Blog Posts

Three Dimensional Geometry

馃搹 Chapter 11: Three Dimensional Geometry

1. Introduction

This chapter extends coordinate geometry to three dimensions using the vector approach developed in Chapter 10. We describe lines and planes in space using direction cosines, direction ratios, and vector/Cartesian equations.

2. Direction Cosines and Direction Ratios of a Line

Direction cosines (l, m, n): The cosines of the angles made by a line with the positive x, y, z axes.

l² + m² + n² = 1

Direction ratios (a, b, c): Numbers proportional to the direction cosines.

l = a/√(a²+b²+c²), m = b/√(a²+b²+c²), n = c/√(a²+b²+c²)

Direction ratios of a line joining two points A(x₁,y₁,z₁) and B(x₂,y₂,z₂):

(x₂−x₁, y₂−y₁, z₂−z₁)

3. Equation of a Line in Space

(a) Line through a point, parallel to a given vector

Vector form:

r = a + 位b

where a is the position vector of the given point, b is the direction vector, and 位 is a scalar (parameter).

Cartesian form: If the line passes through (x₁, y₁, z₁) with direction ratios a, b, c:

(x−x₁)/a = (y−y₁)/b = (z−z₁)/c

(b) Line through two points

Vector form:

r = a + 位(b − a)

Cartesian form: through (x₁,y₁,z₁) and (x₂,y₂,z₂):

(x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁)

4. Angle Between Two Lines

If two lines have direction ratios a₁, b₁, c₁ and a₂, b₂, c₂, the angle 胃 between them is given by:

cos 胃 = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁²+b₁²+c₁²) · √(a₂²+b₂²+c₂²)]

Condition Relation
Lines perpendiculara₁a₂ + b₁b₂ + c₁c₂ = 0
Lines parallela₁/a₂ = b₁/b₂ = c₁/c₂

5. Shortest Distance Between Two Lines

Skew lines: Lines that neither intersect nor are parallel — they exist in different planes. The shortest distance between them is measured along the common perpendicular.

For lines r = a₁ + 位b₁ and r = a₂ + 渭b₂, the shortest distance is:

d = |(b₁ × b₂) · (a₂ − a₁)| / |b₁ × b₂|

Distance between two parallel lines r = a₁ + 位b and r = a₂ + 渭b:

d = |b × (a₂ − a₁)| / |b|

6. Plane: Equations and Concepts

(a) Equation of a plane in normal form

Vector form: If n̂ is the unit normal vector to the plane and d is the distance from the origin:

r · n̂ = d

Cartesian form: If l, m, n are direction cosines of the normal:

lx + my + nz = d

(b) Equation of a plane through a point and perpendicular to a given vector

Vector form:

(r − a) · N = 0

Cartesian form: through (x₁, y₁, z₁) with normal direction ratios A, B, C:

A(x−x₁) + B(y−y₁) + C(z−z₁) = 0

(c) Equation of a plane through three non-collinear points

For points A(a), B(b), C(c), the vector equation is:

(r − a) · [(b − a) × (c − a)] = 0

(d) Intercept form of a plane

If a plane cuts the x, y, z axes at distances a, b, c from the origin:

x/a + y/b + z/c = 1

7. Angle Between Two Planes

If two planes have normal direction ratios A₁,B₁,C₁ and A₂,B₂,C₂:

cos 胃 = |A₁A₂+B₁B₂+C₁C₂| / [√(A₁²+B₁²+C₁²)·√(A₂²+B₂²+C₂²)]

Condition Relation
Planes perpendicularA₁A₂ + B₁B₂ + C₁C₂ = 0
Planes parallelA₁/A₂ = B₁/B₂ = C₁/C₂

8. Angle Between a Line and a Plane

If a line has direction ratios a, b, c and the plane has normal direction ratios A, B, C, the angle 蠁 between the line and the plane satisfies:

sin 蠁 = |aA+bB+cC| / [√(a²+b²+c²)·√(A²+B²+C²)]

9. Distance of a Point from a Plane

Vector form: Distance from point with position vector a to plane r · n̂ = d:

Distance = |a · n̂ − d|

Cartesian form: Distance of (x₁, y₁, z₁) from plane Ax + By + Cz = D:

Distance = |Ax₁ + By₁ + Cz₁ − D| / √(A²+B²+C²)

10. Important Formula Summary

Concept Formula
Line (Cartesian, point + DRs)(x−x₁)/a = (y−y₁)/b = (z−z₁)/c
Line through two points(x−x₁)/(x₂−x₁) = (y−y₁)/(y₂−y₁) = (z−z₁)/(z₂−z₁)
Shortest distance (skew lines)d = |(b₁×b₂)·(a₂−a₁)| / |b₁×b₂|
Plane (point + normal)A(x−x₁)+B(y−y₁)+C(z−z₁) = 0
Plane intercept formx/a + y/b + z/c = 1
Distance of point from plane|Ax₁+By₁+Cz₁−D| / √(A²+B²+C²)

11. Solved Examples

✏️ Example 1: Find the direction cosines of the line joining A(1, 2, 3) and B(4, 6, 15).

Solution: Direction ratios = (4−1, 6−2, 15−3) = (3, 4, 12)

Magnitude = √(9+16+144) = √169 = 13

Direction cosines = (3/13, 4/13, 12/13)

✏️ Example 2: Find the equation of the plane passing through (1, 0, −2) and perpendicular to the vector 3i + j − k.

Solution: Using A(x−x₁)+B(y−y₁)+C(z−z₁)=0 with A=3, B=1, C=−1:

3(x−1) + 1(y−0) − 1(z+2) = 0

3x + y − z − 5 = 0

✏️ Example 3: Find the distance of the point (2, 5, −3) from the plane x + 2y − 2z = 9.

Solution: Distance = |1(2)+2(5)−2(−3)−9| / √(1²+2²+(−2)²)

= |2+10+6−9| / √9 = |9| / 3 = 3

Distance = 3 units

12. Quick Revision Points

  • Direction cosines satisfy l² + m² + n² = 1; direction ratios are just proportional to them.
  • Two lines are perpendicular if a₁a₂+b₁b₂+c₁c₂ = 0, parallel if ratios are equal.
  • Skew lines don't intersect and aren't parallel — shortest distance uses the cross product formula.
  • A plane needs either a point + normal, or three non-collinear points, to be fully defined.
  • Distance of a point from a plane always uses the normal's direction ratios in the denominator.
  • Angle between line and plane uses sin (not cos, unlike line-line and plane-plane angles).

Comments

Post a Comment