☰ Year plan · Pure Mathematics

Term 1 review: binomial expansions, complex numbers, and vectors, lines and planes

Week 15 · Term 1 review · 3 lessons
3

Review: vectors in space, lines and planes

Pure Mathematics Sec 3 Term 1, Units 1–4, Review
GoalRevise vectors in space, the scalar and vector products, lines and planes, and choose the right tool for each question.
New words
normal vector · a vector perpendicular to a plane; it gives the plane’s equation and its distancesdirection vector · a vector parallel to a line; it gives the line’s equations and its anglesskew lines · lines that are not parallel and never meet
Explanation

Weeks 8 to 14 start with points and vectors and end with lines and planes, and the same few ideas return. Vectors: the norm, the unit vector A/‖A‖ and the direction cosines. The scalar product A · B gives angles and the test for perpendicular vectors; the vector product A × B gives a vector perpendicular to both, areas, and parallel vectors when it is zero; the scalar triple product gives volumes and coplanar vectors. A line is a point and a direction vector, and a plane is a point and a normal vector. For two lines use the angle formula with their directions, then equate coordinates with different parameters to decide between meeting and skew. For a plane, find a normal: given directly, from AB × AC, or from d₁ × d₂. Distances use a perpendicular: point to line through the foot of the perpendicular or ‖AB × d‖/‖d‖, and point to plane through |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²). A line meets a plane after putting its general point into the equation; two planes meet in a line with direction n₁ × n₂. Always check the answer in the original data.

tasktool
norm, unit vector, direction cosines‖A‖ = √(x² + y² + z²); A/‖A‖ = (ℓ, m, n); ℓ² + m² + n² = 1
angle between vectors; perpendicularcos θ = A · B/(‖A‖‖B‖); A · B = 0
a vector perpendicular to A and B; areasA × B (3 × 3 determinant); triangle area = ½‖A × B‖
volume; coplanar vectors|A · (B × C)|; A · (B × C) = 0
a liner = A + t d; x = x₁ + at …; (x − x₁)/a = (y − y₁)/b = (z − z₁)/c
two linescos θ = |d₁ · d₂|/(‖d₁‖‖d₂‖); equate coordinates with s and t: meet or skew
a planen · r = n · A; ax + by + cz + d = 0; n = AB × AC, or d₁ × d₂
distancespoint–line: ‖AB × d‖/‖d‖; point–plane: |ax₁ + by₁ + cz₁ + d|/√(a² + b² + c²)
planes and lines meetingline–plane: (n · d)t = k − n · A; planes: d = n₁ × n₂; angle between planes from n₁ · n₂
Worked examples
Triangle P(1, 0, 2), Q(3, 2, 3), R(0, 3, 3). PQ = (2, 2, 1) and PR = (−1, 3, 1). The angle at P: PQ · PR = 5, norms 3 and √11, so cos P = 5/(3√11) and P ≈ 59.8°. PQ × PR = (2 × 1 − 1 × 3, −(2 × 1 − 1 × (−1)), 2 × 3 − 2 × (−1)) = (−1, −3, 8) of size √74, so the area is √74/2 ≈ 4.30. The plane through the three points has normal (−1, −3, 8): −x − 3y + 8z = −1 + 16 = 15, that is x + 3y − 8z = −15. Check Q: 3 + 6 − 24 = −15 and R: 0 + 9 − 24 = −15.
Line and plane. The line through A(0, 1, −2) with direction (2, −1, 2) meets the plane 3x + 2y − z = 10: its point is (2t, 1 − t, −2 + 2t), so 6t + 2 − 2t + 2 − 2t = 2t + 4 = 10 and t = 3. The point is (6, −2, 4). The parallel plane 3x + 2y − z = −4 is at distance |10 − (−4)|/√14 = 14/√14 = √14 ≈ 3.74 from it.
Two lines and a plane. r₁ = (1, 2, 3) + s(1, −1, 2) and r₂ = (5, 1, 6) + t(2, 1, −1) meet at s = 2, t = −1, at the point (3, 0, 7). A normal of the plane containing both is (1, −1, 2) × (2, 1, −1) = ((−1)(−1) − 2 × 1, −(1 × (−1) − 2 × 2), 1 × 1 − (−1) × 2) = (−1, 5, 3). The plane is −x + 5y + 3z = −1 + 10 + 9 = 18, that is x − 5y − 3z = −18. Check (5, 1, 6): 5 − 5 − 18 = −18.
Class activity

Plan a question trail. Pick the data ‘three points A, B, C’. List five different questions you can ask about them, each with its tool, from this unit (angle, area, plane, distance from the origin to the plane, a line through two points). Solve one with numbers of your choice.

Practice
1
Find the direction cosines of the vector from P(1, −2, 3) to Q(5, 2, 5).
2
Find a unit vector perpendicular to both (2, 0, 1) and (1, 3, −1).
3
Write the line through (3, 1, −2) and (1, 4, −2) in parametric and Cartesian form.
4
Find the distance from A(−3, −1, −1) to the line r = (0, 1, 0) + t(2, 2, −1).
5
Find the plane through (1, −1, 2) with normal vector (3, −2, 6), its distance from the origin, and the distance of (4, 1, 0) from it.