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CBSE • Class XI • Mathematics • Ch 11
Estimated Time: 45 Mins
Study Progress: In Progress

Introduction to Three Dimensional Geometry

In Class 11 Mathematics, "Introduction to Three-Dimensional Geometry" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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📦 Have You Ever Wondered?

How do 3D game engines pinpoint a flying helicopter in virtual space using $(x, y, z)$ coordinates, and how does the universe split into eight octants...

How do 3D game engines pinpoint a flying helicopter in virtual space using $(x, y, z)$ coordinates, and how does the universe split into eight octants? 3D geometry extends coordinate systems into the physical world.

Why This Chapter Matters

In Class 11 Mathematics, "Introduction to Three-Dimensional Geometry" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • 2D Cartesian coordinate plane.
  • Distance formula and section formula in 2D.
  • Right-hand rule.

What You Will Learn (Core Objectives)

  • Identify Coordinate Axes and Coordinate Planes ($xy, yz, zx$ planes) dividing space into 8 Octants.
  • Determine coordinates of points in three-dimensional space.
  • Apply the 3D Distance Formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$.
  • Apply the 3D Section Formula for internal and external division.
  • Find coordinates of the Centroid of a triangle in 3D space.

Chapter Roadmap & Progression

1 1. Coordinates & The Eight Octants
2 2. 3D Distance Formula
3 3. Section Formula & Centroid

Complete Concept Guide (100% Curriculum Coverage)

1. Coordinates & The Eight Octants

Three mutually perpendicular axes ($X, Y, Z$) meet at origin $O(0, 0, 0)$. The three coordinate planes ($XY, YZ, ZX$) partition space into Eight Octants.
• Points on $XY$-plane have $z = 0$.
• Points on $X$-axis have coordinates $(x, 0, 0)$.

2. 3D Distance Formula

The Euclidean distance between $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is: $$\mathbf{PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}}$$

3. Section Formula & Centroid

Point $R$ dividing $PQ$ in ratio $m : n$ internally: $$\mathbf{R = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right)}$$ Midpoint: $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$.
Centroid of triangle: $\mathbf{G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)}$.

Visual Learning & Conceptual Map

Introduction to Three-Dimensional Geometry Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. Coordinates & The Eight Octants • 2. 3D Distance Formula

Chapter Summary & 10 Key Takeaways

Takeaway 1
Eight Octants: Spatial regions determined by sign combinations of $(\pm x, \pm y, \pm z)$.
Takeaway 2
Coordinate Planes: Fundamental reference surfaces ($z=0, x=0, y=0$).
Takeaway 3
Pythagorean 3D Extension: Adding squared $z$-coordinate delta to distance formula.
Takeaway 4
Spatial Midpoint: Coordinate-wise arithmetic average of endpoints.
Takeaway 5
Centroid: Balancing center of mass in 3D triangle.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
In which octant do the points $(-3, 1, 2)$ and $(2, -4, -7)$ lie?
Reveal Answer & Explanation
Answer: $(-3, 1, 2)$ has $x<0, y>0, z>0 \implies$ Octant II. $(2, -4, -7)$ has $x>0, y<0, z<0 \implies$ Octant VIII.
Octant II and Octant VIII.
2
Find the distance between the points $P(1, -3, 4)$ and $Q(-4, 1, 2)$.
Reveal Answer & Explanation
Answer: $d = \sqrt{(-4-1)^2 + (1 - (-3))^2 + (2-4)^2} = \sqrt{(-5)^2 + 4^2 + (-2)^2} = \sqrt{25 + 16 + 4} = \sqrt{45} = 3\sqrt{5}\text{ units}$.
3√5 units.
3
Find the coordinates of the point which divides the line segment joining $(-2, 3, 5)$ and $(1, -4, 6)$ in the ratio $2 : 3$ internally.
Reveal Answer & Explanation
Answer: $x = \frac{2(1) + 3(-2)}{2+3} = -\frac{4}{5}$; $y = \frac{2(-4) + 3(3)}{5} = \frac{1}{5}$; $z = \frac{2(6) + 3(5)}{5} = \frac{27}{5}$. Point is $(-\frac{4}{5}, \frac{1}{5}, \frac{27}{5})$.
(-4/5, 1/5, 27/5).
4
Find the centroid of a triangle with vertices $(3, -5, 7)$, $(-1, 7, -6)$, and $(1, 1, 2)$.
Reveal Answer & Explanation
Answer: $G = (\frac{3 - 1 + 1}{3}, \frac{-5 + 7 + 1}{3}, \frac{7 - 6 + 2}{3}) = (\frac{3}{3}, \frac{3}{3}, \frac{3}{3}) = (1, 1, 1)$.
(1, 1, 1).
5
What are the coordinates of the projection of point $(4, 7, 9)$ onto the $xy$-plane?
Reveal Answer & Explanation
Answer: On the $xy$-plane, $z = 0$. The projection is $(4, 7, 0)$.
(4, 7, 0).
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