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Home Class 12 Three Dimensional Geometry

Chapter 11: Three Dimensional Geometry

Overview

Prepare for Class 12 three-dimensional geometry questions on direction cosines and ratios, equations of lines, angles, and shortest distance between skew lines. Worked examples and CBSE PYQs support revision.

Complete Board Exam Focused Notes with Lines, Shortest Distance & PYQs

2026?27 Syllabus Focus

The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.

  • Direction ratios and direction cosines of the line joining two points.
  • Vector and Cartesian equations of a line, and the angle between two lines.
  • Skew lines and the shortest distance between two lines.

CBSE Mathematics (Code 041) curriculum, 2026?27

Last 24-Hour Checklist

Direction Cosines & Direction Ratios

Direction Cosines (DCs): If a line makes angles $α, β, γ$ with x, y, z axes respectively, then $l = \cosα, m = \cosβ, n = \cosγ$ are called direction cosines.
Relation: $l^2 + m^2 + n^2 = 1$
Direction Ratios (DRs): Any three numbers $a, b, c$ proportional to direction cosines $l, m, n$.
$\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k$

Finding DCs from DRs: $$l = \pm \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \pm \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \pm \frac{c}{\sqrt{a^2+b^2+c^2}}$$

Direction Ratios of line joining two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$:

$a = x_2 - x_1, \quad b = y_2 - y_1, \quad c = z_2 - z_1$

Equation of a Line in Space

1. Line through a Point and Parallel to a Vector

Vector Form

$\vec{r} = \vec{a} + λ\vec{b}$

$\vec{a}$: Position vector of point

$\vec{b}$: Parallel vector (Direction)

Cartesian Form

$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$

$(x_1, y_1, z_1)$: Point

$a, b, c$: Direction Ratios

2. Line Passing Through Two Points

Vector Form

$\vec{r} = \vec{a} + λ(\vec{b} - \vec{a})$

Cartesian Form

$\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$

Angle and Distance Between Lines

Angle Between Two Lines:
If lines are parallel to vectors $\vec{b_1}$ and $\vec{b_2}$: $$\cosθ = | \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|} |$$ Conditions:
Perpendicular: $\vec{b_1} \cdot \vec{b_2} = 0$ (or $a_1a_2 + b_1b_2 + c_1c_2 = 0$)
Parallel: $\vec{b_1} = λ\vec{b_2}$ (or $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$)

Shortest Distance (Most Important 5M Topic)

Geometry of Skew Lines & Common Perpendicular

Plane containing $L_1$ $L_1 (along \vec{b_1})$ $A(\vec{a_1})$ Plane containing $L_2$ $L_2 (along \vec{b_2})$ $B(\vec{a_2})$ $\vec{a_2} - \vec{a_1}$ Shortest Distance $d$ $\perp (\vec{b_1} × \vec{b_2})$

The shortest distance is the projection of $(\vec{a_2} - \vec{a_1})$ along the unit normal $\frac{\vec{b_1} × \vec{b_2}}{|\vec{b_1} × \vec{b_2}|}$.

Distance between Skew Lines

Vector Form: Lines $\vec{r} = \vec{a_1} + λ\vec{b_1}$ and $\vec{r} = \vec{a_2} + μ\vec{b_2}$

$$d = | \frac{(\vec{b_1} × \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} × \vec{b_2}|} |$$

Cartesian Form:

$d = | \frac{\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1c_2 - b_2c_1)^2 + (c_1a_2 - c_2a_1)^2 + (a_1b_2 - a_2b_1)^2}} |$

