What is Three Dimensional Geometry Class 12: Complete Guide

Introduction to Three Dimensional Geometry in Class 12

Three Dimensional Geometry is a crucial chapter in the Class 12 NCERT Mathematics syllabus. It extends the concepts of coordinate geometry from two dimensions to three dimensions by introducing the $x$-, $y$-, and $z$-axes. This chapter helps students visualize and analyze geometric shapes in space, such as points, lines, and planes.

Key concepts include:

• Representation of points in 3D space as $(x, y, z)$
• Understanding coordinate axes and planes
• Introduction to vectors and their role in 3D geometry

This foundation allows students to solve problems related to distances, angles, and intersections in three-dimensional space, which are essential for higher studies in mathematics, engineering, and sciences.

Coordinate System and Representation of Points

In three dimensional geometry, every point in space is represented by an ordered triplet $(x, y, z)$ where:

• $x$ is the coordinate along the X-axis
• $y$ is the coordinate along the Y-axis
• $z$ is the coordinate along the Z-axis

These axes are mutually perpendicular, intersecting at the origin $O(0,0,0)$.

Important terms:

• Coordinate Planes: XY-plane ($z=0$), YZ-plane ($x=0$), ZX-plane ($y=0$)
• Distance between two points: For points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$,

$$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Worked Example:

Find the distance between points $A(3, -2, 5)$ and $B(1, 2, -1)$.

$$d = \sqrt{(1 - 3)^2 + (2 + 2)^2 + (-1 - 5)^2} = \sqrt{(-2)^2 + 4^2 + (-6)^2} = \sqrt{4 + 16 + 36} = \sqrt{56} = 2\sqrt{14}$$

Direction Cosines and Direction Ratios of a Line

Direction cosines and direction ratios help describe the orientation of a line in 3D space.

• Direction Ratios (DRs): Any three proportional numbers $(a, b, c)$ that represent the direction of a line.
• Direction Cosines (DCs): The cosines of the angles that the line makes with the coordinate axes, denoted as $(l, m, n)$.

These satisfy the relation:

$$l^2 + m^2 + n^2 = 1$$

where

$$l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}}$$

Use: Direction cosines are used to find angles between lines and planes.

Example:

If a line has direction ratios $(2, -3, 6)$, find its direction cosines.

Calculate magnitude:

$$\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

So,

$$l = \frac{2}{7}, \quad m = \frac{-3}{7}, \quad n = \frac{6}{7}$$

Equations of a Line in Three Dimensions

In 3D geometry, a line can be represented in several forms:

1. Vector form:

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

where $\vec{a}$ is the position vector of a point on the line, $\vec{b}$ is the direction vector, and $\lambda$ is a scalar.

2. Parametric form:

If the line passes through point $P(x_1, y_1, z_1)$ and has direction ratios $(a, b, c)$,

$$x = x_1 + \lambda a, \quad y = y_1 + \lambda b, \quad z = z_1 + \lambda c$$

3. Symmetric form:

If $a, b, c \neq 0$,

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

Worked Example:

Find the symmetric form of the line passing through $A(1, 2, 3)$ with direction ratios $(2, -1, 4)$.

Answer:

$$\frac{x - 1}{2} = \frac{y - 2}{-1} = \frac{z - 3}{4}$$

Equations of Planes and Their Properties

A plane in 3D space can be defined by the equation:

$$Ax + By + Cz + D = 0$$

where $(A, B, C)$ is the normal vector perpendicular to the plane.

Key points:

• The normal vector determines the plane's orientation.
• The distance $d$ from the origin to the plane is:

$$d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$$

• The angle $\theta$ between two planes with normal vectors $\vec{n_1}$ and $\vec{n_2}$ is given by:

$$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}||\vec{n_2}|} = \frac{|A_1A_2 + B_1B_2 + C_1C_2|}{\sqrt{A_1^2 + B_1^2 + C_1^2}\sqrt{A_2^2 + B_2^2 + C_2^2}}$$

Example:

Find the distance of the plane $2x - 3y + 6z - 12 = 0$ from the origin.

$$d = \frac{| -12 |}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{12}{\sqrt{4 + 9 + 36}} = \frac{12}{7} \approx 1.71$$

Shortest Distance Between Two Skew Lines

Skew lines are lines that do not intersect and are not parallel. Finding the shortest distance between them is important in 3D geometry.

If two lines are given by:

$$\vec{r} = \vec{a_1} + \lambda \vec{b_1}$$

$$\vec{r} = \vec{a_2} + \mu \vec{b_2}$$

The shortest distance $d$ between these lines is:

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

where $\times$ denotes the cross product.

Example:

Find the shortest distance between the lines:

$$\vec{r} = (1, 2, 3) + \lambda (1, -1, 2)$$

and

$$\vec{r} = (3, 1, 0) + \mu (2, 1, -1)$$

Calculate:

• $\vec{a_2} - \vec{a_1} = (3 - 1, 1 - 2, 0 - 3) = (2, -1, -3)$
• $\vec{b_1} = (1, -1, 2)$
• $\vec{b_2} = (2, 1, -1)$

Cross product:

$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{vmatrix} = (-1)(-1) - (2)(1) \hat{i} - (1)(-1) - (2)(2) \hat{j} + (1)(1) - (-1)(2) \hat{k}$$

Simplify to get the vector and then apply the formula to find $d$.

Comparison of 2D and 3D Geometry Concepts

Understanding the differences between two-dimensional and three-dimensional geometry helps clarify the scope of this chapter.

This comparison highlights the added complexity and applications of 3D geometry in Class 12 mathematics.