Question Simulator

AS & A Level Further Mathematics (9231)

Vectors Banker

9 questions · about 5% of the marks we indexed · examined in 9 of 9 sessions.

Questions by year

2026

  • 2026 May/June · Paper 12 · Q5 (a, b)11 marksAngles between lines & planes · Distances (point, plane, skew lines)

2025

  • 2025 Oct/Nov · Paper 12 · Q6 (a, b, c)14 marksAngles between lines & planes · Distances (point, plane, skew lines)
  • 2025 May/June · Paper 12 · Q6 (a, b, c)16 marksAngles between lines & planes · Distances (point, plane, skew lines) · Equations of planes

2024

  • 2024 Oct/Nov · Paper 12 · Q2 (a, b)7 marksAngles between lines & planes · Equations of planes
  • 2024 May/June · Paper 12 · Q5 (a, b, c)12 marksDistances (point, plane, skew lines) · Equations of planes

2023

  • 2023 Oct/Nov · Paper 12 · Q5 (a, b, c, d)13 marksDistances (point, plane, skew lines) · Equations of planes · Lines & planes: intersection, parallel
  • 2023 May/June · Paper 12 · Q7 (a, b, c, d)14 marksAngles between lines & planes · Distances (point, plane, skew lines) · Equations of planes · Lines & planes: intersection, parallel

2022

  • 2022 Oct/Nov · Paper 12 · Q6 (a, b(i), b(ii), c)15 marksAngles between lines & planes · Distances (point, plane, skew lines) · Equations of planes
  • 2022 May/June · Paper 12 · Q2 (a, b, c)10 marksDistances (point, plane, skew lines) · Equations of planes · Lines & planes: intersection, parallel

References only, not question text. Open Official paper ↗ where Cambridge publishes the paper, or see the official Cambridge past-papers page; for other papers, ask your school. “Paper 42” means Paper 4, variant 2.

Practice questions

Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.

