amazon.co.ukFurther Vectors
3D vector equations of lines, scalar and vector products, angles between lines and planes, intersections and shortest distances - Edexcel 9FM0 Core Pure 2 vectors revision with worked examples and mark-scheme logic.
🎯 Learning Objectives
- Find the vector equation of a line in three dimensions and convert it to Cartesian (symmetric) form (AO1) - Edexcel 9FM0/02 Core Pure 2; OCR MEI H645 Further Pure
- Use the scalar (dot) product to test perpendicularity and to find angles between two lines, and between a line and a plane (AO1/AO2)
- Compute the vector (cross) product and use it to find a vector perpendicular to two given vectors and the area of a triangle (AO1/AO2)
- Find the intersection of two lines and of a line and a plane, and describe the relative positions of planes (AO2/AO3)
- Find the shortest distance from a point to a line and from a point to a plane (AO2)
- Write the equation of a plane from a normal vector and a point, and find the angle between two planes (AO1/AO2)
📋 Key Definitions and Core Concepts
Vector equation of a line: r = a + t d, where a is the position vector of any point on the line and d is a direction vector; t is a scalar parameter.
Cartesian equation of a line: (x − x₁)/l = (y − y₁)/m = (z − z₁)/n for direction vector (l, m, n); obtained by eliminating the parameter t.
Scalar (dot) product: a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ; a and b are perpendicular exactly when a·b = 0.
Vector (cross) product: a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁), a vector perpendicular to both a and b with magnitude |a × b| = |a||b| sin θ.
Normal vector: A vector perpendicular to a plane; a plane with normal n passing through the point with position vector p has equation r·n = p·n.
Skew lines: Lines that are not parallel and do not intersect - they lie in different planes, so no consistent solution for the two parameters exists.
Shortest distance from a point to a line: The perpendicular distance |AP × d|/|d|, where A is any point on the line with direction d and P is the given point.
Shortest distance from a point to a plane: The perpendicular distance |n·p₀ − c|/|n|, where the plane is r·n = c and p₀ is the position vector of the point.
📌 Key Points
a·b = |a||b| cos θ; a·b = 0 means perpendicular; d = k b means parallel.
Angle between line and plane: sin α = |d·n|/(|d||n|); between two planes: cos β = |n₁·n₂|/(|n₁||n₂|).
a × b is perpendicular to both a and b; |a × b| is the area of the parallelogram they span.
Area of triangle with sides AB and AC: ½|AB × AC|.
Distance point-to-line: |AP × d|/|d|; point-to-plane: |n·p₀ − c|/|n|.
Two non-parallel planes meet in a line; its direction is the cross product of the two normals.
Lines in Three Dimensions
Vector equation. A line in 3D is described by r = a + t d, where a is the position vector of any point on the line, d is a direction vector, and t is a scalar parameter. Equivalently, the line through points A and B is r = a + t(b − a). Any non-zero multiple of a direction vector works, so there are infinitely many correct equations of the same line - examiners accept any valid point and direction. A point P lies on the line exactly when some t satisfies p = a + t d; solve one component and check the other two.
Cartesian (symmetric) form. Writing d = (l, m, n) and eliminating t gives (x − x₁)/l = (y − y₁)/m = (z − z₁)/n, where (x₁, y₁, z₁) is the point a. Converting between the two forms is routine exam material: from the symmetric form read off the point and direction; from the vector equation set components equal and eliminate t. Note the symmetric form breaks if any direction component is zero - that coordinate is then simply constant.
Worked example. The line through A(2, −1, 4) and B(5, 2, 10): direction d = AB = (3, 3, 6) = 3(1, 1, 2). A clean equation is r = (2, −1, 4) + t(1, 1, 2); the Cartesian form is (x − 2)/1 = (y + 1)/1 = (z − 4)/2.
Relative position of two lines. Given r = a₁ + t d₁ and r = a₂ + s d₂, use different parameters (t and s) and compare components. If d₁ = k d₂ the lines are parallel - then check whether a₂ − a₁ is also parallel to d₁: if yes the lines coincide, if no they never meet. If a consistent pair (t, s) solves all three component equations, the lines intersect at that point. If no consistent solution exists and the directions are not parallel, the lines are skew - the classic 3D-only category that a full geometric check must rule out before claiming intersection.
Common trap
Using the same parameter letter for both lines forces you to solve for one t fitting both lines simultaneously - almost never intended. Write r = a₁ + t d₁ and r = a₂ + s d₂, equate, solve for t and s, then verify the third component equation before concluding.
