What this chapter is about
Three dimensional geometry extends the coordinate methods you learnt for the plane into the space around you. Instead of two axes (x and y), you now work with three mutually perpendicular axes (x, y and z). Every point in space has an ordered triple (x, y, z), and you can describe lines and planes using algebra.
This chapter teaches you how to write the equation of a line passing through given points or parallel to a given direction, and how to write the equation of a plane. You learn to find the angle between two lines, between two planes, or between a line and a plane. You also learn to calculate the shortest distance between two lines that do not meet (skew lines) and the perpendicular distance from a point to a plane.
After studying this chapter, you should be able to convert between different forms of line and plane equations, solve problems involving angles and distances in space, and visualise geometric relationships in three dimensions.
Key ideas
- A line in space is fixed by a point it passes through and a direction; direction ratios (a, b, c) or direction cosines (l, m, n) describe that direction.
- Direction cosines satisfy l² + m² + n² = 1; direction ratios are any set proportional to the direction cosines.
- Two lines in space are either parallel, intersecting, or skew (neither parallel nor meeting).
- A plane in space is fixed by a point and a normal vector, or by three non-collinear points.
- The angle between two lines equals the angle between their direction vectors; the angle between two planes equals the angle between their normals.
- The shortest distance between skew lines is measured along the common perpendicular to both.
Formulas and facts to remember
Direction cosines relation: l² + m² + n² = 1, where l = cos α, m = cos β, n = cos γ are the cosines of angles the line makes with the x, y, z axes.
Equation of a line through point (x₁, y₁, z₁) with direction ratios a, b, c (Cartesian form): (x − x₁)/a = (y − y₁)/b = (z − z₁)/c
Vector form of a line: r = a + λb, where a is the position vector of a point on the line and b is the direction vector.
Equation of a line through two points (x₁, y₁, z₁) and (x₂, y₂, z₂): (x − x₁)/(x₂ − x₁) = (y − y₁)/(y₂ − y₁) = (z − z₁)/(z₂ − z₁)
Angle θ between two lines with direction ratios (a₁, b₁, c₁) and (a₂, b₂, c₂): cos θ = |a₁a₂ + b₁b₂ + c₁c₂| / [√(a₁² + b₁² + c₁²) × √(a₂² + b₂² + c₂²)]
General equation of a plane: ax + by + cz + d = 0, where (a, b, c) is normal to the plane.
Equation of plane through (x₁, y₁, z₁) with normal (a, b, c): a(x − x₁) + b(y − y₁) + c(z − z₁) = 0
Distance from point (x₁, y₁, z₁) to plane ax + by + cz + d = 0: Distance = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²)
Shortest distance between skew lines r = a₁ + λb₁ and r = a₂ + μb₂: Distance = |(a₂ − a₁) · (b₁ × b₂)| / |b₁ × b₂|
Worked examples
Example 1: Equation of a line
Find the Cartesian equation of the line passing through (2, −1, 3) and parallel to the vector 4i − 2j + 5k.
Solution: The point is (2, −1, 3) and the direction ratios are (4, −2, 5). Using the formula: (x − 2)/4 = (y + 1)/(−2) = (z − 3)/5
This is the required equation.
Example 2: Angle between two planes
Find the angle between the planes 2x − y + z = 7 and x + y − 2z = 3.
Solution: Normal to first plane: n₁ = (2, −1, 1). Normal to second plane: n₂ = (1, 1, −2).
cos θ = |n₁ · n₂| / (|n₁| × |n₂|)
n₁ · n₂ = 2(1) + (−1)(1) + (1)(−2) = 2 − 1 − 2 = −1
- n₁| = √(4 + 1 + 1) = √6: n₂| = √(1 + 1 + 4) = √6
cos θ = |−1| / (√6 × √6) = 1/6
θ = cos⁻¹(1/6)
Example 3: Distance from a point to a plane
Find the perpendicular distance from the point (3, −2, 1) to the plane x − 2y + 2z = 5.
Solution: Rewrite the plane as x − 2y + 2z − 5 = 0. Here a = 1, b = −2, c = 2, d = −5.
Distance = |1(3) + (−2)(−2) + 2(1) − 5| / √(1 + 4 + 4) = |3 + 4 + 2 − 5| / √9 = |4| / 3 = 4/3 units
Common mistakes
Forgetting to take the absolute value when finding angle or distance → the formulas use modulus to ensure a non-negative result.
Confusing direction ratios with direction cosines → direction cosines must satisfy l² + m² + n² = 1; direction ratios need not.
Using the wrong normal vector for a plane → in ax + by + cz + d = 0, the normal is (a, b, c), not (a, b, d).
Sign errors when subtracting coordinates in the two-point form of a line → keep (x₂ − x₁), (y₂ − y₁), (z₂ − z₁) in order.
Applying the skew-line distance formula to parallel lines → for parallel lines, the cross product b₁ × b₂ is zero, so a different method (point-to-line distance) is needed.
Quick revision
- Direction cosines: l² + m² + n² = 1; direction ratios are proportional to them.
- Line through (x₁, y₁, z₁) with direction (a, b, c): (x − x₁)/a = (y − y₁)/b = (z − z₁)/c.
- Plane with normal (a, b, c): ax + by + cz + d = 0.
- Angle between planes = angle between their normals.
- Distance from point to plane = |ax₁ + by₁ + cz₁ + d| / √(a² + b² + c²).
- Skew-line shortest distance uses the scalar triple product divided by |b₁ × b₂|.