Q1Distances (point, plane, skew lines) · Core · Paper 1 style · 5 marks
  1. (a) Find the perpendicular distance from the point P(3, −1, 4) to the plane Π with equation 2x − 2y + z = 5. [2]
  2. (b) Find the coordinates of the foot of the perpendicular from P to Π. [3]
Show answer
  1. (a) 7/3
    M1: |2(3) − 2(−1) + 4 − 5| / √(4 + 4 + 1)A1: 7/3
  2. (b) (13/9, 5/9, 29/9)
    M1: point on the normal through P: (3 + 2t, −1 − 2t, 4 + t)M1: substitute into Π: 9t + 7 = 0, so t = −7/9A1: foot = (13/9, 5/9, 29/9)
Q2Distances (point, plane, skew lines) · Core · Paper 1 style · 4 marks
  1. (a) Show that the planes Π₁: x + 2y − 2z = 4 and Π₂: −2x − 4y + 4z = 22 are parallel. [1]
  2. (b) Find the distance between Π₁ and Π₂. [3]
Show answer
  1. (a) The normal of Π₂ is −2 times the normal of Π₁, so the normals are parallel and so are the planes.
    B1: normal of Π₂ = −2(i + 2j − 2k)
  2. (b) 5
    M1: write Π₂ as x + 2y − 2z = −11M1: distances from origin 4/3 and −11/3 (same unit normal, opposite sides)A1: (4 − (−11))/3 = 5
Q3Distances (point, plane, skew lines) · Core · Paper 1 style · 4 marks
  1. (a) The plane Π has equation r · (6i + 2j − 3k) = 14. Find the perpendicular distance of Π from the origin. [1]
  2. (b) Find the perpendicular distance from the point A(1, 4, −2) to Π, and state whether A is on the same side of Π as the origin. [3]
Show answer
  1. (a) 2
    B1: |n| = √(36 + 4 + 9) = 7, distance 14/7 = 2
  2. (b) 6/7; A and the origin are on opposite sides of Π
    M1: |6(1) + 2(4) − 3(−2) − 14| / 7A1: 6/7B1: r·n = 20 at A (> 14) but 0 at O (< 14), so A and O are on opposite sides
Q4Distances (point, plane, skew lines) · Standard · Paper 1 style · 5 marks
  1. (a) The lines l₁ and l₂ have equationsl₁: r = i + 2j − k + λ(2i + j + 3k)l₂: r = 4i − j + 2k + μ(i − 2j + k)Find a vector that is perpendicular to both l₁ and l₂. [2]
  2. (b) Find the shortest distance between l₁ and l₂, giving your answer in exact form. [3]
Show answer
  1. (a) 7i + j − 5k (or any multiple)
    M1: cross product of the directions (2, 1, 3) × (1, −2, 1)A1: (7, 1, −5)
  2. (b) √3/5 (≈ 0.346)
    M1: vector joining the lines: (3, −3, 3)M1: |(3, −3, 3) · (7, 1, −5)| / √75A1: 3/√75 = √3/5
Q5Distances (point, plane, skew lines) · Standard · Paper 1 style · 6 marks
  1. (a) The line l has equation r = i + 2k + t(2i + j − 2k) and the point P has coordinates (5, 1, −2).Find the coordinates of the point N on l such that PN is perpendicular to l. [4]
  2. (b) Hence find the perpendicular distance from P to l. [2]
Show answer
  1. (a) (43/9, 17/9, −16/9)
    M1: PN = (1 + 2t − 5, t − 1, 2 − 2t + 2) = (2t − 4, t − 1, 4 − 2t)M1: PN · (2, 1, −2) = 0A1: 9t − 17 = 0, t = 17/9A1: N = (43/9, 17/9, −16/9)
  2. (b) 2√2/3 ≈ 0.943
    M1: |PN|² = 8/9A1: |PN| = 0.943
Q6Distances (point, plane, skew lines) · Standard · Paper 1 style · 6 marks
  1. (a) The points A, B and C have coordinates A(1, 2, 0), B(3, −1, 2) and C(0, 4, 1). Find an equation of the plane ABC in the form ax + by + cz = d. [4]
  2. (b) The point D has coordinates (6, 5, −3). Find the perpendicular distance from D to the plane ABC. [2]
Show answer
  1. (a) 7x + 4y − z = 15
    M1: AB = (2, −3, 2), AC = (−1, 2, 1)M1: AB × ACA1: normal (7, 4, −1) (or multiple)A1: 7x + 4y − z = 15
