Introduction to Three Dimensional Geometry Class 11 PDF: Complete Guide

Basics of Three Dimensional Geometry for Class 11 Students

Three dimensional geometry extends the concepts of coordinate geometry into space with three axes: $x$, $y$, and $z$. Each point in space is represented by coordinates $(x, y, z)$, which specify its position relative to these axes.

Key concepts include:

• Coordinate Axes: The three mutually perpendicular lines intersecting at the origin.
• Coordinate of a Point: Given as $(x, y, z)$.
• Distance Formula: The distance between points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ is

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

Understanding these basics is essential before moving to more complex topics like lines and planes in 3D.

Understanding Distance and Section Formulas in 3D Geometry

The distance formula helps find the length between two points in space, as shown above.

The section formula is used to find 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 the segment joining $A(2, -1, 3)$ and $B(4, 3, 5)$ in the ratio 2:3.

Solution: $$ R = \left( \frac{2 \times 4 + 3 \times 2}{5}, \frac{2 \times 3 + 3 \times (-1)}{5}, \frac{2 \times 5 + 3 \times 3}{5} \right) = ( \frac{8+6}{5}, \frac{6-3}{5}, \frac{10+9}{5} ) = (2.8, 0.6, 3.8) $$

Direction Ratios and Direction Cosines Explained

Direction ratios (DRs) and direction cosines (DCs) describe the orientation of a line in 3D space.

• Direction Ratios: Any three numbers proportional to the direction cosines of a line.
• Direction Cosines: Cosines of the angles ($\alpha$, $\beta$, $\gamma$) that the line makes with the $x$, $y$, and $z$ axes respectively.

If $l, m, n$ are the direction cosines, then

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

Relation with direction ratios $a, b, c$:

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

These concepts help in defining the equation of lines and planes in 3D.

Equations of Lines in Three Dimensional Geometry

There are two main forms to represent a line in 3D:

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. Cartesian Form: $$ \frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n} $$ where $(x_1, y_1, z_1)$ is a point on the line and $l, m, n$ are direction ratios.

Example: Find the equation of the line passing through $(1, 2, 3)$ and parallel to the vector $\vec{b} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.

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

Planes in 3D: Equation and Properties

A plane in three dimensional geometry can be defined by the equation:

$$ ax + by + cz + d = 0 $$ where $a$, $b$, and $c$ are the direction ratios of the normal to the plane, and $d$ is a constant.

Key points:

• The vector $\vec{n} = a\hat{i} + b\hat{j} + c\hat{k}$ is perpendicular to the plane.
• To find the distance of a point $(x_0, y_0, z_0)$ from the plane,

$$ \text{Distance} = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}} $$

Example: Find the distance of point $(3, -2, 1)$ from the plane $2x - 3y + 6z - 4 = 0$.

Solution: $$ \frac{|2(3) - 3(-2) + 6(1) - 4|}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{|6 + 6 + 6 - 4|}{\sqrt{4 + 9 + 36}} = \frac{14}{7} = 2 $$

Comparing 2D and 3D Coordinate Geometry

Understanding the difference between two and three dimensional geometry is crucial for Class 11 students. Here's a quick comparison:

This table helps clarify the extension of 2D concepts into the third dimension.