{3},\dfrac{z_1+z_2+z_3}{3}\right)$.

Show that the points $A(0,7,10)$, $B(-1,6,6)$, $C(-4,9,6)$ are the vertices of a right-angled isosceles triangle.

Step 1. Apply the distance formula to each pair. $AB^2 = (0-(-1))^2+(7-6)^2+(10-6)^2 = 1+1+16 = 18$.

Step 2. Similarly, $BC^2 = (-1-(-4))^2+(6-9)^2+(6-6)^2 = 9+9+0 = 18$, and $CA^2 = (-4-0)^2+(9-7)^2+(6-10)^2 = 16+4+16 = 36$.

Step 3. Compare the three squared lengths. Since $AB^2=BC^2=18$, sides $AB$ and $BC$ are equal, so $\triangle ABC$ is isosceles. Also $AB^2+BC^2 = 18+18 = 36 = CA^2$, so by the converse of the Pythagoras theorem the angle opposite $CA$ — angle $B$ — is a right angle.

Answer: $\triangle ABC$ is right-angled at $B$ and isosceles, with $AB=BC=\sqrt{18}$ and $CA=6$.

Competitive stretch

Find the point dividing $A(-2,3,5)$ and $B(1,-4,6)$ internally in the ratio $2:3$. Here $m:n=2:3$, so $x=\dfrac{2(1)+3(-2)}{5}=-\dfrac{4}{5}$, $y=\dfrac{2(-4)+3(3)}{5}=\dfrac{1}{5}$, $z=\dfrac{2(6)+3(5)}{5}=\dfrac{27}{5}$. The point is $\left(-\dfrac{4}{5},\dfrac{1}{5},\dfrac{27}{5}\right)$ — the same section formula that produced the vertical drop $NQ$ in the derivation above, now used to split a segment rather than measure it. Notice each coordinate is computed independently by the identical rule: multiply the far point's coordinate by $m$, the near point's by $n$, add, and divide by $m+n$ — there is nothing extra to learn for the $z$-coordinate once the $x$- and $y$-coordinates are understood.

• Dropping the $z$-term out of habit. Students fluent in 2-D formulas sometimes forget the third coordinate — check that $x$, $y$ and $z$ all appear in every step.
• Misreading the octant from the coordinate signs. Three coordinates give eight sign combinations, not four; one careless sign flip puts a point in the wrong octant.
• Swapping $m$ and $n$ in the section formula. The formula is not symmetric in $P$ and $Q$: $m$ pairs with $Q$'s coordinates, $n$ with $P$'s, in the order the ratio $m:n$ was written.
• Reaching for vectors or direction cosines. This Section-B unit is worked entirely with coordinates; direction cosines, direction ratios and vector equations belong to Class XII.

Ask the tutor: If keeping track of all three coordinates through a distance or section-formula calculation feels fiddly, ask the AI tutor to walk through one coordinate — $x$, then $y$, then $z$ — at a time.