Lesson 22: Vectors in 3D: Lines and Planes (HL) HL ONLY

In 3D, vectors stop being bookkeeping and become architecture (AA Topic 3, HL): lines and planes are the objects, and the questions are relational — do they meet, at what angle, how far apart? The tools: the vector equation of a line r = a + λd; the equation of a plane via its normal n; the scalar product for angles; and the vector (cross) product a × b — new at HL — which manufactures a vector perpendicular to both inputs, with |a × b| = |a||b| sin θ equal to the parallelogram's area.

Lines in 3D

A line through A(a) with direction d: r = a + λd, equivalently the parametric system x = a₁ + λd₁, y = a₂ + λd₂, z = a₃ + λd₃, or Cartesian symmetric form (x − a₁)/d₁ = (y − a₂)/d₂ = (z − a₃)/d₃. To test whether a point P lies on the line, find λ from one component and verify it in the other two — consistency across all three is the test.

Planes: the normal runs everything

A plane through A with normal n satisfies n · (r − a) = 0, i.e. r · n = a · n = d, i.e. in Cartesian form n₁x + n₂y + n₃z = d. The coefficients of x, y, z ARE the normal — read it straight off any Cartesian equation. To find a plane through three points, take two edge vectors and let the cross product produce the normal. Two planes are parallel iff their normals are parallel; a line is parallel to a plane iff d · n = 0.

Worked Example 1 — plane from three points. Find the Cartesian equation of the plane through A(1, 0, 2), B(3, 1, 0), C(0, 2, 1).

Solution. Edge vectors: AB = (2, 1, −2), AC = (−1, 2, −1). Normal n = AB × AC = |i j k; 2 1 −2; −1 2 −1| = ( (1)(−1) − (−2)(2), (−2)(−1) − (2)(−1), (2)(2) − (1)(−1) ) = (−1 + 4, 2 + 2, 4 + 1) = (3, 4, 5). Plane: 3x + 4y + 5z = d with d from A: 3 + 0 + 10 = 13. Equation: 3x + 4y + 5z = 13. Verify B: 9 + 4 + 0 = 13 ✓; C: 0 + 8 + 5 = 13 ✓. The cross product's component order is the arithmetic risk — verify the normal is perpendicular to *both* edge vectors (dot products 6 + 4 − 10 = 0 ✓ and −3 + 8 − 5 = 0 ✓) before proceeding.

Intersections

Line–plane: substitute the parametric line into the plane equation, solve for λ, read off the point (one solution = piercing; no solution = parallel and outside; identity = line lies in the plane). Line–line: equate components and solve two of the three equations for λ, μ; the third equation must *agree* — if not, the lines are skew (or parallel, checked via directions). Plane–plane: the intersection is a line whose direction is n₁ × n₂.

Worked Example 2 — line meets plane (past-paper style). Find the intersection of the line r = (0, 1, 2) + λ(1, 2, −1) with the plane x + 2y − z = 5.

Solution. Points on the line: (λ, 1 + 2λ, 2 − λ). Substitute: λ + 2(1 + 2λ) − (2 − λ) = 5 → λ + 2 + 4λ − 2 + λ = 5 → 6λ = 5 → λ = 5/6. Point: (5/6, 1 + 5/3, 2 − 5/6) = (5/6, 8/3, 7/6). Verify in the plane: 5/6 + 16/3 − 7/6 = 5/6 + 32/6 − 7/6 = 30/6 = 5 ✓.

Angles and distances

Angle between lines: from their direction vectors (dot product, take the acute value). Angle between a line and a plane: 90° minus the angle between the line and the normal — equivalently sin(angle) = |d · n|/(|d||n|). Angle between planes: angle between normals. Distance from point P to plane n₁x + n₂y + n₃z = d: |n₁x₀ + n₂y₀ + n₃z₀ − d| / √(n₁² + n₂² + n₃²) — the formula booklet carries it. Distance between skew lines with points a₁, a₂ and directions d₁, d₂: |(a₂ − a₁) · (d₁ × d₂)| / |d₁ × d₂| — the component of the separation along the common perpendicular.

Worked Example 3 — skew lines and their distance. Show that L₁: r = (1, 0, 0) + λ(1, 1, 0) and L₂: r = (0, 1, 3) + μ(0, 1, 1) are skew, and find the distance between them.

