Chapter 11: Introduction to Three-Dimensional Geometry
11.1 Introduction
In two-dimensional geometry, we use two intersecting coordinate axes (X-axis and Y-axis) to locate a point in a plane. To locate a point in space, we require a system of three intersecting, mutually perpendicular lines, forming the three-dimensional coordinate system.
11.2 Coordinate Axes and Coordinate Planes
Let X’OX, Y’OY, and Z’OZ be three mutually perpendicular lines passing through the origin O. These three lines form the coordinate axes (X-axis, Y-axis, Z-axis). The planes determined by pairs of axes are called Coordinate Planes:
- XY-plane: The plane determined by the X-axis and Y-axis. Every point on this plane has a z-coordinate of zero.
- YZ-plane: Determined by Y and Z axes. The x-coordinate is zero.
- ZX-plane: Determined by Z and X axes. The y-coordinate is zero.
These three planes divide space into eight parts called octants.
Coordinates of a Point in Space
The coordinates of a point \( P \) in 3D space are always written in the form of an ordered triplet \( (x, y, z) \). Here, \( x, y, z \) represent the perpendicular distances from the YZ, ZX, and XY planes respectively.
11.3 Distance between Two Points
The distance \( PQ \) between two points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) in space is given by the 3D distance formula: \[ PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
This is a direct derivation from the 2D distance formula by applying the Pythagorean Theorem twice in space.
11.4 Section Formula
The coordinates of a point \( R \) that divides the line segment joining points \( P(x_1, y_1, z_1) \) and \( Q(x_2, y_2, z_2) \) internally in the ratio \( m:n \) are given by: \[ \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n} \right) \]
If point \( R \) divides the segment externally in the ratio \( m:n \), then the coordinates are: \[ \left( \frac{mx_2 - nx_1}{m-n}, \frac{my_2 - ny_1}{m-n}, \frac{mz_2 - nz_1}{m-n} \right) \]
Mid-point Formula: If \( R \) is the midpoint of segment \( PQ \), the ratio is \( 1:1 \), and the coordinates are: \[ \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \]
Competency-Based Questions
Question 1:
Show that the points \( P(-2, 3, 5) \), \( Q(1, 2, 3) \), and \( R(7, 0, -1) \) are collinear.
Answer 1:
Points are collinear if they lie on the same straight line. We can check this by calculating the distances between the points. If the sum of any two distances is equal to the third distance, the points are collinear.
Calculate \( PQ \):
\[ PQ = \sqrt{(1 - (-2))^2 + (2 - 3)^2 + (3 - 5)^2} \]
\[ PQ = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \]
Calculate \( QR \):
\[ QR = \sqrt{(7 - 1)^2 + (0 - 2)^2 + (-1 - 3)^2} \]
\[ QR = \sqrt{6^2 + (-2)^2 + (-4)^2} = \sqrt{36 + 4 + 16} = \sqrt{56} = 2\sqrt{14} \]
Calculate \( PR \):
\[ PR = \sqrt{(7 - (-2))^2 + (0 - 3)^2 + (-1 - 5)^2} \]
\[ PR = \sqrt{9^2 + (-3)^2 + (-6)^2} = \sqrt{81 + 9 + 36} = \sqrt{126} = 3\sqrt{14} \]
Notice that \( PQ + QR = \sqrt{14} + 2\sqrt{14} = 3\sqrt{14} \).
Since \( PQ + QR = PR \), the points \( P, Q, \) and \( R \) are collinear.
Conclusion: The given points are collinear.
Question 2:
Find the coordinates of the point which divides the line segment joining the points \( (1, -2, 3) \) and \( (3, 4, -5) \) in the ratio \( 2:3 \) internally.
Answer 2:
Let the given points be \( A(1, -2, 3) \) and \( B(3, 4, -5) \).
The ratio is \( m:n = 2:3 \).
Using the internal section formula, the coordinates \( (x,y,z) \) of the point \( P \) are:
\[ x = \frac{mx_2 + nx_1}{m+n} = \frac{2(3) + 3(1)}{2+3} = \frac{6 + 3}{5} = \frac{9}{5} \]
\[ y = \frac{my_2 + ny_1}{m+n} = \frac{2(4) + 3(-2)}{2+3} = \frac{8 - 6}{5} = \frac{2}{5} \]
\[ z = \frac{mz_2 + nz_1}{m+n} = \frac{2(-5) + 3(3)}{2+3} = \frac{-10 + 9}{5} = \frac{-1}{5} \]
Conclusion: The coordinates of the point are \( \left(\frac{9}{5}, \frac{2}{5}, \frac{-1}{5}\right) \).