Exam Weightage & Blueprint
Total: ~11-13 Marks3D 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
Relation: $l^2 + m^2 + n^2 = 1$
$\frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k$
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
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)
$a = 1 - (-2) = 3$
$b = 2 - 4 = -2$
$c = 3 - (-5) = 8$
$\sqrt{a^2+b^2+c^2} = \sqrt{9 + 4 + 64} = \sqrt{77}$
DCs are $\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}$
Q2. Find the shortest distance between the lines $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(3\hat{i} - 5\hat{j} + 2\hat{k})$. (5 Marks)
$\vec{a_1} = \hat{i} + \hat{j}, \quad \vec{b_1} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{a_2} = 2\hat{i} + \hat{j} - \hat{k}, \quad \vec{b_2} = 3\hat{i} - 5\hat{j} + 2\hat{k}$
$\vec{a_2} - \vec{a_1} = \hat{i} - \hat{k}$
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{vmatrix}$
$= \hat{i}(-2+5) - \hat{j}(4-3) + \hat{k}(-10+3) = 3\hat{i} - \hat{j} - 7\hat{k}$
$|\vec{b_1} \times \vec{b_2}| = \sqrt{9 + 1 + 49} = \sqrt{59}$
$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = (3\hat{i} - \hat{j} - 7\hat{k}) \cdot (\hat{i} - \hat{k})$
$= 3(1) + (-1)(0) + (-7)(-1) = 3 + 7 = 10$
$d = \left| \frac{10}{\sqrt{59}} \right| = \frac{10}{\sqrt{59}}$ units.
Previous Year Questions (PYQs)
(A) $k > 0$ (B) $0 < k < 1$ (C) $k = \pm \frac{1}{\sqrt{3}}$ (D) $k = \frac{1}{3}$
Ans: (C). Since $l^2+m^2+n^2=1 \Rightarrow 3k^2=1 \Rightarrow k = \pm \frac{1}{\sqrt{3}}$.
Hint: Direction of required line is $\vec{b} = \vec{b_1} \times \vec{b_2}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix}$. Then use point-parallel form.
Solution:
Convert to standard form:
$\vec{r} = (\hat{i}-2\hat{j}+3\hat{k}) + t(-\hat{i}+\hat{j}-2\hat{k})$
$\vec{r} = (\hat{i}-\hat{j}-\hat{k}) + s(\hat{i}+2\hat{j}-2\hat{k})$
Here $\vec{a_1} = \hat{i}-2\hat{j}+3\hat{k}, \vec{b_1} = -\hat{i}+\hat{j}-2\hat{k}$
$\vec{a_2} = \hat{i}-\hat{j}-\hat{k}, \vec{b_2} = \hat{i}+2\hat{j}-2\hat{k}$
Calculate $\vec{a_2}-\vec{a_1} = \hat{j}-4\hat{k}$ and $\vec{b_1} \times \vec{b_2} = 2\hat{i}-4\hat{j}-3\hat{k}$.
Distance $d = |\frac{8}{\sqrt{29}}|$.
Exam Strategy & Mistake Bank
Common Mistakes
Scoring Tips
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: $\cos\theta = |\frac{\vec{b_1} \cdot \vec{b_2}}{|\vec{b_1}||\vec{b_2}}|}$ or $|\frac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{\dots}\sqrt{\dots}}|$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.
Q2. Find the equation of the line passing through $(1, 2, 3)$ and parallel to vector $3\hat{i} + 2\hat{j} - 2\hat{k}$.
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 $\frac{x+1}{1} = \frac{y-4}{1} = \frac{z-5}{2}$.
Q4. Find the coordinates of the foot of the perpendicular drawn from the point $(0, 2, 3)$ to the line $\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3}$.
Level 3: Advanced (5 Marks Each)
Q5. Find the shortest distance between the lines $\vec{r} = (6\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$ and $\vec{r} = (-4\hat{i} - \hat{k}) + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$.
Quick Revision Tips (1 Day Before Exam)
Must Revise Topics
Formulas to Memorize:
- Direction Cosines relation
- Shortest Distance (Skew Lines)
- Angle between lines
- Equation of line (2 forms)
Common Question Types:
- Find Shortest Distance (5M)
- Find Foot of Perpendicular (3M/5M)
- Find Angle (2M)
- Convert Cartesian to Vector (1M)
Time Management Per Question
1 Mark (MCQ) = 1-2 minutes2 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.