Distance between Parallel Lines

Lines: $\vec{r} = \vec{a_1} + λ\vec{b}$ and $\vec{r} = \vec{a_2} + μ\vec{b}$

$$d = | \frac{\vec{b} × (\vec{a_2} - \vec{a_1})}{|\vec{b}|} |$$

Foot of Perpendicular & Image of a Point in a 3D Line

⚡ High-Yield Foot of Perpendicular, Image & Skew Lines Rules:
  • Algorithm: Foot of Perpendicular from Point $Q(x_0, y_0, z_0)$ to Line $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$:
    • Step 1 (General Point): Any point on the line has coordinates $P(x_1 + a\lambda, y_1 + b\lambda, z_1 + c\lambda)$.
    • Step 2 (Perpendicular Condition): Form direction ratios of $\vec{QP} = (x_1+a\lambda - x_0, y_1+b\lambda - y_0, z_1+c\lambda - z_0)$. Since $\vec{QP} \perp \text{Line} \implies a(x_1+a\lambda - x_0) + b(y_1+b\lambda - y_0) + c(z_1+c\lambda - z_0) = 0$.
    • Step 3: Solve this linear equation for $\lambda$. Substitute $\lambda$ into $P$ to obtain the Foot of the Perpendicular $F$.
    • Perpendicular Distance: Distance $= |\vec{QF}| = \sqrt{(x_F - x_0)^2 + (y_F - y_0)^2 + (z_F - z_0)^2}$.
    • Image Point $Q'$: Since $F$ is the midpoint of segment $QQ'$, the Image is: $$Q' = 2F - Q = (2x_F - x_0, 2y_F - y_0, 2z_F - z_0)$$
  • Condition for Two Lines to Intersect (Coplanar Lines):
    Lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ intersect if and only if their shortest distance is $0$: $$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = 0 \iff \begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$
  • Shortest Distance Between Parallel Lines:
    Lines $\vec{r} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r} = \vec{a_2} + \mu\vec{b}$ share the same direction vector $\vec{b}$: $$d = \left| \frac{\vec{b} \times (\vec{a_2} - \vec{a_1})}{|\vec{b}|} \right|$$
  • 📌 Syllabus Clarification: All equations of planes (vector/Cartesian plane equations, angle between two planes, distance of a point from a plane, NCERT Ex 11.3) have been completely removed from the CBSE Class 12 syllabus. 3D geometry tests lines in space, direction ratios, shortest distance, and foot/image coordinates.
Solve NCERT Ex 11.2

Solved Examples (Board Marking Scheme)

Q1. Find the direction cosines of the line passing through the points $(-2, 4, -5)$ and $(1, 2, 3)$. (2 Marks)

Step 1: Find Direction Ratios 1 Mark

$a = 1 - (-2) = 3$

$b = 2 - 4 = -2$

$c = 3 - (-5) = 8$

Step 2: Calculate Magnitude & DCs 1 Mark

$\sqrt{a^2+b^2+c^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$

DCs are $\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$

Q2. Find the shortest distance between the lines $\vec{r} = \mathbf{i} + \mathbf{j} + λ(2\mathbf{i} - \mathbf{j} + \mathbf{k})$ and $\vec{r} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} + μ(3\mathbf{i} - 5\mathbf{j} + 2\mathbf{k})$. (5 Marks)

Step 1: Identify Vectors 1 Mark

$\vec{a_1} = \mathbf{i} + \mathbf{j}, \quad \vec{b_1} = 2\mathbf{i} - \mathbf{j} + \mathbf{k}$

$\vec{a_2} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}, \quad \vec{b_2} = 3\mathbf{i} - 5\mathbf{j} + 2\mathbf{k}$

$\vec{a_2} - \vec{a_1} = \mathbf{i} - \mathbf{k}$

Step 2: Calculate $\vec{b_1} × \vec{b_2}$ 2 Marks

$\vec{b_1} × \vec{b_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix}$

$= \mathbf{i}(-2+5) - \mathbf{j}(4-3) + \mathbf{k}(-10+3) = 3\mathbf{i} - \mathbf{j} - 7\mathbf{k}$

$|\vec{b_1} × \vec{b_2}| = \sqrt{9 + 1 + 49} = \sqrt{59}$

Step 3: Calculate Dot Product 1 Mark

$(\vec{b_1} × \vec{b_2}) · (\vec{a_2} - \vec{a_1}) = (3\mathbf{i} - \mathbf{j} - 7\mathbf{k}) · (\mathbf{i} - \mathbf{k})$

$= 3(1) + (-1)(0) + (-7)(-1) = 3 + 7 = 10$

Step 4: Final Formula 1 Mark

$d = | \frac{10}{\sqrt{59}} | = \frac{10}{\sqrt{59}}$ units.

