☰ 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
3

A plane through three points, and a plane containing two lines

Pure Mathematics Sec 3 Term 1, Unit 4, Lessons 1–2
GoalFind the equation of the plane through three points, and of the plane containing two intersecting lines.
New words
non-collinear points Β· three points not on one line; they fix exactly one planenormal from a cross product Β· n = AB Γ— AC is perpendicular to both AB and AC, so to the planecoplanar Β· lying in one plane
Explanation

Three points A, B, C that are not on one line fix exactly one plane. The vectors AB and AC lie in the plane. Their vector product n = AB Γ— AC is perpendicular to both, so it is a normal vector. If AB Γ— AC is the zero vector, the points are on one line, and then there is no single plane. Now use the last lesson: with n and the point A, write n Β· r = n Β· A. Always check that B and C satisfy the final equation. A plane may also be given by two intersecting lines: first find their point of intersection P as in Week 12. Their direction vectors d₁ and dβ‚‚ lie in the plane, so n = d₁ Γ— dβ‚‚ is a normal vector, and the plane is n Β· r = n Β· P. If the two lines are skew, no plane contains both. The table shows the two methods side by side.

then use any known point of the plane, and check the other points in the final equation
giventwo vectors in the planenormal vector n
three points A, B, CAB and ACn = AB Γ— AC
two intersecting linesd₁ and dβ‚‚n = d₁ Γ— dβ‚‚
Worked examples
Plane through A(2, 1, 0), B(0, 3, 1) and C(1, 1, 3). AB = (βˆ’2, 2, 1), AC = (βˆ’1, 0, 3). n = AB Γ— AC = (2 Γ— 3 βˆ’ 1 Γ— 0, βˆ’(βˆ’2 Γ— 3 βˆ’ 1 Γ— (βˆ’1)), βˆ’2 Γ— 0 βˆ’ 2 Γ— (βˆ’1)) = (6, 5, 2). n Β· A = 12 + 5 + 0 = 17, so 6x + 5y + 2z = 17. Check B: 0 + 15 + 2 = 17. Check C: 6 + 5 + 6 = 17.
Are (1, 2, 3), (3, 4, 5) and (5, 6, 7) on one plane in a unique way? AB = (2, 2, 2) and AC = (4, 4, 4) are parallel, so AB Γ— AC is the zero vector: the three points are on one line and many planes pass through them.
Plane containing the lines r₁ = (0, 1, 2) + s(1, 2, βˆ’1) and rβ‚‚ = (3, 4, βˆ’2) + t(βˆ’1, 1, 2). They meet at s = 2, t = 1, at the point P(2, 5, 0). n = d₁ Γ— dβ‚‚ = (2 Γ— 2 βˆ’ (βˆ’1) Γ— 1, βˆ’(1 Γ— 2 βˆ’ (βˆ’1)(βˆ’1)), 1 Γ— 1 βˆ’ 2 Γ— (βˆ’1)) = (5, βˆ’1, 3). n Β· P = 10 βˆ’ 5 + 0 = 5, so 5x βˆ’ y + 3z = 5. Check: the point (0, 1, 2) gives βˆ’1 + 6 = 5, and (3, 4, βˆ’2) gives 15 βˆ’ 4 βˆ’ 6 = 5.
Choosing the point: with the first plane, any of A, B or C gives the same right-hand side, 17. If the right-hand sides differ, something is wrong: recheck the cross product.
Class activity

Three pins and a card. Stand three pins of different heights in a lump of modelling clay and lay a stiff card on their tops: the card rests on all three, so three points fix a plane. Now add a fourth pin of any height. Does the card touch it too? (Usually not: four points are coplanar only in special cases, as in the test of Week 11.)

Practice
1
Find the plane through A(1, 0, 1), B(2, 2, 1) and C(0, 1, 3).
2
Find the plane through the origin, A(1, 2, 3) and B(2, βˆ’1, 0).
3
Does D(βˆ’1, βˆ’1, 3) lie in the plane of question 1?
4
Find the plane containing (1, 1, 0) + s(1, 0, 1) and (2, 0, 3) + t(0, 1, βˆ’2), which meet at (2, 1, 1).
5
Why does no plane contain both (0, 0, 0) + s(1, 0, 0) and (0, 1, 0) + t(0, 0, 1)?