Solution. The directions (1, 1, 0) and (0, 1, 1) are not scalar multiples, so the lines are not parallel. Test for an intersection: L₁'s points are (1 + λ, λ, 0); L₂'s are (0, 1 + μ, 3 + μ). From the x-components, 1 + λ = 0 forces λ = −1, giving the L₁ point (0, −1, 0). On L₂ the z-component vanishes when 3 + μ = 0, i.e. μ = −3, giving the point (0, −2, 0) — a different point. No single pair (λ, μ) satisfies all three components, so the lines never meet: they are skew. (This third-component check is not a formality; it is the only thing standing between you and a false "intersection".) Distance: the common perpendicular direction is d₁ × d₂ = (1·1 − 0·1, 0·0 − 1·1, 1·1 − 1·0) = (1, −1, 1), with magnitude √3. The separation of the base points is a₂ − a₁ = (−1, 1, 3). Distance = |(−1)(1) + (1)(−1) + (3)(1)| / √3 = |1|/√3 = 1/√3 ≈ 0.577. Sanity check: the two near-miss points (0, −1, 0) on L₁ and (0, −2, 0) on L₂ differ by only 1 in the y-coordinate, so a closest distance below 1 is entirely plausible ✓.

Real-world modelling: flight paths and collision logic

Air-traffic and drone problems are line–line problems: model each craft's path as r = a + tv (t = time now, not a free parameter — the same t must work for both crafts at a collision, which is stricter than the lines merely intersecting). Closest-approach distance is the skew-line distance; time of closest approach comes from minimising the separation — a Paper 3 favourite because it fuses vectors with calculus or completing the square.

Common mistakes

• Cross product component errors — the middle component carries the sign flip (the cofactor pattern +, −, +); verify perpendicularity afterwards.
• Line–line "intersection" declared after solving only two equations — the third-component consistency check is where skew hides.
• Angle between line and plane taken as the angle with the normal — it is its complement.
• Point-to-plane distance with the wrong constant sign — the plane must be in the form n₁x + n₂y + n₃z = d with d on the correct side before substituting.
• λ treated as time or distance — λ scales with |d|; distance along the line is |λ| × |d|.
• Normal read from a plane given in a non-standard arrangement — rearrange to n₁x + n₂y + n₃z = d first.

IB Exam Experience

Question 1 [2 marks]. Find the Cartesian equation of the plane through (2, −1, 3) with normal (1, 4, −2). *Marking notes:* M1 for n · (r − a) = 0 expanded; A1: x + 4y − 2z = −8 (since 2 − 4 − 6 = −8). Sign of the constant is the mark.

Question 2 [3 marks]. The line r = (1, 2, 0) + λ(2, −1, 3) meets the plane 3x − y + 2z = 10 at P. Find P. *Marking notes:* M1 for substituting (1 + 2λ, 2 − λ, 3λ): 3(1 + 2λ) − (2 − λ) + 2(3λ) = 10 → 3 + 6λ − 2 + λ + 6λ = 10 → 13λ = 9; A1: λ = 9/13; A1: P = (31/13, 17/13, 27/13). Fractions are expected — reaching for decimals here loses exactness marks.

Question 3 [4 marks]. Find the angle between the planes x + 2y + 2z = 5 and 2x − y + 2z = 3, in degrees. *Marking notes:* M1 for reading the normals (1, 2, 2) and (2, −1, 2) off the equations; M1: dot = 2 − 2 + 4 = 4, magnitudes 3 and 3; A1: cos θ = 4/9; A1: θ ≈ 63.6° (acute value reported — plane angles are quoted in [0°, 90°]).

Extended problem (Paper 2/3 style) [10 marks]. Two drones fly in straight lines at constant speeds: A: r = (2, 0, 10) + t(3, 4, −1) and B: r = (3, 18, −2) + s(1, −2, 2), where t and s are times in seconds from launch (both launched together) and coordinates are in metres. (a) Find the point where the two flight paths cross, and the time at which each drone reaches it. [5] (b) Do the drones collide? Justify your answer precisely. [2] (c) Find the distance between the drones at time 5 seconds. [3] *Marking notes:* (a) M1 for equating components: 2 + 3t = 3 + s; 4t = 18 − 2s; 10 − t = −2 + 2s. M1 for solving two of them: from the first, s = 3t − 1; substituting into the second: 4t = 18 − 2(3t − 1) = 20 − 6t, so t = 2 and s = 5. M1 for the third-component check (z: A gives 10 − 2 = 8; B gives −2 + 2(5) = 8 ✓ — the lines genuinely intersect); A1 for the crossing point (8, 8, 8); A1 for the times: A reaches it at t = 2 s, B at s = 5 s. (b) A1: no collision; A1 for the justification — collision requires the same point at the same time; here the point is shared but the times differ by 3 seconds. (A shared point alone is the trap answer.) (c) M1: A(5) = (17, 20, 5) and B(5) = (8, 8, 8); M1: separation vector (9, 12, −3); A1: distance √234 ≈ 15.3 m. Examiner's note: parts (a) and (b) separate candidates who compute from candidates who reason — the parameters t and s are the same physical time here only because the launches were simultaneous; say so.

Exploration (IA) idea: take two real flight/drone/GPS tracks (or model two delivery drones over your neighbourhood with map coordinates), compute closest approach and its time, and evaluate the flat-earth/constant-velocity assumptions — vectors earning their keep on real data.