What is Introduction to Three Dimensional Geometry Class 11: A Clear Guide

Understanding Three Dimensional Geometry in Class 11

Three Dimensional Geometry extends the study of geometry into three dimensions: length, breadth, and height. Unlike 2D geometry which deals with points on a plane, 3D geometry studies points in space using three coordinates: $x$, $y$, and $z$. This chapter in Class 11 NCERT Mathematics introduces these basic concepts to help students visualize and solve problems involving spatial figures.

Key elements include:

• Point: A location in space represented as $(x, y, z)$.
• Line: A straight path extending in both directions, defined by direction ratios.
• Plane: A flat surface extending infinitely in two dimensions within 3D space.

This foundation is crucial for understanding shapes like cubes, spheres, and cylinders, and for solving problems involving distances and angles in space.

Coordinate System and Representation of Points in 3D

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

• $x$ is the coordinate on the X-axis,
• $y$ is the coordinate on the Y-axis,
• $z$ is the coordinate on the Z-axis.

The three axes are mutually perpendicular and intersect at the origin $(0,0,0)$.

For example, the point $P(2, -3, 5)$ lies 2 units along X, -3 units along Y, and 5 units along Z from the origin.

This coordinate system helps in locating points precisely in space and forms the basis for further calculations like distance and section formulas.

Distance Formula Between Two Points in 3D Space

The distance between two points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in three dimensional space is given by the formula:

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

Example: Find the distance between $A(1, 2, 3)$ and $B(4, 6, 8)$.

Calculation:

$$ AB = \sqrt{(4-1)^2 + (6-2)^2 + (8-3)^2} = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2} $$

This formula is essential for solving problems related to lengths in 3D geometry.

Section Formula: Finding Coordinates of a Dividing Point

The section formula helps find the coordinates of a point dividing a line segment joining two points in a given ratio.

If point $R$ divides the line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ in the ratio $m:n$, then:

$$ R = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right) $$

Example: Find the coordinates of the point dividing $A(2, 3, 4)$ and $B(8, 5, 6)$ in the ratio 2:3.

Calculation:

$$ R = \left( \frac{2 \times 8 + 3 \times 2}{5}, \frac{2 \times 5 + 3 \times 3}{5}, \frac{2 \times 6 + 3 \times 4}{5} \right) = (4.8, 3.8, 4.8) $$

This formula is widely used in coordinate geometry problems.

Direction Ratios and Direction Cosines of a Line

Direction ratios (DRs) are three numbers proportional to the direction cosines of a line in 3D space. They describe the orientation of the line.

• If a line makes angles $\alpha$, $\beta$, and $\gamma$ with the X, Y, and Z axes respectively, then its direction cosines are:

$$ (l, m, n) = (\cos \alpha, \cos \beta, \cos \gamma) $$

• Direction ratios $(a, b, c)$ satisfy:

$$ \frac{a}{l} = \frac{b}{m} = \frac{c}{n} $$

These help in defining the vector form and equations of a line in 3D geometry.

Example: If a line has direction ratios 3, 4, 12, find its direction cosines.

Calculation:

$$ \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 $$

So,

$$ (l, m, n) = \left( \frac{3}{13}, \frac{4}{13}, \frac{12}{13} \right) $$

Comparing Two Dimensional and Three Dimensional Geometry

Understanding the difference between 2D and 3D geometry is important for grasping three dimensional geometry concepts.

This comparison helps Class 11 students appreciate the added complexity and utility of 3D geometry.