  2. (b) 25√66/33 ≈ 6.15
    M1: |7(6) + 4(5) − (−3) − 15| / √66A1: 50/√66
Q7Distances (point, plane, skew lines) · Standard · Paper 1 style · 5 marks
  1. (a) The line l has equation r = i + 2j + 3k + t(i − j + 2k) and the plane Π has equation 2x + y − 2z = 1. Show that l is not parallel to Π. [1]
  2. (b) Find the coordinates of the two points on l whose perpendicular distance from Π is 3. [4]
Show answer
  1. (a) Direction · normal = 2 − 1 − 4 = −3 ≠ 0, so l is not parallel to Π.
    B1: (1, −1, 2)·(2, 1, −2) = −3 ≠ 0
  2. (b) (−3, 6, −5) and (3, 0, 7)
    M1: distance of (1 + t, 2 − t, 3 + 2t) from Π: |2(1 + t) + (2 − t) − 2(3 + 2t) − 1| / 3A1: |−3 − 3t| / 3 = |1 + t|M1: |1 + t| = 3, t = 2 or t = −4A1: (3, 0, 7) and (−3, 6, −5)
Q8Distances (point, plane, skew lines) · Standard · Paper 1 style · 6 marks
  1. (a) In a textile factory in Faisalabad, two straight pipes are modelled as lines. Coordinates are in metres, with the z-axis vertical.Pipe A passes through (0, 0, 3) in the direction 4i + 3j.Pipe B passes through (2, 10, 1) in the direction −j + k.Find the shortest distance between the two pipes, correct to 3 significant figures. [5]
  2. (b) State one assumption made by modelling each pipe as a line. [1]
Show answer
  1. (a) 4.06 m
    M1: directions (4, 3, 0) and (0, −1, 1); common perpendicular = (4, 3, 0) × (0, −1, 1)A1: (3, −4, −4)M1: joining vector (2, 10, −2)M1: |(2, 10, −2)·(3, −4, −4)| / √41A1: 26/√41 = 4.06 m
  2. (b) The thickness (radius) of each pipe is ignored.
    B1: pipes have negligible width / are perfectly straight
Q9Distances (point, plane, skew lines) · Challenge · Paper 1 style · 8 marks
  1. (a) The lines l₁: r = 2i + k + λ(i − j + 2k) and l₂: r = 3j − 2k + μ(2i + j − k) are skew. The point P on l₁ and the point Q on l₂ are such that PQ is perpendicular to both lines.Find the coordinates of P and Q. [6]
  2. (b) Find the shortest distance between l₁ and l₂. [2]
Show answer
  1. (a) P(6/35, 64/35, −93/35), Q(−2/35, 104/35, −69/35)
    M1: general points; QP = P − Q = (2 + λ − 2μ, −3 − λ − μ, 3 + 2λ + μ)M1: QP · (1, −1, 2) = 0 and QP · (2, 1, −1) = 0A1: 6λ + μ = −11 and λ + 6μ = −2 (or equivalent)M1: solve simultaneouslyA1: λ = −64/35, μ = −1/35A1: P = (6/35, 64/35, −93/35), Q = (−2/35, 104/35, −69/35)
  2. (b) 8√35/35 ≈ 1.35
    M1: |PQ| = √(64/35), or |(a₂ − a₁)·(b₁ × b₂)|/|b₁ × b₂| with b₁ × b₂ = (−1, 5, 3)A1: 8√35/35
Q10Distances (point, plane, skew lines) · Challenge · Paper 1 style · 8 marks
  1. (a) A flat mirror in a physics laboratory lies in the plane Π: 2x − y + 2z = 3 (units are metres). A small light source is at the point S(5, 2, 6).Find the perpendicular distance from S to Π. [2]
  2. (b) Find the coordinates of the image S′ of S in the mirror (the reflection of S in Π). [3]
  3. (c) A camera is at C(1, −3, 2). Find the distance travelled by light that goes from S to the mirror and is reflected to C, using the shortest such path. [3]
Show answer
  1. (a) 17/3 m
    M1: |10 − 2 + 12 − 3| / √9A1: 17/3 m
  2. (b) (−23/9, 52/9, −14/9)
    M1: line through S along the normal: (5 + 2t, 2 − t, 6 + 2t)M1: meets Π when 9t + 17 = 0, t = −17/9; foot F = (11/9, 35/9, 20/9)A1: S′ = S + 2(F − S) = (−23/9, 52/9, −14/9)
