Home Class 12 Three Dimensional Geometry

Chapter 11: Three Dimensional Geometry

Overview

This page provides an Overview of Chapter Wise Notes for CBSE Class 12. Key topics covered: Exam Weightage & Blueprint, Last 24-Hour Checklist, Direction Cosines & Direction Ratios, Equation of a Line in Space, 2. Line Passing Through Two Points, Angle and Distance Between Lines, Solved Examples (Board Marking Scheme), Previous Year Questions (PYQs). All content is aligned with the NCERT syllabus for Board Exam preparation.

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

Exam Weightage & Blueprint

Total: ~11-13 Marks

3D Geometry is a high-weightage chapter often combined with Vectors. The "Shortest Distance between two lines" is a guaranteed 5-mark question in almost every board exam.

Question Type Marks Frequency Focus Topic
MCQ 1 Very High Direction Cosines, Angle between lines
Short Answer (2M) 2 High Equation of Line (Vector/Cartesian)
Short Answer (3M) 3 Medium Angle between lines, Perpendicular lines
Long Answer (5M) 5 Very High Shortest Distance between Skew Lines

Last 24-Hour Checklist

Direction Cosines & Direction Ratios

Direction Cosines (DCs): If a line makes angles $\alpha, \beta, \gamma$ with x, y, z axes respectively, then $l = \cos\alpha, m = \cos\beta, n = \cos\gamma$ 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

$\vec{r} = \vec{a} + \lambda\vec{b}$

$\vec{a}$: 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

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

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 $\vec{b_2}$: $$\cos\theta = \left| \frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}|} \right|$$ Conditions:
Perpendicular: $\vec{b_1} \cdot \vec{b_2} = 0$ (or $a_1a_2 + b_1b_2 + c_1c_2 = 0$)
Parallel: $\vec{b_1} = \lambda\vec{b_2}$ (or $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$)

Shortest Distance (Most Important 5M Topic)

Distance between Skew Lines

Lines: $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$

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

Distance between Parallel Lines

Lines: $\vec{r} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r} = \vec{a_2} + \mu\vec{b}$

$$d = \left| \frac{\vec{b} \times (\vec{a_2} - \vec{a_1})}{|\vec{b}|} \right|$$

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 $\vec{r} = \mathbf{i} + \mathbf{j} + \lambda(2\mathbf{i} - \mathbf{j} + \mathbf{k})$ and $\vec{r} = 2\mathbf{i} + \mathbf{j} - \mathbf{k} + \mu(3\mathbf{i} - 5\mathbf{j} + 2\mathbf{k})$. (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 $\vec{b_1} \times \vec{b_2}$ 2 Marks

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

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

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

Step 3: Calculate Dot Product 1 Mark

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

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

Step 4: Final Formula 1 Mark

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

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 $l^2+m^2+n^2=1 \Rightarrow 3k^2=1 \Rightarrow k = \pm \frac{1}{\sqrt{3}}$.
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 $\vec{b} = \vec{b_1} \times \vec{b_2}$.
$\vec{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$. 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 $\vec{b}_1 \times \vec{b}_2 = 2\mathbf{i}-4\mathbf{j}-3\mathbf{k}$.
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 $\vec{b_1} \times \vec{b_2}$.

Scoring Tips

Tip 1: Always check if lines are parallel ($\vec{b_1} = \lambda\vec{b_2}$) 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: $\vec{r} = \vec{a} + \lambda\vec{b}$

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: $\cos\theta = \left| \frac{\vec{b}_1 \cdot \vec{b}_2}{|\vec{b}_1||\vec{b}_2|} \right|$ or $\left| \frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} \right|$

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

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

Practice Problems (Self-Assessment)

Level 1: Basic (1-2 Marks Each)

Q1. If a line makes angles $90^\circ, 135^\circ, 45^\circ$ 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: $\vec{r} = (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) + \lambda(3\mathbf{i}+2\mathbf{j}-2\mathbf{k})$.

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

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

Level 3: Advanced (5 Marks Each)

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

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.
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