Mathematics Notes for Class 12 Three Dimensional-Geometry

Three Dimensional Geometry

Welcome to the chapter on Three Dimensional Geometry for Class 12. In this chapter, you will learn about the basics of 3D geometry, direction cosines and ratios, equations of lines and planes in space, and the shortest distance between lines. By the end of this chapter, you will be able to solve problems involving points, lines, and planes in three dimensions.

Key Concepts

Three Dimensional Geometry: The study of points, lines, and planes in space.
Coordinates: A point in space is represented as (x, y, z).
Direction Cosines (l, m, n): Cosines of the angles a line makes with the x, y, and z axes.
Direction Ratios: Any set of three numbers proportional to the direction cosines.

Distance Between Two Points

The distance between points \( A(x_1, y_1, z_1) \) and \( B(x_2, y_2, z_2) \) is:

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

Direction Cosines and Ratios

If a line makes angles α, β, γ with the x, y, z axes, then its direction cosines are l = cosα, m = cosβ, n = cosγ.
Direction cosines satisfy \( l^2 + m^2 + n^2 = 1 \).
Direction ratios are any numbers proportional to l, m, n.

Equation of a Line in Space

The equation of a line passing through point \( A(x_1, y_1, z_1) \) and having direction ratios a, b, c is:

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

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 a direction vector, and λ is a parameter.

Equation of a Plane

The general equation of a plane is:

\( ax + by + cz + d = 0 \)

a, b, c are direction ratios of the normal to the plane.
If the plane passes through point \( (x_1, y_1, z_1) \), then the equation is:
\( a(x-x_1) + b(y-y_1) + c(z-z_1) = 0 \)

Angle Between Two Lines

If two lines have direction ratios \( (a_1, b_1, c_1) \) and \( (a_2, b_2, c_2) \), the angle θ between them is given by:

\( \cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \)

Angle Between Two Planes

The angle between two planes is the angle between their normals.

\( \cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \)

Shortest Distance Between Two Skew Lines

The shortest distance (SD) between two skew lines is given by:

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

where \( \vec{a_1}, \vec{a_2} \) are position vectors of points on the lines, and \( \vec{b_1}, \vec{b_2} \) are direction vectors.

Practice Questions

Find the distance between the points (1, 2, 3) and (4, 6, 8).
Write the vector and Cartesian equations of the line passing through (2, -1, 3) and parallel to the vector \( 3\hat{i} + 2\hat{j} - \hat{k} \).
Find the equation of the plane passing through (1, 2, 3) and perpendicular to the vector \( 2\hat{i} + \hat{j} + 2\hat{k} \).
If two lines have direction ratios (1, 2, 2) and (2, 1, -2), find the angle between them.
Find the shortest distance between the lines \( \vec{r} = \hat{i} + 2\hat{j} + 3\hat{k} + \lambda (2\hat{i} - \hat{j} + \hat{k}) \) and \( \vec{r} = 2\hat{i} - \hat{j} + \mu (\hat{i} + 2\hat{j} + 2\hat{k}) \).

Challenge Yourself

Prove that the lines \( \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1} \) and \( \frac{x-2}{1} = \frac{y+1}{2} = \frac{z}{2} \) are skew and find the shortest distance between them.
Find the equation of the plane passing through three points: (1, 0, 0), (0, 1, 0), and (0, 0, 1).

Did You Know?

Three dimensional geometry is used in computer graphics, engineering, and architecture.
The shortest distance between two skew lines is always perpendicular to both lines.

Glossary

Skew Lines: Lines that do not intersect and are not parallel.
Direction Cosines: Cosines of the angles a line makes with the axes.
Plane: A flat surface extending in all directions.
Vector Equation: An equation representing a line or plane using vectors.

Answers to Practice Questions

Distance = \( \sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} \)
Vector: \( \vec{r} = 2\hat{i} - \hat{j} + 3\hat{k} + \lambda (3\hat{i} + 2\hat{j} - \hat{k}) \)
Cartesian: \( \frac{x-2}{3} = \frac{y+1}{2} = \frac{z-3}{-1} \)
Equation: \( 2(x-1) + (y-2) + 2(z-3) = 0 \) or \( 2x + y + 2z = 9 \)
\( \cos\theta = \frac{1 \times 2 + 2 \times 1 + 2 \times (-2)}{\sqrt{1^2+2^2+2^2} \sqrt{2^2+1^2+(-2)^2}} = \frac{2+2-4}{3 \times 3} = 0 \). So, the lines are perpendicular.
Use the shortest distance formula for skew lines (substitute the vectors and compute).

Practice 3D geometry to master spatial reasoning and solve advanced geometry problems!

Mathematics Class 12 - Three Dimensional-Geometry Notes