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

Three Dimensional Geometry

In Class 12 Mathematics, "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 air traffic controllers prevent two jet airliners flying at different altitudes from colliding, or determine the exact shortest flight path between two commercial flight routes? Skew lines and 3D spatial geometry govern commercial airspace.

Why This Chapter Matters

In Class 12 Mathematics, "Three Dimensional Geometry" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • 3D coordinates from Class 11.
  • Vectors from Chapter 10.
  • Linear systems.

What You Will Learn (Core Objectives)

  • Find Direction Cosines ($l, m, n$) and Direction Ratios ($a, b, c$) of a line.
  • Write Vector and Cartesian equations of a line passing through a point and parallel to a vector: $\vec{r} = \vec{a} + \lambda\vec{b}$.
  • Write equation of a line passing through two given points: $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
  • Calculate the Shortest Distance between two Skew Lines: $d = \left|\frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|}\right|$.
  • Calculate shortest distance between parallel lines.

Chapter Roadmap & Progression

1 1. Direction Cosines & Equation of...
2 2. Shortest Distance Between Skew L...

Complete Concept Guide (100% Curriculum Coverage)

1. Direction Cosines & Equation of a Line

A line has direction ratios $a, b, c$. Direction cosines: $l = \frac{a}{\sqrt{a^2+b^2+c^2}}$, etc.
• Line through Point $\vec{a}$ parallel to $\vec{b}$: $$\mathbf{\vec{r} = \vec{a} + \lambda\vec{b}} \iff \mathbf{\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}}$$
• Line through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$: $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$.

2. Shortest Distance Between Skew Lines

Skew Lines are non-intersecting, non-parallel lines in 3D space. For lines $\vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ and $\vec{r} = \vec{a}_2 + \mu\vec{b}_2$: $$\mathbf{d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|}$$ Lines intersect if and only if $d = 0$ (i.e. $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = 0$).

Three Dimensional Geometry - Key Conceptual & Analytical Model

Three Dimensional Geometry - Mathematical Architecture Axiomatic & Matrix Foundations Equivalence theorems & algebraic proofs Calculus & 3D Vector Geometry Differential optimization & spatial lines High-Stakes Examination & Engineering Mastery CBSE Class 12 Board criteria, JEE Advanced problem frameworks & applications

Chapter Summary & 10 Key Takeaways

Takeaway 1
Direction Cosines: Projections of unit vector along $X, Y, Z$ axes ($l^2+m^2+n^2=1$).
Takeaway 2
Cartesian Line Form: Symmetric ratio $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$.
Takeaway 3
Skew Lines: Lines in different planes that neither intersect nor are parallel.
Takeaway 4
Shortest Distance Formula: Projection of difference vector along common perpendicular cross-product.
Takeaway 5
Intersection Test: Vanishing shortest distance scalar triple product.

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
Find the vector and Cartesian equations of the line through the point $(5, 2, -4)$ and which is parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$.
Reveal Answer & Explanation
Answer: Vector equation: $\vec{r} = (5\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(3\hat{i} + 2\hat{j} - 8\hat{k})$. Cartesian equation: $\frac{x - 5}{3} = \frac{y - 2}{2} = \frac{z + 4}{-8}$.
r = (5i+2j-4k) + λ(3i+2j-8k); (x-5)/3 = (y-2)/2 = (z+4)/-8.
2
Find the angle between the pair of lines: $\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4}$ and $\frac{x+1}{1} = \frac{y-4}{1} = \frac{z-5}{2}$.
Reveal Answer & Explanation
Answer: Direction vectors: $\vec{b}_1 = 3\hat{i} + 5\hat{j} + 4\hat{k}$, $\vec{b}_2 = \hat{i} + \hat{j} + 2\hat{k}$. $\cos\theta = \frac{3(1) + 5(1) + 4(2)}{\sqrt{9+25+16}\sqrt{1+1+4}} = \frac{16}{\sqrt{50}\sqrt{6}} = \frac{16}{5\sqrt{2}\sqrt{6}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}} \implies \theta = \cos^{-1}\left(\frac{8}{5\sqrt{3}}\right)$.
θ = cos^-1(8 / 5√3).
3
Find the shortest distance between the lines: $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$.
Reveal Answer & Explanation
Answer: $\vec{a}_2 - \vec{a}_1 = \hat{i} - 3\hat{j} - 2\hat{k}$. $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 3\hat{k}$. $|\vec{b}_1 \times \vec{b}_2| = \sqrt{9 + 0 + 9} = 3\sqrt{2}$. Dot product: $(-3)(1) + 0(-3) + 3(-2) = -3 - 6 = -9$. $d = \frac{|-9|}{3\sqrt{2}} = \frac{9}{3\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}\text{ units}$.
3√2 / 2 units.
4
What are the direction cosines of the $Z$-axis?
Reveal Answer & Explanation
Answer: The $Z$-axis makes $90^\circ, 90^\circ, 0^\circ$ with $X, Y, Z$ axes. Direction cosines are $(\cos 90^\circ, \cos 90^\circ, \cos 0^\circ) = (0, 0, 1)$.
(0, 0, 1).
5
Show that the lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.
Reveal Answer & Explanation
Answer: $a_1 a_2 + b_1 b_2 + c_1 c_2 = 7(1) + (-5)(2) + 1(3) = 7 - 10 + 3 = 0$. Since the sum of products of direction ratios is zero, the lines are perpendicular.
Sum of products of direction ratios is 0; perpendicular.
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