Solve NCERT Misc Exercise

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): If the direction cosines of a line are $k, k, k$, then:
(A) $k > 0$   (B) $0 < k < 1$   (C) $k=\pm \frac{1}{\sqrt{3}}$   (D) $k=\frac{1}{3}$
Ans: (C). Since $l^2+m^2+n^2=1 \Rightarrow 3k^2=1 \Rightarrow k = ± \frac{1}{\sqrt{3}}$.
2022 (3 Marks): Find the equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.
Hint: Direction of required line is $\vec{b} = \vec{b_1} × \vec{b_2}$.
$\vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$. Then use point-parallel form.
2020 (5 Marks): Find the shortest distance between the lines whose vector equations are $\vec{r} = (1-t)\mathbf{i} + (t-2)\mathbf{j} + (3-2t)\mathbf{k}$ and $\vec{r} = (s+1)\mathbf{i} + (2s-1)\mathbf{j} - (2s+1)\mathbf{k}$.
Solution:
Convert to standard form:
$\vec{r} = (\mathbf{i}-2\mathbf{j}+3\mathbf{k}) + t(-\mathbf{i}+\mathbf{j}-2\mathbf{k})$
$\vec{r} = (\mathbf{i}-\mathbf{j}-\mathbf{k}) + s(\mathbf{i}+2\mathbf{j}-2\mathbf{k})$
Here $\vec{a}_1 = \mathbf{i}-2\mathbf{j}+3\mathbf{k}, \vec{b}_1 = -\mathbf{i}+\mathbf{j}-2\mathbf{k}$
$\vec{a}_2 = \mathbf{i}-\mathbf{j}-\mathbf{k}, \vec{b}_2 = \mathbf{i}+2\mathbf{j}-2\mathbf{k}$
Calculate $\vec{a}_2-\vec{a}_1 = \mathbf{j}-4\mathbf{k}$ and $\vec{b}_1 × \vec{b}_2 = 2\mathbf{i}-4\mathbf{j}-3\mathbf{k}$.
Distance $d = |\frac{8}{\sqrt{29}}|$.

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: In Cartesian form $\frac{x-x_1}{a}$, ensuring coefficient of $x, y, z$ is 1. If given $\frac{2x-1}{3}$, rewrite as $\frac{x-1/2}{3/2}$.
Mistake 2: Forgetting the modulus in the shortest distance formula. Distance cannot be negative!
Mistake 3: Confusing Direction Ratios (DRs) with Direction Cosines (DCs). Remember $l^2+m^2+n^2=1$ for DCs.
Mistake 4: Calculation error in determinant expansion for cross product $\vec{b_1} × \vec{b_2}$.

Scoring Tips

Tip 1: Always check if lines are parallel ($\vec{b_1} = λ\vec{b_2}$) before applying the skew lines formula.
Tip 2: For 5-mark questions on Shortest Distance, write the formula clearly at the start. It carries 1 mark.
Tip 3: If finding foot of perpendicular, assume a general point on the line and use dot product = 0.
Tip 4: Convert Cartesian to Vector form if you find vector operations easier (or vice versa).

Formula Sheet (Must Remember!)

Direction Cosines & Lines

1. Relation: $l^2 + m^2 + n^2 = 1$

2. From DRs: $l = \pm \frac{a}{\sqrt{a^2+b^2+c^2}}$, etc.

3. Vector Eq: $\vec{r} = \vec{a} + λ\vec{b}$

4. Cartesian Eq: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$

5. Two Points: $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$

Angle & Distance

6. Angle: $\cosθ = | \frac{\vec{b}_1 · \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} |$ or $| \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} |$

7. Perpendicular: $\vec{b_1} \cdot \vec{b_2} = 0$ or $a_1a_2 + b_1b_2 + c_1c_2 = 0$

8. Parallel: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

9. Shortest Distance (Skew): $d = |\frac{(\vec{b_1} × \vec{b_2}) · (\vec{a_2} - \vec{a_1})}{|\vec{b_1} × \vec{b_2}|}|$

10. Distance (Parallel): $d = |\frac{\vec{b} × (\vec{a_2} - \vec{a_1})}{|\vec{b}|}|$

Practice Problems (Self-Assessment)

Level 1: Basic (1-2 Marks Each)

Q1. If a line makes angles $90^°, 135^°, 45^°$ with the x, y and z axes respectively, find its direction cosines.

Answer: $0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Q2. Find the equation of the line passing through $(1, 2, 3)$ and parallel to vector $3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}$.

Answer: $\vec{r} = (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) + λ(3\mathbf{i}+2\mathbf{j}-2\mathbf{k})$.

Level 2: Intermediate (3-4 Marks Each)

Q3. Find the angle between the lines $\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4}$ and $\frac{x+1}{1} = \frac{y-4}{1} = \frac{z-5}{2}$.

Answer: $\cosθ = \frac{3(1)+5(1)+4(2)}{\sqrt{9+25+16}\sqrt{1+1+4}} = \frac{16}{\sqrt{50}\sqrt{6}} = \frac{16}{10\sqrt{3}}$. $θ = \cos^{-1}(\frac{8}{5\sqrt{3}})$.

