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Introduction to Three Dimensional Geometry

Chapter 11Notes

CBSE Class 11 Mathematics · NCERT Mathematics

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Shishya's notes

What this chapter is about

This chapter extends the coordinate geometry you learnt in two dimensions to three-dimensional space. In Class 10 and earlier Class 11 work, you located points on a plane using two numbers (x, y). Real objects — a bird in the sky, a corner of a room, a satellite above Earth — need three numbers because they occupy space, not just a flat surface. Here you learn how to describe such positions precisely.

You will study the three mutually perpendicular coordinate axes (x, y, z), understand the eight regions (octants) they create, and learn to find the distance between any two points in space. You will also locate a point that divides a line segment joining two given points in a specified ratio. These skills form the foundation for vectors, three-dimensional lines and planes in later mathematics.

After completing this chapter, you should be able to plot a point in space, calculate the distance between two points, find the coordinates of a point dividing a segment internally or externally, and determine the centroid of a triangle whose vertices are given in three dimensions.

Key ideas

  • Three mutually perpendicular axes — the x-axis, y-axis and z-axis — meet at the origin O, dividing space into eight octants.
  • Any point P in space is represented by an ordered triple (x, y, z), giving its signed distances from the three coordinate planes.
  • The coordinate planes are the xy-plane (z = 0), the yz-plane (x = 0) and the zx-plane (y = 0).
  • Distance in three dimensions follows from the Pythagorean theorem extended: √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].
  • The section formula finds a point dividing a segment in a given ratio, using weighted averages of the coordinates.
  • Signs of coordinates decide which octant a point lies in, just as signs decide quadrants in two dimensions.

Formulas and facts to remember

Distance between two points If P = (x₁, y₁, z₁) and Q = (x₂, y₂, z₂), then PQ = √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²]. Meaning: extend the 2-D distance formula by adding the square of the difference in z-coordinates.

Distance from the origin OP = √(x² + y² + z²) for the point P = (x, y, z).

Section formula (internal division) If R divides PQ internally in the ratio m : n, then R = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n), (mz₂ + nz₁)/(m + n)). Meaning: each coordinate of R is the weighted average, with weights m and n.

Section formula (external division) Replace + with − in denominators and cross-terms: R = ((mx₂ − nx₁)/(m − n), (my₂ − ny₁)/(m − n), (mz₂ − nz₁)/(m − n)), provided m ≠ n.

Midpoint When m = n = 1, the midpoint is ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2).

Centroid of a triangle For vertices A(x₁, y₁, z₁), B(x₂, y₂, z₂), C(x₃, y₃, z₃), the centroid is G = ((x₁ + x₂ + x₃)/3, (y₁ + y₂ + y₃)/3, (z₁ + z₂ + z₃)/3).

Worked examples

Example 1 — Finding distance in space A drone hovers at point A(2, 3, 6) metres from the corner of a warehouse, and a sensor is at B(6, 0, 2) metres from the same corner. Find the straight-line distance AB.

Solution: AB = √[(6 − 2)² + (0 − 3)² + (2 − 6)²] = √[4² + (−3)² + (−4)²] = √[16 + 9 + 16] = √41 metres ≈ 6.40 metres.

Example 2 — Internal division Point R divides the segment joining P(1, −2, 4) and Q(7, 4, −2) internally in the ratio 2 : 1. Find R.

Solution: Using m = 2, n = 1: x-coordinate = (2 × 7 + 1 × 1)/(2 + 1) = (14 + 1)/3 = 15/3 = 5. y-coordinate = (2 × 4 + 1 × (−2))/3 = (8 − 2)/3 = 6/3 = 2. z-coordinate = (2 × (−2) + 1 × 4)/3 = (−4 + 4)/3 = 0/3 = 0. Therefore R = (5, 2, 0).

Example 3 — Centroid of a triangle A triangular metal plate has vertices A(0, 0, 0), B(6, 0, 0) and C(3, 4, 5) in centimetres. Locate the centroid.

Solution: G = ((0 + 6 + 3)/3, (0 + 0 + 4)/3, (0 + 0 + 5)/3) = (9/3, 4/3, 5/3) = (3, 4/3, 5/3) cm ≈ (3, 1.33, 1.67) cm.

Common mistakes

Forgetting to square all three differences → always include (z₂ − z₁)² inside the square root.

Using the wrong sign in external division → remember denominators become (m − n), and the cross-term uses a minus sign.

Placing a point in the wrong octant because of sign errors → check that each coordinate's sign matches its position relative to the coordinate planes.

Taking the centroid formula with factor 1/2 instead of 1/3 → the centroid divides each median in ratio 2 : 1 from a vertex, so divide by 3.

Writing (x, y) instead of (x, y, z) when a third coordinate is zero → always write all three coordinates; if z = 0, write it as 0.

Quick revision

  • Three axes, eight octants; every point needs three coordinates (x, y, z).
  • Distance formula: √[(x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²].
  • Section formula: weighted averages with m and n; plus for internal, minus for external.
  • Midpoint: average each coordinate.
  • Centroid: (sum of x)/3, (sum of y)/3, (sum of z)/3.

Written by Shishya's AI on 26 Sept 2026 from the chapter's title and class level, in Shishya's own words — not a copy or summary of the textbook. Read the official chapter for the book's own text, activities and exercises.