Introduction to 3D Geometry

In Class 10 you fixed the position of a point in a plane with two numbers $(x, y)$. To locate a point in space you need a third measurement — how far the point is above or below the plane — so every point now carries an ordered triple $(x, y, z)$.

Take three mutually perpendicular lines meeting at a single point $O$, the origin. These are the coordinate axes: the $x$-axis, the $y$-axis and the $z$-axis. Each axis is a number line, positive on one side of $O$ and negative on the other.

The three coordinate planes: Taken two at a time, the axes determine three planes that slice space apart.

Reading the coordinates of a point: For a point $P(x, y, z)$, the number $x$ is the perpendicular distance of $P$ from the $YZ$-plane, $y$ from the $ZX$-plane and $z$ from the $XY$-plane — each taken with the appropriate sign. A point lies on an axis when its other two coordinates are zero (e.g. $(5, 0, 0)$ is on the $x$-axis) and on a coordinate plane when one coordinate is zero.

The eight octants: The three coordinate planes cut space into eight compartments called octants. The first octant is where all three coordinates are positive; the others follow the sign pattern below.

Deeper Insight — three dimensions as two dimensions plus a height: The whole framework of 3D geometry is built by bolting one more perpendicular axis onto the familiar 2D plane, and almost every formula you meet later is the 2D version with a single extra term tagged on. Notice the symmetry running through the tables: the first four octants are exactly the four quadrants of the $XY$-plane "lifted" into positive $z$, and octants V–VIII are their mirror images below the $XY$-plane — so the top four all carry $z $ 0>3431$ and the bottom four $z $ 0<3464$. This is why a coordinate being zero is so informative: one zero pins the point to a coordinate plane, two zeros pin it to an axis, and three zeros give the origin itself. Holding this "plane plus height" picture firmly in mind makes the distance and section formulas in the next two topics feel like old friends rather than new rules.

The distance between two points in space follows directly from Pythagoras’ theorem applied twice. If $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ are any two points, the distance $PQ$ is the length of the straight segment joining them.

This is exactly the 2D distance formula with one extra squared term — the $z$-difference — added under the root. Compare the two side by side.

Distance from the origin: Putting $P = O(0, 0, 0)$ gives a handy special case.

What the formula lets you prove: Because distance is unambiguous, you can settle a great many geometric questions purely by computing lengths. Three points are collinear when the sum of two of the three pairwise distances equals the third. A triangle is isosceles if two sides are equal, equilateral if all three are equal, and right-angled when the three side-lengths satisfy $a^2 + b^2 = c^2$.

Deeper Insight — why one extra term is all you need: The 3D distance formula is not a new idea so much as Pythagoras’ theorem used twice. First imagine the box (rectangular parallelepiped) whose opposite corners are $P$ and $Q$ and whose edges are parallel to the axes; the diagonal of its base has length $\sqrt{(\Delta x)^2 + (\Delta y)^2}$ by the ordinary plane theorem. That base diagonal and the vertical edge of length $|\Delta z|$ are themselves perpendicular, so a second application of Pythagoras gives the space diagonal $\sqrt{(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2}$. This "Pythagoras twice" argument is why every coordinate dimension contributes its own squared difference, and it generalises cleanly to any number of dimensions. The practical payoff is huge: shape problems that would be hard to picture in space — is this triangle right-angled? are these four points the vertices of a square? — collapse into routine arithmetic the moment you compute a few distances.

Suppose a point $R$ lies on the segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ and divides it in the ratio $m : n$. The section formula gives the coordinates of $R$ — again the plane version with a third coordinate handled identically.

Internal division (point $R$ lies between $P$ and $Q$):

External division (point $R$ lies on the line but outside the segment): replace $n$ by $-n$.

Midpoint (the special internal case $m = n = 1$):

Centroid of a triangle with vertices $A(x_1,y_1,z_1)$, $B(x_2,y_2,z_2)$, $C(x_3,y_3,z_3)$ — the average of the three vertices:

Deeper Insight — one weighted average, four results: Every formula above is really the same operation — a weighted average of coordinates — wearing different clothes. Internal division weights $Q$ by $m$ and $P$ by $n$ (notice the cross-pairing: the $m$ that measures the distance toward $Q$ multiplies $P$’s neighbour $x_2$), and dividing by $m+n$ keeps the result a genuine average. The midpoint is just the symmetric case $1:1$, and external division is the same algebra with one weight made negative, which is why the sign flips to $m - n$. Even the centroid fits the pattern: it sits two-thirds of the way along each median, dividing it $2:1$ from the vertex, and that ratio is exactly what produces the clean average of all three vertices. Seeing these as one idea means you only have to remember the internal-division formula carefully; the others are quick specialisations. A practical warning: keep the order $(P, Q)$ and the ratio $(m, n)$ consistent, because swapping them silently changes which point you compute.