  3. (c) 10.1 m
    B1: C is on the same side of Π as S (2(1) − (−3) + 2(2) − 3 = 6 > 0 and 2(5) − 2 + 2(6) − 3 = 17 > 0)M1: shortest path length = |S′C| = √((−23/9 − 1)² + (52/9 + 3)² + (−14/9 − 2)²)A1: 10.1 m
Q11Distances (point, plane, skew lines) · Challenge · Paper 1 style · 7 marks
  1. (a) The lines l₁ and l₂ have equationsl₁: r = i + 2k + s(i + j)l₂: r = pj + k + t(j + k)where p is a constant. Find a vector perpendicular to both lines. [2]
  2. (b) The shortest distance between l₁ and l₂ is 2√3. Find the two possible values of p. [4]
  3. (c) Explain why the lines are skew for both values of p. [1]
Show answer
  1. (a) i − j + k (or any multiple)
    M1: (1, 1, 0) × (0, 1, 1)A1: (1, −1, 1)
  2. (b) p = 4 or p = −8
    M1: vector between the lines (−1, p, −1)M1: |(−1, p, −1)·(1, −1, 1)| / √3 = 2√3A1: |−p − 2| = 6A1: p = 4 or p = −8
  3. (c) The directions are not parallel and the shortest distance is 2√3 ≠ 0, so the lines do not meet.
    B1: non-parallel directions and non-zero distance
Q12Equations of planes · Core · Paper 1 style · 3 marks
  1. (a) Find, in the form ax + by + cz = d, an equation of the plane through the point A(2, −3, 1) that is perpendicular to the vector 4i + j − 5k. [2]
  2. (b) Determine whether the point B(1, 6, 2) lies in this plane. [1]
Show answer
  1. (a) 4x + y − 5z = 0
    M1: d = 4(2) + 1(−3) − 5(1)A1: 4x + y − 5z = 0
  2. (b) Yes: 4 + 6 − 10 = 0
    B1: substitute: 4(1) + 6 − 5(2) = 0 = d, so B lies in the plane
Q13Equations of planes · Core · Paper 1 style · 5 marks
  1. (a) The points A(2, 1, −1), B(0, 3, 1) and C(1, −2, 4) lie in the plane Π. Find a vector normal to Π. [3]
  2. (b) Hence find the Cartesian equation of Π. [2]
Show answer
  1. (a) 2i + j + k (or any multiple)
    M1: AB = (−2, 2, 2), AC = (−1, −3, 5)M1: AB × AC = (16, 8, 8)A1: normal (2, 1, 1)
  2. (b) 2x + y + z = 4
    M1: substitute AA1: 2x + y + z = 4
Q14Equations of planes · Core · Paper 1 style · 4 marks
  1. (a) The plane Π has vector equation r = i + 2j + s(i + 3k) + t(2j − k). Find the equation of Π in the form ax + by + cz = d. [3]
  2. (b) Find the point where Π meets the z-axis. [1]
Show answer
  1. (a) 6x − y − 2z = 4
    M1: n = (1, 0, 3) × (0, 2, −1) = (−6, 1, 2)A1: normal (6, −1, −2)A1: 6x − y − 2z = 4
  2. (b) (0, 0, −2)
    B1: x = y = 0 gives −2z = 4
Q15Equations of planes · Standard · Paper 1 style · 6 marks
  1. (a) The line l has equation r = i + 3k + λ(2i − j + k) and the point P has coordinates (4, 2, −1). Show that P does not lie on l. [2]
  2. (b) Find an equation of the plane that contains l and P, in the form ax + by + cz = d. [4]
Show answer
  1. (a) No single λ fits all three coordinates, so P is not on l.
    M1: 1 + 2λ = 4 gives λ = 3/2; then y = −3/2 ≠ 2A1: contradiction, so P is not on l
  2. (b) 2x + 11y + 7z = 23
    B1: vector from (1, 0, 3) to P: (3, 2, −4)M1: normal = (2, −1, 1) × (3, 2, −4)A1: (2, 11, 7) (or multiple)A1: 2x + 11y + 7z = 23
Q16Equations of planes · Standard · Paper 1 style · 6 marks
  1. (a) The lines l₁: r = 3i − j + 2k + λ(i + 2j − k) and l₂: r = −i + 6j − 9k + μ(2i − j + 3k) intersect. Find the coordinates of their point of intersection. [3]
