☰ Year plan · Pure Mathematics

Lines and planes meeting, angles between planes, and distances

Week 14 · Where a line meets a plane, the angle between planes, parallel and perpendicular planes, the line of intersection of two planes, distances from a point to a plane and between parallel planes, the intercept form · 3 lessons
2

The line of intersection of two planes, and the distance from a point to a plane

Pure Mathematics Sec 3 Term 1, Unit 4, Lesson 2
GoalFind the line in which two planes meet, and the distance from a point to a plane.
New words
line of intersection · the line common to two planes that are not parallel; its direction is n₁ × n₂distance from a point to a plane · the length of the perpendicular from the point to the planeprojection on the normal · the distance is the size of the projection of BA on n, where B is a point of the plane
Explanation

Two planes that are not parallel meet in a line. To find it, use two facts. The direction: the line lies in both planes, so it is perpendicular to both normals, and d = n₁ × n₂. A point: choose a convenient coordinate, say z = 0 (or any one that makes the equations easy), and solve the two equations for the other two unknowns. The line is r = P + t d. Check that P satisfies both equations and that n₁ · d = n₂ · d = 0. Now the distance from a point A(x₁, y₁, z₁) to a plane. Take any point B in the plane with normal n. The distance is the size of the projection of the vector BA on n, which is |BA · n|/‖n‖. In Cartesian form for the plane ax + by + cz + d = 0 this becomes |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²). The numerator is the equation of the plane with the coordinates of the point put in; the denominator is the size of the normal. In vector form n · r = k, the same distance is |n · A − k|/‖n‖. A zero value means the point is on the plane.

ƒFormula
distance = |ax₁ + by₁ + cz₁ + d| ÷ √(a² + b² + c²)
for the point (x₁, y₁, z₁) and the plane ax + by + cz + d = 0; the numerator is the plane’s equation with the point put in
Worked examples
Planes x + y + z = 6 and 2x − y + z = 3. d = (1, 1, 1) × (2, −1, 1) = (1 × 1 − 1 × (−1), −(1 × 1 − 1 × 2), 1 × (−1) − 1 × 2) = (2, 1, −3). For z = 0: x + y = 6 and 2x − y = 3 give x = 3, y = 3. The line is r = (3, 3, 0) + t(2, 1, −3). Check t = 1: (5, 4, −3) gives 5 + 4 − 3 = 6 and 10 − 4 − 3 = 3.
Distance from the origin to 2x − y + 2z = 12, that is 2x − y + 2z − 12 = 0: |−12|/√(4 + 1 + 4) = 12/3 = 4.
Distance from (2, −1, 3) to 2x + 3y + 6z = 5: the numerator is |4 − 3 + 18 − 5| = 14 and √(4 + 9 + 36) = 7, so the distance is 2.
In vector form. The plane (2, −2, 1) · r = 4 contains B(1, 0, 2) (2 + 0 + 2 = 4). For A(5, 1, −3): n · A = 10 − 2 − 3 = 5, so the distance is |5 − 4|/3 = 1/3.
Class activity

Measure to a wall. Choose a wall as the plane and a point in the room. Find the shortest distance with a tape measure held perpendicular to the wall. Does the formula give the same, if you set the wall as x = 0 and the point as (x₁, y₁, z₁)? (The answer is |x₁|, because here n = (1, 0, 0).)

Practice
1
Find the line of intersection of x + 2y − z = 3 and 2x − y + z = 0.
2
Find the distance from (1, 2, 1) to the plane 3x − 4y + 12z + 6 = 0.
3
Find the distance from (3, −2, 1) to x + 2y − 2z + 1 = 0.
4
Find the distance from the origin to the plane through (2, 1, −1) with normal vector (2, −1, −2).
5
Find the line of intersection of 2x + y − z = 1 and x − y + 2z = 5.