Scalar Product and Angles
Definition and uses. The scalar product of a = (a₁, a₂, a₃) and b = (b₁, b₂, b₃) is a·b = a₁b₁ + a₂b₂ + a₃b₃ = |a||b| cos θ, where θ is the angle between the vectors. It returns a scalar, not a vector. Three consequences do most of the exam work: a·b = 0 (with neither vector zero) means perpendicular; cos θ = (a·b)/(|a||b|) finds the angle; and a·a = |a|² gives lengths quickly. It is also distributive and commutative, so you can expand expressions like (a + b)·(a − b) = |a|² − |b|².
Angle between two lines. Take the angle between their direction vectors. Convention: when cos θ comes out negative the angle between the directions is obtuse, but the acute angle between the lines is 180° − θ, i.e. use |cos θ|. Worked example: a = (2, −1, 3), b = (1, 4, −2). Then a·b = 2 − 4 − 6 = −8, |a| = √14, |b| = √21, so cos θ = −8/√294 = −8/(7√6) = −4√6/21 ≈ −0.4666, giving θ ≈ 117.8°; the acute angle between the lines is ≈ 62.2°.
Angle between a line and a plane. For line direction d and plane normal n, the angle α between the line and the plane satisfies sin α = |d·n|/(|d||n|) - sine, not cosine, because α is the complement of the angle between d and n. Worked example: d = (2, −1, 2), n = (1, 1, 4): d·n = 2 − 1 + 8 = 9, |d| = 3, |n| = √18 = 3√2, so sin α = 9/(9√2) = 1/√2 and α = 45°. If d·n = 0 the line is parallel to the plane (or lies in it).
Angle between two planes. Planes have normal vectors, and the angle between planes is the angle between their normals: cos β = |n₁·n₂|/(|n₁||n₂|). Worked example: n₁ = (1, 2, 2), n₂ = (2, 1, −2): n₁·n₂ = 2 + 2 − 4 = 0, so the planes meet at 90°. Keep normal vectors to integer or simple surd components by clearing fractions early - it keeps every later cosine clean.
Exam technique
Angle questions: state which vectors you are using (directions or normals), quote the scalar product, then the cosine/sine, then the angle to 1 d.p. or in exact form. The sine-versus-cosine distinction between line-plane and plane-plane angles is a guaranteed mark-loser if muddled.
Cross Product, Intersections and Distances
The vector (cross) product. For a = (a₁, a₂, a₃), b = (b₁, b₂, b₃): a × b = (a₂b₃ − a₃b₂, a₃b₁ − a₁b₃, a₁b₂ − a₂b₁), computed componentwise or via the 3x3 determinant pattern with i, j, k in the top row. Key properties: a × b is perpendicular to both a and b; |a × b| = |a||b| sin θ is the area of the parallelogram spanned by a and b, so the triangle with sides AB and AC has area ½|AB × AC|; and a × b = −b × a (anticommutative), with a × a = 0. Worked example: a = (2, −1, 3), b = (1, 4, −2): a × b = ((−1)(−2) − 3×4, 3×1 − 2×(−2), 2×4 − (−1)×1) = (−10, 7, 9), |a × b| = √230 ≈ 15.17 (check: 294 − 64 = 230 from |a|²|b|² − (a·b)² ✓).
Planes. A plane with normal n through point p is r·n = p·n - substitute once to get the constant. Example: normal (1, 2, 2) through (2, 1, −1): r·(1, 2, 2) = 2 + 2 − 2 = 5. A plane given in Cartesian form ax + by + cz = d has normal (a, b, c).
Intersections. Line meets plane: substitute the line into r·n = c and solve for t. Worked example: r = (1, 0, 2) + t(2, 1, −1) with plane r·(1, 2, 2) = 16: ((1, 0, 2) + t(2, 1, −1))·(1, 2, 2) = 5 + 2t = 16, so t = 11/2 and the point is (12, 11/2, −7/2). Two non-parallel planes meet in a line whose direction is n₁ × n₂ (find one point on it by inspection, e.g. setting z = 0). Three planes: unique point (three consistent equations, non-zero determinant in the 3x3 sense), a common line (sheaf), or no common point (triangular prism / two parallel planes).
