Vectors

Further Maths · Further Pure Mathematics 1. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.

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The idea

This topic is the geometry of lines and planes in three dimensions. A plane is fixed by one point on it and its normal, a vector perpendicular to the plane. In ax + by + cz = d the normal is the vector (a, b, c).

The vector product a × b gives a vector perpendicular to both a and b. Use it to find the normal of a plane from two directions in the plane, the direction of the line where two planes meet, and the direction of the common perpendicular of two skew lines.

The scalar product is used for angles and for testing whether a line is parallel to a plane. Distances come from projecting onto a normal.

Rules to remember

Plane: r·n = p is the same as ax + by + cz = d with n = (a, b, c) and d = a·n for any point a on the plane. For r = a + λb + μc the normal is b × c
a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁), and |a × b| = |a||b| sin θ
Line with direction d and plane with normal n: if d·n ≠ 0 they meet at one point; if d·n = 0 the line is parallel to the plane, or lies in it if one of its points is on the plane
Angle between a line and a plane: sin φ = |d·n|/(|d||n|). Angle between two planes: cos φ = |n₁·n₂|/(|n₁||n₂|)
Distance from (x₁, y₁, z₁) to ax + by + cz = d is |ax₁ + by₁ + cz₁ − d|/√(a² + b² + c²)
The line of intersection of two planes has direction n₁ × n₂
Skew lines r = a + sb and r = c + td: common perpendicular has direction b × d, and shortest distance = |(c − a)·(b × d)|/|b × d|

Common mistake

For the angle between a line and a plane students use cos, which gives the angle between the line and the normal. Use sin φ = |d·n|/(|d||n|), or subtract the cos answer from 90°.

Worked example

Find the shortest distance between the skew lines r = (1, 0, 2) + s(1, 1, 0) and r = (0, 3, 1) + t(0, 1, 1).

b × d = (1, 1, 0) × (0, 1, 1) = (1 × 1 − 0 × 1, 0 × 0 − 1 × 1, 1 × 1 − 1 × 0) = (1, −1, 1), and |b × d| = √3.
c − a = (0, 3, 1) − (1, 0, 2) = (−1, 3, −1).
(c − a)·(b × d) = (−1)(1) + (3)(−1) + (−1)(1) = −5.
Distance = |−5|/√3 = 5/√3 = 5√3/3.

Answer: 5√3/3 (about 2.89)

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