Chapter 11: Three-dimensional Geometry
In Class XI, we studied the basic concepts of three-dimensional geometry like coordinates of a point, distance formulas, section formulas. In this chapter, we extend that to lines.
Equation of a Line in Space
A line is uniquely determined if:
- It passes through a given point and has given direction (parallel to a given vector), or
- It passes through two given points.
Line through a Given Point and Parallel to a Given Vector
Vector Form: $\vec{r} = \vec{a} + \lambda\vec{b}$, where $\vec{a}$ is the position vector of the point and $\vec{b}$ is the direction vector. $\lambda$ is a scalar. Cartesian Form: $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
Line passing through Two Given Points
Vector Form: $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$ Cartesian Form: $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$
Shortest Distance Between Two Lines
If two lines in space intersect at a point, then the shortest distance between them is zero. Also, if two lines in space are parallel, then the shortest distance between them will be the perpendicular distance. If lines are neither parallel nor intersecting (called skew lines), the shortest distance between lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ is: $$ d = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|} \right| $$
Competency-Based Questions
1. [CBSE 2022] Find the shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(2\hat{i} + \hat{j} + 2\hat{k})$.
Solution:
$\vec{a_1} = \hat{i} + 2\hat{j} + \hat{k}, \quad \vec{b_1} = \hat{i} - \hat{j} + \hat{k}$
$\vec{a_2} = 2\hat{i} - \hat{j} - \hat{k}, \quad \vec{b_2} = 2\hat{i} + \hat{j} + 2\hat{k}$
$\vec{a_2} - \vec{a_1} = \hat{i} - 3\hat{j} - 2\hat{k}$
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix}$
$= \hat{i}(-2 - 1) - \hat{j}(2 - 2) + \hat{k}(1 - (-2)) = -3\hat{i} - 0\hat{j} + 3\hat{k}$.
$|\vec{b_1} \times \vec{b_2}| = \sqrt{(-3)^2 + 0^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$.
Now, $(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} - \vec{a_1}) = (-3\hat{i} + 3\hat{k}) \cdot (\hat{i} - 3\hat{j} - 2\hat{k})$
$= -3(1) + 0(-3) + 3(-2) = -3 - 6 = -9$.
Shortest distance, $d = \left| \frac{-9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$ units.
2. [Sample Paper 2023] Find the vector and cartesian equations of the line that passes through the points $(3, -2, -5)$ and $(3, -2, 6)$.
Solution:
Let $\vec{a} = 3\hat{i} - 2\hat{j} - 5\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + 6\hat{k}$.
The equation of line passing through two points $\vec{a}$ and $\vec{b}$ is $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$.
$\vec{b} - \vec{a} = (3 - 3)\hat{i} + (-2 - (-2))\hat{j} + (6 - (-5))\hat{k} = 0\hat{i} + 0\hat{j} + 11\hat{k}$.
Vector form: $\vec{r} = (3\hat{i} - 2\hat{j} - 5\hat{k}) + \lambda(11\hat{k})$.
Cartesian form:
$\frac{x - 3}{0} = \frac{y - (-2)}{0} = \frac{z - (-5)}{11}$.
$\frac{x - 3}{0} = \frac{y + 2}{0} = \frac{z + 5}{11}$.
3. [CBSE 2018] 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.
Solution:
The direction ratios of the first line are $a_1 = 7, b_1 = -5, c_1 = 1$.
The direction ratios of the second line are $a_2 = 1, b_2 = 2, c_2 = 3$.
Two lines are perpendicular if their direction vectors are perpendicular, i.e., their dot product is zero ($a_1a_2 + b_1b_2 + c_1c_2 = 0$).
$a_1a_2 + b_1b_2 + c_1c_2 = 7(1) + (-5)(2) + 1(3) = 7 - 10 + 3 = 10 - 10 = 0$.
Since the dot product is zero, the lines are perpendicular to each other.