  2. (b) Find the Cartesian equation of the plane containing l₁ and l₂. [3]
Show answer
  1. (a) (5, 3, 0)
    M1: equate components: 3 + λ = −1 + 2μ, −1 + 2λ = 6 − μM1: solve: λ = 2, μ = 3A1: check third component; point (5, 3, 0)
  2. (b) x − y − z = 2
    M1: normal = (1, 2, −1) × (2, −1, 3)A1: (5, −5, −5), i.e. normal (1, −1, −1)A1: x − y − z = 2
Q17Equations of planes · Standard · Paper 1 style · 6 marks
  1. (a) The line l has equation r = 2j − k + λ(3i + j + 2k) and the line m has direction vector i − 2j.Write down a vector equation, in the form r = a + λb + μc, of the plane Π that contains l and is parallel to m. [1]
  2. (b) Find the Cartesian equation of Π. [3]
  3. (c) The line m passes through the point (2, 1, 5). Find the distance between m and Π. [2]
Show answer
  1. (a) r = 2j − k + λ(3i + j + 2k) + μ(i − 2j)
    B1
  2. (b) 4x + 2y − 7z = 11
    M1: n = (3, 1, 2) × (1, −2, 0)A1: (4, 2, −7)A1: 4x + 2y − 7z = 11
  3. (c) 12√69/23 ≈ 4.33
    M1: |−25 − 11| / √69A1: 4.33
Q18Equations of planes · Challenge · Paper 1 style · 8 marks
  1. (a) The lines l₁: r = i + j + λ(2i + j − k) and l₂: r = 3i − 2j + 4k + μ(−4i − 2j + 2k) are parallel. Explain why they are parallel and show that they are distinct. [2]
  2. (b) Find the Cartesian equation of the plane containing both lines. [3]
  3. (c) Find the distance between the two lines. [3]
Show answer
  1. (a) Directions are multiples ((−4, −2, 2) = −2(2, 1, −1)); (3, −2, 4) does not lie on l₁, so they are distinct.
    B1: direction of l₂ = −2 × direction of l₁B1: 1 + 2λ = 3 gives λ = 1, then y = 2 ≠ −2, so (3, −2, 4) is not on l₁
  2. (b) x − 10y − 8z = −9
    M1: use b and a₂ − a₁ = (2, −3, 4)M1: (2, 1, −1) × (2, −3, 4) = (1, −10, −8)A1: x − 10y − 8z = −9
  3. (c) 5.24
    M1: distance = |(a₂ − a₁) × b| / |b|M1: |(1, −10, −8)| / √6 = √165/√6A1: 5.24
Q19Equations of planes · Challenge · Paper 1 style · 7 marks
  1. (a) The planes Π₁: x + y − z = 2 and Π₂: 2x − y + 3z = 1 meet in the line l. Find a vector equation of l. [4]
  2. (b) Find the Cartesian equation of the plane Π₃ which contains l and passes through the point A(2, 0, 4). [3]
Show answer
  1. (a) r = i + j + t(2i − 5j − 3k) (or equivalent)
    M1: direction = (1, 1, −1) × (2, −1, 3)A1: (2, −5, −3)M1: find a common point, e.g. z = 0 gives x + y = 2, 2x − y = 1A1: r = i + j + t(2i − 5j − 3k)
  2. (b) 23x + 11y − 3z = 34
    M1: normal = (direction of l) × (A − (1, 1, 0)) = (2, −5, −3) × (1, −1, 4), or use Π₁ + kΠ₂A1: normal (23, 11, −3)A1: 23x + 11y − 3z = 34
Q20Angles between lines & planes · Core · Paper 1 style · 3 marks
  1. (a) Find the acute angle between the planes 2x − y + 2z = 7 and x + 4y − z = 3, giving your answer in degrees to 1 decimal place. [3]
Show answer
  1. (a) 71.7°
    M1: cos θ = |(2, −1, 2)·(1, 4, −1)| / (3 × √18)A1: |2 − 4 − 2| / (3√18) = 4/(9√2)A1: θ = 71.7°
Q21Angles between lines & planes · Core · Paper 1 style · 3 marks
  1. (a) Find the acute angle between the line r = 2i − k + λ(3i + j − 2k) and the plane x − 2y + 2z = 5. Give your answer in degrees to 1 decimal place. [3]
Show answer
  1. (a) 15.5°
    M1: sin θ = |b·n| / (|b||n|) (or find the angle with the normal and subtract from 90°)A1: |3 − 2 − 4| / (√14 × 3) = 3/(3√14)A1: θ = 15.5°