Shortest distances. Point P to line (A on line, direction d): |AP × d|/|d|, or find the foot of the perpendicular by projecting. Worked example: P(3, 2, 1), line r = (1, 0, 2) + t(1, 2, 2): AP = (2, 2, −1), AP × d = (6, −5, 2), so distance = √(36 + 25 + 4)/√9 = √65/3 ≈ 2.69 (alternative: foot = A + (AP·d/|d|²)d = (13/9, 8/9, 26/9), and |P − foot| = √65/3 ✓). Point p₀ to plane r·n = c: |n·p₀ − c|/|n|. Worked example: point (1, 2, 3), plane r·(2, 3, 6) = 21: |2 + 6 + 18 − 21|/7 = 5/7.
Mark-scheme habit
Distance questions award method marks for the perpendicularity argument, not just the number: state the formula you are using, define A and d (or n and c), show the substitution, then give the exact distance. Exact surd answers (√65/3, 5/7) are expected - decimals alone lose the accuracy mark.
💡 Worked Example
Exam-style: The line l₁ has equation r = (2, 1, 3) + t(1, −1, 2) and the line l₂ has equation r = (1, 2, 1) + s(−1, 3, 1).
(a) Show that l₁ and l₂ intersect and find the coordinates of the point of intersection. (5 marks)
(b) Find the acute angle between l₁ and l₂. (4 marks)
Mark-scheme logic: (a) Equating components: 2 + t = 1 − s, 1 − t = 2 + 3s, 3 + 2t = 1 + s. From the first, t = −1 − s. Substituting into the second: 1 + 1 + s = 2 + 3s, so s = 0 and t = −1. Checking the third equation: 3 + 2(−1) = 1 and 1 + 0 = 1 ✓, consistent - so the lines intersect. Substituting t = −1 into l₁: (2 − 1, 1 + 1, 3 − 2) = (1, 2, 1).
(b) Direction vectors d₁ = (1, −1, 2), d₂ = (−1, 3, 1). d₁·d₂ = −1 − 3 + 2 = −2; |d₁| = √6, |d₂| = √11. cos θ = −2/√66 < 0, so the acute angle has cos α = 2/√66 ≈ 0.2462, giving α ≈ 75.7°.
❓ Practice Questions & Mark Schemes
a = (2, −1, 3) and b = (1, 4, −2). Find a·b and the angle between a and b. (5 marks)
Answer: a·b = 2 − 4 − 6 = −8. |a| = √14, |b| = √21. cos θ = −8/(√14 × √21) = −8/√294 = −8/(7√6) = −4√6/21 ≈ −0.4666, so θ ≈ 117.8° (obtuse; the acute angle between the directions is ≈ 62.2°).
For the same vectors, find a × b and |a × b|. (4 marks)
Answer: a × b = ((−1)(−2) − 3×4, 3×1 − 2×(−2), 2×4 − (−1)×1) = (2 − 12, 3 + 4, 8 + 1) = (−10, 7, 9). |a × b| = √(100 + 49 + 81) = √230 ≈ 15.17. (Check: |a × b|² = |a|²|b|² − (a·b)² = 294 − 64 = 230 ✓.)
Find the shortest distance from the point P(3, 2, 1) to the line with equation r = (1, 0, 2) + t(1, 2, 2). (5 marks)
Answer: AP = P − A = (2, 2, −1). AP × d = (2×2 − (−1)×2, (−1)×1 − 2×2, 2×2 − 2×1) = (6, −5, 2). Distance = |AP × d|/|d| = √(36 + 25 + 4)/√9 = √65/3 ≈ 2.69. (Alternative: the foot of the perpendicular is A + (AP·d/|d|²)d = (1, 0, 2) + (4/9)(1, 2, 2) = (13/9, 8/9, 26/9), and |P − foot| = √65/3 ✓.)
Find the shortest distance from the point (1, 2, 3) to the plane with equation r·(2, 3, 6) = 21. (4 marks)
Answer: Distance = |n·p₀ − c|/|n| = |2×1 + 3×2 + 6×3 − 21|/√(4 + 9 + 36) = |5|/7 = 5/7.
The line l has equation r = (1, 0, 2) + t(2, 1, −1) and the plane Π has equation r·(1, 2, 2) = 16. Find the coordinates of the point where l meets Π. (5 marks)
Answer: Substitute the line into the plane equation: ((1, 0, 2) + t(2, 1, −1))·(1, 2, 2) = (1 + 0 + 4) + t(2 + 2 − 2) = 5 + 2t. Setting 5 + 2t = 16 gives t = 11/2. The point is (1, 0, 2) + (11/2)(2, 1, −1) = (12, 11/2, −7/2).