While plane geometry deals with two-dimensional figures, three-dimensional (3D) geometry extends our understanding to space. Every point in 3D space is located by three coordinates (x, y, z), forming an ordered triple. This chapter introduces coordinate axes in space, the distance formula, and the section formula in three dimensions.
- Coordinate Axes and Planes:
- Three mutually perpendicular lines intersect at the origin O, forming three axes: X-axis, Y-axis, Z-axis. These divide space into eight octants. Three coordinate planes are formed:
- XY-plane: z = 0
- YZ-plane: x = 0
- XZ-plane: y = 0
- A point P(x, y, z) has:
- x = perpendicular distance from YZ-plane
- y = perpendicular distance from XZ-plane
- z = perpendicular distance from XY-plane
Distance Formula in 3D:
The distance between P(x1, y1, z1) and Q(x2, y2, z2) is:
PQ = sqrt[(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2]
Distance from Origin:
|OP| = √(x2 + y2 + z2)
- Section Formula:
- The coordinates of the point R that divides the line segment joining P(x1, y1, z1) and Q(x2, y2, z2) in the ratio m:n are:
- Internally: R = [(mx2 + nx1)/(m+n), (my2 + ny1)/(m+n), (mz2 + nz1)/(m+n)]
- Externally: Replace + in denominator and in numerator with -.
Midpoint Formula:
Midpoint M = [(x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2]
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Worked Examples
Find the distance between A(1, 2, 3) and B(4, 6, 3).
AB = sqrt[(4-1)2 + (6-2)2 + (3-3)2] = sqrt[9 + 16 + 0] = √(25) = 5
Find the distance of the point (2, -3, 4) from the origin.
|OP| = √(4 + 9 + 16) = √(29)
Find the midpoint of the segment joining (1, 3, 5) and (7, -1, 3).
M = [(1+7)/2, (3-1)/2, (5+3)/2] = (4, 1, 4)
Find the coordinates of the point that divides A(2, 1, 4) and B(8, 7, 10) internally in ratio 2:1.
R = [(2×8 + 1×2)/(2+1), (2×7 + 1×1)/3, (2×10 + 1×4)/3]
= [(16+2)/3, (14+1)/3, (20+4)/3] = (6, 5, 8)
Show that the points A(3, -2, 4), B(1, 1, 1), and C(-1, 4, -2) are collinear.
AB = √(4+9+9) = √(22), BC = √(4+9+9) = √(22), AC = √(16+36+36) = √(88) = 2sqrt(22).
Since AC = AB + BC, the points are collinear.
Find the point on the y-axis equidistant from A(3, 1, 2) and B(-1, 2, 3).
A point on y-axis is P(0, y, 0). PA2 = 9 + (y-1)2 + 4; PB2 = 1 + (y-2)2 + 9.
Setting equal: 13 + y2 - 2y + 1 = 10 + y2 - 4y + 4. 2y = 0, y = 0. Point: (0, 0, 0)
Which octant does the point (-2, 3, -5) lie in?
x < 0, y > 0, z < 0 — this is Octant VI (second octant of the lower half).
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Common mistakes
- The coordinates of a point on the XZ-plane are (x, 0, z), NOT (x, y, 0). Remember, the coordinate that is zero identifies the plane the point lies on.
- In the section formula, apply the ratio m:n consistently — do not swap m and n.
- Distance is always positive; never forget to take the square root.
Summary
3D geometry extends 2D concepts by adding a z-coordinate. The distance formula, section formula, and midpoint formula all generalise naturally from 2D to 3D.