Q22Angles between lines & planes · Standard · Paper 1 style · 5 marks
  1. (a) The lines l₁ and l₂ have equationsl₁: r = i − 2j + 3k + λ(2i + 2j + k)l₂: r = 4j − k + μ(4i − j + k)Show that l₁ and l₂ are skew. [3]
  2. (b) Find the acute angle between the directions of l₁ and l₂. [2]
Show answer
  1. (a) Not parallel (directions not multiples) and the equations for a common point have no solution, so skew.
    B1: directions not parallelM1: 1 + 2λ = 4μ, −2 + 2λ = 4 − μ; solve λ = 23/10, μ = 7/5A1: z-components then differ, so no intersection (or (a₂ − a₁)·(b₁ × b₂) ≠ 0)
  2. (b) 56.6°
    M1: cos θ = |8 − 2 + 1| / (3 × √18) = 7/(3√18)A1: 56.6°
Q23Angles between lines & planes · Standard · Paper 1 style · 6 marks
  1. (a) The line l: r = 2i + j − 3k + t(i + 2j + 2k) meets the plane Π: 3x − y + z = 4 at the point X. Find the coordinates of X. [3]
  2. (b) Find the acute angle between l and Π. [3]
Show answer
  1. (a) (8/3, 7/3, −5/3)
    M1: substitute (2 + t, 1 + 2t, −3 + 2t)M1: 3(2 + t) − (1 + 2t) + (−3 + 2t) = 4, so 3t + 2 = 4A1: t = 2/3, X = (8/3, 7/3, −5/3)
  2. (b) 17.5°
    M1: sin θ = |(1, 2, 2)·(3, −1, 1)| / (3√11)A1: = 3/(3√11) = 1/√11A1: 17.5°
Q24Angles between lines & planes · Standard · Paper 1 style · 6 marks
  1. (a) A sloping roof panel on a house in Murree is modelled as part of a plane through the points A(0, 0, 3), B(8, 0, 3) and C(4, 5, 6), in metres, where the z-axis is vertical and the ground is z = 0.Find an equation of the plane of the roof panel in the form ax + by + cz = d. [4]
  2. (b) Find the angle between the roof panel and the horizontal ground. [2]
Show answer
  1. (a) 3y − 5z = −15
    M1: AB = (8, 0, 0), AC = (4, 5, 3)M1: AB × AC = (0, −24, 40)A1: normal (0, 3, −5)A1: 3y − 5z = −15
  2. (b) 31.0°
    M1: cos θ = |(0, 3, −5)·(0, 0, 1)| / √34 = 5/√34A1: 31.0°
Q25Angles between lines & planes · Challenge · Paper 1 style · 7 marks
  1. (a) The planes Π₁: 2x + y + z = 3 and Π₂: x + ky − z = 5, where k is a constant, meet at an angle of 60°. Show that k satisfies k² − 4k + 4 = 0, and hence find k. [4]
  2. (b) For this value of k, find a vector equation of the line of intersection of Π₁ and Π₂. [3]
Show answer
  1. (a) k = 2
    M1: cos 60° = |2 + k − 1| / (√6 √(2 + k²))M1: square: 4(1 + k)² = 6(2 + k²)A1: 2k² − 8k + 8 = 0, i.e. k² − 4k + 4 = 0 (AG)A1: k = 2
  2. (b) r = (1/3)i + (7/3)j + t(−i + j + k) (or equivalent)
    M1: direction = (2, 1, 1) × (1, 2, −1) = (−3, 3, 3), i.e. (−1, 1, 1)M1: common point, e.g. z = 0: 2x + y = 3, x + 2y = 5A1: (1/3, 7/3, 0); r = (1/3)i + (7/3)j + t(−i + j + k)
Q26Angles between lines & planes · Challenge · Paper 1 style · 7 marks
  1. (a) The planes Π₁: x − y + 2z = 3 and Π₂: 3x + y − z = 4 meet in the line l. Find a vector equation of l. [4]
  2. (b) Find the acute angle between l and the plane Π₃: 2x + 2y + z = 9. [3]
Show answer
  1. (a) r = i + 4j + 3k + t(−i + 7j + 4k) (or equivalent)
    M1: direction = (1, −1, 2) × (3, 1, −1)A1: (−1, 7, 4)M1: common point, e.g. x = 1 gives −y + 2z = 2, y − z = 1A1: (1, 4, 3); r = i + 4j + 3k + t(−i + 7j + 4k)
  2. (b) 41.0°
    M1: sin θ = |(−1, 7, 4)·(2, 2, 1)| / (√66 × 3)A1: = 16/(3√66)A1: 41.0°
Q27Lines & planes: intersection, parallel · Core · Paper 1 style · 3 marks
