☰ Year plan · Pure Mathematics

Perpendiculars to a line, distances from a line and the equation of a plane

Week 13 Β· A line meeting a given line at right angles, the distance from a point to a line, the equation of a plane in vector, standard and general form, a plane through three points or containing two lines Β· 3 lessons
1

A line through a point meeting a given line at right angles, and the distance from a point to a line

Pure Mathematics Sec 3 Term 1, Unit 4, Lessons 1–2
GoalFind the foot of the perpendicular from a point to a line, the line through the point that meets the line at right angles, and the distance.
New words
foot of the perpendicular Β· the point C of the line closest to A; AC is perpendicular to the direction of the linedistance from a point to a line Β· the length AC of the perpendicular segment, the shortest distanceperpendicular line through A Β· the line through A and the foot C, written r = A + sΒ·AC
Explanation

Let A be a point and β„“ the line r = B + t d. Of all points of β„“, one is closest to A: the foot of the perpendicular, C. The segment AC is perpendicular to β„“, so AC Β· d = 0. Method 1. (1) Write a general point of β„“ as C = B + t d, with t still unknown. (2) Form AC = C βˆ’ A, whose components contain t. (3) Put AC Β· d = 0: this is a linear equation in t. (4) Solve it, put t back to get C and AC. Then the distance from A to β„“ is β€–ACβ€–, and the line through A that meets β„“ at right angles is r = A + sΒ·AC (in space there is just one line through A that meets β„“ at right angles; many other lines through A are perpendicular to β„“ in direction but never meet it). Method 2 uses the vector product and needs no unknown. Take any point B of β„“. The parallelogram on BA and d has area β€–BA Γ— dβ€–, and its base is β€–dβ€–, so its height, the distance from A to β„“, is β€–BA Γ— dβ€–/β€–dβ€–. This gives the distance but not the foot. Check any answer by two tests: AC Β· d must be 0 and C must be on β„“.

β„“AC
Worked examples
Find the foot and the distance from A(2, βˆ’3, 0) to the line r = (1, 0, 2) + t(1, 2, 2). A general point is C = (1 + t, 2t, 2 + 2t), so AC = (t βˆ’ 1, 2t + 3, 2t + 2). AC Β· d = (t βˆ’ 1) + 2(2t + 3) + 2(2t + 2) = 9t + 9 = 0, so t = βˆ’1. Then C = (0, βˆ’2, 0) and AC = (βˆ’2, 1, 0). Check: AC Β· d = βˆ’2 + 2 + 0 = 0. The distance is √(4 + 1) = √5, and the perpendicular line is r = (2, βˆ’3, 0) + s(βˆ’2, 1, 0).
The same distance by the vector product. Take B = (1, 0, 2), so BA = (1, βˆ’3, βˆ’2). BA Γ— d = (βˆ’3 Γ— 2 βˆ’ (βˆ’2) Γ— 2, βˆ’(1 Γ— 2 βˆ’ (βˆ’2) Γ— 1), 1 Γ— 2 βˆ’ (βˆ’3) Γ— 1) = (βˆ’2, βˆ’4, 5), of size √45. Dividing by β€–dβ€– = 3 gives √45/3 = √5. The two methods agree.
Distance from P(βˆ’2, 1, 4) to the line r = (1, 1, 1) + t(2, 2, βˆ’1) by the vector product. BP = (βˆ’3, 0, 3). BP Γ— d = (0 Γ— (βˆ’1) βˆ’ 3 Γ— 2, βˆ’((βˆ’3)(βˆ’1) βˆ’ 3 Γ— 2), (βˆ’3) Γ— 2 βˆ’ 0 Γ— 2) = (βˆ’6, 3, βˆ’6), of size 9. Divide by β€–dβ€– = 3: the distance is 3.
If A is on the line, the distance is 0 and the foot is A itself, so AC is the zero vector and the method has nothing to find. Before you start, check whether some value of t gives A.
Class activity

Shortest path. Put a pencil on a table as the line and a coin as the point A. Use a ruler to find the shortest distance from the coin to the pencil and notice where the ruler touches the pencil: that is the foot, and the ruler is perpendicular to the pencil.

Practice
1
Find the foot of the perpendicular from A(4, 1, 4) to the line r = (1, 1, 1) + t(1, 2, 2), and the distance.
2
Find the line through A(5, βˆ’2, βˆ’2) that meets r = (2, βˆ’1, 3) + t(2, 2, βˆ’1) at right angles.
3
Use the vector product to find the distance from A(4, βˆ’2, 5) to the line through the origin with direction (1, βˆ’2, 2).
4
Find the distance from the point (5, 7, 2) to the x-axis, the line through the origin with direction (1, 0, 0).
5
Why is AC Β· d = 0 the right condition for the foot?