Q4. Find the coordinates of the foot of the perpendicular drawn from the point $(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.

Hint: General point $P(5λ-3, 2λ+1, 3λ-4)$. Direction of perpendicular is $\vec{AP}$. Dot product with line direction $(5, 2, 3)$ is 0. Solve for $λ$.

Level 3: Advanced (5 Marks Each)

Q5. Find the shortest distance between the lines $\vec{r} = (6\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) + λ(\mathbf{i} - 2\mathbf{j} + 2\mathbf{k})$ and $\vec{r} = (-4\mathbf{i} - \mathbf{k}) + μ(3\mathbf{i} - 2\mathbf{j} - 2\mathbf{k})$.

Answer: 9 units.

Quick Revision Tips (1 Day Before Exam)

Must Revise Topics

Formulas to Memorize:

  1. Direction Cosines relation
  2. Shortest Distance (Skew Lines)
  3. Angle between lines
  4. Equation of line (2 forms)

Common Question Types:

  1. Find Shortest Distance (5M)
  2. Find Foot of Perpendicular (3M/5M)
  3. Find Angle (2M)
  4. Convert Cartesian to Vector (1M)

Time Management Per Question

1 Mark (MCQ) = 1-2 minutes
2 Marks = 3-4 minutes
3 Marks = 5-6 minutes
5 Marks = 8-10 minutes (Shortest Distance calculation takes time!)

Exam Day Checklist

Before Starting 3D Questions:

  • Read question carefully - Is it Vector or Cartesian form?
  • Check if lines are parallel or skew.
  • Verify if equation is in standard form ($\frac{x-x_1}{a}$).

While Solving:

  • Write formula first.
  • Show determinant expansion for cross product.
  • Don't forget modulus for distance/angle.
  • Box the final answer with units.

Chapter 11 Board Mastery Test

Practice test • 10 marks • 25 min

Q1 (1 Mark): If a line makes angles $90^°, 135^°, 45^°$ with the $x, y$ and $z$ axes respectively, find its direction cosines.
Q2 (2 Marks): Find the vector and cartesian equations of the line that passes through the point $(5, 2, -4)$ and is parallel to the vector $3\hat{i} + 2\hat{j} - 8\hat{k}$.
Q3 (3 Marks): Find the angle between the pair of lines: $\vec{r} = (3\hat{i} + 2\hat{j} - 4\hat{k}) + λ(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = (5\hat{i} - 2\hat{j}) + μ(3\hat{i} + 2\hat{j} + 6\hat{k})$.
Q4 (4 Marks): Find the shortest distance between the lines: $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + λ(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + μ(2\hat{i} + \hat{j} + 2\hat{k})$.
PRACTICE THIS CHAPTER
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Chapter 11 Study Ecosystem: Complete Preparation Hub

Strengthen your conceptual grasp of Three-Dimensional Geometry by progressing seamlessly through our chapter notes, step-by-step NCERT exercise solutions, authentic past 10-year board questions, and timed diagnostic assessments:

Textbook Solutions NCERT Exercise Practice →

Full step-by-step textbook solutions with formulas and explanations.

 Board Questions CBSE Board PYQs (2015-2024) →

Past 10 years board examination problems sorted by topic.

 Online Assessment Chapter Diagnostic Test →

Timed online test with instant grading and step-by-step solutions.

Frequently Asked Questions: Class 12 Three-Dimensional Geometry

1. What is the weightage of Chapter 11 Three-Dimensional Geometry in CBSE Class 12? +

Together with Vectors, it forms Unit IV (14 marks). Chapter 11 frequently features a 5-mark subjective question or a 4-mark case study.

2. What is the formula for the shortest distance between two skew lines? +

For lines r = a1 + lambda b1 and r = a2 + mu b2, the shortest distance d = | (b1 x b2) . (a2 - a1) | / |b1 x b2|.

3. What is the relation between direction cosines l, m, n? +

Direction cosines satisfy l^2 + m^2 + n^2 = 1. If direction ratios are a, b, c, then l = a / sqrt(a^2 + b^2 + c^2).

4. Are equations of planes included in the CBSE Class 12 syllabus? +

No, equations of planes have been removed from the current CBSE board syllabus. The syllabus focuses strictly on direction cosines/ratios, equations of lines, and shortest distance between lines.

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