  1. (a) Find the coordinates of the point where the line r = i − 2j + 4k + λ(2i + 3j − k) meets the plane x + y + 2z = 3. [3]
Show answer
  1. (a) (−5/3, −6, 16/3)
    M1: substitute (1 + 2λ, −2 + 3λ, 4 − λ)M1: (1 + 2λ) + (−2 + 3λ) + 2(4 − λ) = 3, so 3λ + 7 = 3A1: λ = −4/3; point (−5/3, −6, 16/3)
Q28Lines & planes: intersection, parallel · Standard · Paper 1 style · 5 marks
  1. (a) The line l has equation r = 3i + j − 2k + λ(2i + k) and the plane Π has equation x + 4y − 2z = 5. Show that l is parallel to Π and does not lie in Π. [3]
  2. (b) Find the distance between l and Π. [2]
Show answer
  1. (a) b·n = 2 + 0 − 2 = 0, so l ∥ Π; the point (3, 1, −2) gives 3 + 4 + 4 = 11 ≠ 5, so l is not in Π.
    M1: (2, 0, 1)·(1, 4, −2) = 0A1: parallelB1: (3, 1, −2) does not satisfy the plane equation (11 ≠ 5)
  2. (b) 2√21/7 ≈ 1.31
    M1: |11 − 5| / √21A1: 6/√21 = 2√21/7
Q29Lines & planes: intersection, parallel · Standard · Paper 1 style · 6 marks
  1. (a) Find a vector equation of the line of intersection of the planes x + 2y + z = 4 and 2x − y + 3z = 1. [4]
  2. (b) Find the point where this line meets the plane x + y + z = 6. [2]
Show answer
  1. (a) r = (6/5)i + (7/5)j + t(7i − j − 5k) (or equivalent)
    M1: direction = (1, 2, 1) × (2, −1, 3) = (7, −1, −5)A1M1: common point, e.g. z = 0: x + 2y = 4, 2x − y = 1A1: (6/5, 7/5, 0); r = (6/5)i + (7/5)j + t(7i − j − 5k)
  2. (b) (25, −2, −17)
    M1: substitute general point; t = 17/5A1: (25, −2, −17)
Q30Lines & planes: intersection, parallel · Challenge · Paper 1 style · 8 marks
  1. (a) The line l has equation r = i + pj + 2k + t(2i − j + k), where p is a constant, and the plane Π₁ has equation x + 3y + z = 9.Show that l is parallel to Π₁ for every value of p. [1]
  2. (b) Find the value of p for which l lies in Π₁. [2]
  3. (c) For the value of p found in (b), the plane Π₂ contains l and is perpendicular to Π₁. Find the Cartesian equation of Π₂. [3]
  4. (d) For p = 5, find the distance between l and Π₁. [2]
Show answer
  1. (a) (2, −1, 1)·(1, 3, 1) = 0
    B1: direction · normal = 2 − 3 + 1 = 0
  2. (b) p = 2
    M1: (1, p, 2) in Π₁: 1 + 3p + 2 = 9A1: p = 2
  3. (c) 4x + y − 7z = −8
    M1: normal of Π₂ is perpendicular to both (2, −1, 1) and (1, 3, 1)A1: (2, −1, 1) × (1, 3, 1) = (−4, −1, 7)A1: through (1, 2, 2): 4x + y − 7z = −8
  4. (d) 9/√11 = 9√11/11 ≈ 2.71
    M1: |1 + 15 + 2 − 9| / √11A1: 9/√11

Similar questions

Questions that test the same subtopic in different years, grouped so you can work on one skill at a time.

References only: find each question in its paper (official link where available, otherwise through your school).

Distances (point, plane, skew lines) Banker8 questions · 8/9 sessions
Equations of planes 7 questions · 7/9 sessions
Angles between lines & planes 6 questions · 6/9 sessions
Lines & planes: intersection, parallel 3 questions · 3/9 sessions

Subtopics

Our analysis of how often each subtopic was examined and how many marks it carried, 2022 to May/June 2026.

SubtopicMarks indexedSessions
Distances (point, plane, skew lines) Banker418/9
Equations of planes 327/9
Angles between lines & planes 266/9
Lines & planes: intersection, parallel 133/9

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