Introduction to Three Dimensional Geometry

1. Check this before you revise anything

The section formula is not part of the current book. Coaching material commonly teaches a full "section formula in 3D" here — the point dividing two given points in a ratio $m:n$ — as this chapter's main second topic. CBSE's own formative-only block confirms this is intentional, listing "Section formula" under Introduction to Three-Dimensional Geometry as formative, not summative.

The current 2026-27 book has no such formula anywhere: no derivation, no worked example, no exercise question (confirmed by a full-text search for "section formula", "internal division", and "ratio m" — zero hits). This chapter has only two named exercises, 11.1 and 11.2, not three.

Midpoints and centroids are a different story — they're genuinely used, just never boxed as a named formula. The book's own Example 7 notes that a parallelogram's diagonals bisect each other, and Example 9 finds a triangle's missing vertex from its centroid by averaging coordinates directly.

The Miscellaneous Exercise's own Q1 (a parallelogram's fourth vertex) and Q2 (lengths of medians) need exactly this. This chapter teaches averaging coordinates for a midpoint and a centroid as natural, book-genuine techniques — distinct from, and not to be confused with, the general $m:n$ section formula that stays out of scope.

2. What this chapter covers

3. Coordinate axes, coordinate planes, and octants

Three mutually perpendicular planes intersecting at a point $O$ meet along three mutually perpendicular lines — the x-axis, y-axis, and z-axis — which together form the rectangular coordinate system. The three planes they determine, the XY-plane, YZ-plane, and ZX-plane, are the coordinate planes. $O$ is the origin.

These three planes divide all of space into eight regions called octants, numbered I through VIII by the sign pattern of the coordinates within each:

Worked, mirroring the textbook's own Example 2. Find the octant containing $(-3,1,2)$ and $(-3,1,-2)$. Reading signs against the table: $(-,+,+)$ is octant II; $(-,+,-)$ is octant VI.

4. Coordinates of a point in space

Given a point $P$ in space, drop a perpendicular $PM$ onto the XY-plane, then a perpendicular $ML$ from $M$ onto the x-axis. Writing $OL=x$, $LM=y$, $MP=z$ gives the ordered triplet $(x,y,z)$ — the point's x, y, and z coordinates. Equivalently, $x$, $y$, and $z$ are the perpendicular distances of $P$ from the YZ-plane, ZX-plane, and XY-plane respectively.

This correspondence runs both ways: every point in space corresponds to exactly one ordered triplet of real numbers, and every triplet locates exactly one point.

• The origin is $(0,0,0)$.
• Any point on the x-axis has the form $(x,0,0)$; similarly $(0,y,0)$ on the y-axis and $(0,0,z)$ on the z-axis.
• Any point in the YZ-plane has the form $(0,y,z)$, since its distance from that plane is zero; similarly for the other two coordinate planes.

5. Distance between two points

Deriving the formula, mirroring the book's own construction. For $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$, draw planes through each point parallel to the coordinate planes, forming a rectangular box with $PQ$ as one diagonal. Two applications of the Pythagorean theorem — once in the base rectangle, once for the vertical edge — give $P{Q}^2=P{A}^2+A{N}^2+N{Q}^2$, where $PA=y_2-y_1$, $AN=x_2-x_1$, and $NQ=z_2-z_1$. So:

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

This is exactly the 2D distance formula with one more squared term for the third dimension. Taking $P$ as the origin gives the distance of any point $Q(x,y,z)$ from the origin: $OQ=\sqrt{x^2+y^2+z^2}$.

Worked, mirroring the textbook's own Example 3. Find the distance between $P(1,-3,4)$ and $Q(-4,1,2)$. $PQ=\sqrt{(-4-1{)}^2+(1+3{)}^2+(2-4{)}^2}=\sqrt{25+16+4}=\sqrt{45}=3\sqrt{5}$.

Collinearity, mirroring the book's own Example 4. Three points are collinear exactly when the sum of the two shorter distances between them equals the longest distance. For $P(-2,3,5)$, $Q(1,2,3)$, $R(7,0,-1)$: $PQ=\sqrt{14}$, $QR=2\sqrt{14}$, $PR=3\sqrt{14}$. Since $PQ+QR=PR$, the three points are collinear.

Midpoint, used directly in the book's own Example 7 note and needed in the Miscellaneous Exercise. The midpoint of $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ is the average of their coordinates:

$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\right)$

A parallelogram's diagonals bisect each other, so equating the midpoints of both diagonals is a standard way to find a fourth vertex from three known ones — exactly the technique the book's own Example 7 points to.

Centroid, worked exactly as the book's own Example 9 does it. The centroid of a triangle with vertices $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$ is the average of all three:

$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$

If a triangle's centroid is $(1,1,1)$ and two vertices are $A(3,-5,7)$ and $B(-1,7,-6)$, the third vertex $C(x,y,z)$ satisfies $\frac{3-1+x}{3}=1$, $\frac{-5+7+y}{3}=1$, $\frac{7-6+z}{3}=1$, giving $x=1$, $y=1$, $z=2$, so $C=(1,1,2)$.

Summary

• Three mutually perpendicular planes meet along the x, y, and z-axes at the origin, and divide space into eight octants, each identified by a sign pattern of $(x,y,z)$.
• Every point in space corresponds to exactly one ordered triplet $(x,y,z)$: the perpendicular distances from the YZ, ZX, and XY-planes respectively.
• Distance formula: $PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2}$; distance from the origin: $\sqrt{x^2+y^2+z^2}$.
• Three points are collinear exactly when the sum of the two shorter pairwise distances equals the longest one.
• Midpoint (average of two points) and centroid (average of three points) are genuinely used in this chapter's own worked examples and exercises — the general section formula that divides a segment in an arbitrary ratio $m:n$ is not part of the current book.