OxfordAQA International AS and A-level Mathematics: Vectors and vector equations of lines (9660) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow the OxfordAQA International AS and A-level Mathematics specification (9660, Version 5.2), section P2.10 (first part). The content belongs to International A-level (A2) Unit P2 and is not examined at International AS. Column vectors are written across the line: (3, −4, 12) means a column of 3, −4, 12. See the study guide and revision notes. Other links: course hub, checklist and free 10-minute diagnostics.

Questions

1. Given p = (3, −4, 12) and q = (1, 2, −2):

(a) find |p| [1] (b) find 2p − 3q as a column vector. [2]

2. The vector u = (3, −6, k) is parallel to v = (−2, m, 10). Find m and k. [3]

3. Points A and B have position vectors (2, 5, −1) and (−1, 1, 3).

(a) Find the vector AB and the distance AB. [2] (b) Find the position vector of the midpoint of AB. [2]

4. The points P, Q and R have position vectors (1, 2, 3), (4, −1, 9) and (−2, 5, −3). Show that P, Q and R are collinear, and describe where P lies in relation to Q and R. [5]

5. A is (3, −2, 1) and B is (−1, 4, 5).

(a) Find a vector equation of the line AB. [2] (b) Determine whether C (−3, 7, 4) lies on the line. [2]

6. A line has equation r = (−2, 3) + t(4, 1).

(a) Find its equation in the form y = mx + c. [2] (b) Find the coordinates of the point where it crosses the x-axis. [2]

7. Line l₁: r = (2, −1) + s(3, 2). Line l₂: r = (0, 7) + t(2, −1).

(a) Show that the lines meet, and find the point of intersection. [3] (b) Find the distance from the point of intersection to the point (2, −1). Give the answer as a surd in its simplest form. [2]

8. Line l₁: r = (2, 1, −3) + s(1, 2, 1). Line l₂: r = (5, k, −2) + t(2, −1, 0). The lines intersect. Find k and the position vector of the point of intersection. [6]

9. Show that the lines r = (1, 0, 2) + s(2, 1, −1) and r = (2, 2, 0) + t(1, −1, 2) are skew. [5]

10. Line l₁: r = (1, 2, 3) + s(2, −2, 4). Line l₂: r = (4, −1, 9) + t(−1, 1, −2). Line l₃: r = (0, 1, 1) + u(1, −1, 2).

(a) Show that l₁ and l₂ are the same line. [3] (b) Show that l₃ is parallel to l₁ but is a different line. [2]

11. A, B and C have position vectors (1, 0, 2), (4, 2, 3) and (5, 5, 7). The quadrilateral ABCD is a parallelogram.

(a) Find the position vector of D. [2] (b) Show that the diagonals AC and BD bisect each other. [2] (c) Find the exact perimeter of ABCD. [3]

12. Line l₁: r = (1, −2, 0) + s(2, 1, 2). Line l₂: r = (4, 1, 2) + t(1, −1, 2).

(a) Show that l₁ and l₂ intersect, and find the position vector of the point P where they meet. [5] (b) Find the position vectors of the two points on l₁ that are 9 units from P. [3]

Answers

1. (a) |p| = √(9 + 16 + 144) = √169 = 13 [1] (b) 2p = (6, −8, 24) and 3q = (3, 6, −6) [1]; 2p − 3q = (3, −14, 30) [1] Examiner insight: Subtract every component, and take care with the double negative in 24 − (−6). Setting the working out as two short lines makes a slip easy to spot.

2. Parallel, so v = λu. From the first component, −2 = 3λ, so λ = −2/3 [1]. Then m = −6 × (−2/3) = 4 [1] and 10 = (−2/3)k gives k = −15 [1] Examiner insight: State the scalar multiple you are using. Finding λ from the pair of components with no unknowns, then applying it, is the quickest and least error-prone route.

3. (a) AB = OB − OA = (−1 − 2, 1 − 5, 3 − (−1)) = (−3, −4, 4) [1]; |AB| = √(9 + 16 + 16) = √41 [1] (b) OM = ½(OA + OB) = ½(1, 6, 2) [1] = (½, 3, 1) [1] Examiner insight: “End minus start” gives the correct sign. Leave the distance as √41 unless a decimal is requested.

4. PQ = (4 − 1, −1 − 2, 9 − 3) = (3, −3, 6) [1]; PR = (−2 − 1, 5 − 2, −3 − 3) = (−3, 3, −6) [1]; PR = −PQ, so the vectors are parallel [1]; they share the point P, so P, Q, R lie on one line [1]. Since PR = −PQ, P is the midpoint of QR [1] Examiner insight: A conclusion needs both the multiple and the common point. Write the relationship (PR = −PQ) explicitly instead of leaving it implied.

5. (a) AB = (−1 − 3, 4 − (−2), 5 − 1) = (−4, 6, 4) [1]; r = (3, −2, 1) + t(−4, 6, 4) [1] (b) x: 3 − 4t = −3, so t = 3/2 [1]. Then z = 1 + 4(3/2) = 7, not 4, so C is not on the line [1] Examiner insight: Any point of the line and any multiple of AB gives a correct equation. When you test a point, say which coordinate fails and what values were compared.

6. (a) x = −2 + 4t, so t = (x + 2)/4 [1]; y = 3 + (x + 2)/4, so y = x/4 + 7/2 [1] (b) y = 0 gives 3 + t = 0, t = −3 [1]; x = −2 − 12 = −14, so the point is (−14, 0) [1] Examiner insight: The answer to (b) is a coordinate pair, not just a value of t. Check it in y = x/4 + 7/2: −14/4 + 7/2 = 0.

7. (a) x: 2 + 3s = 2t; y: −1 + 2s = 7 − t [1]. From the second, t = 8 − 2s; then 2 + 3s = 16 − 4s, so s = 2 and t = 4 [1]. Point: (2 + 6, −1 + 4) = (8, 3), and line 2 gives (8, 7 − 4) = (8, 3) [1] (b) Distance = √((8 − 2)² + (3 − (−1))²) = √52 [1] = 2√13 [1] Examiner insight: In 2D, non-parallel directions guarantee a meeting point, but you still solve for it. Simplify a surd by taking out the largest square factor (52 = 4 × 13).

8. x: 2 + s = 5 + 2t; y: 1 + 2s = k − t; z: −3 + s = −2 [1]. The z equation gives s = 1 [1]; the x equation gives 3 = 5 + 2t, so t = −1 [1]. The y equation gives 3 = k + 1, so k = 2 [1]. Point from l₁ with s = 1: (3, 3, −2) [1]. Check on l₂: (5 − 2, 2 + 1, −2) = (3, 3, −2), so the position vector is (3, 3, −2) [1] Examiner insight: Pick the equations without the unknown first. The unknown goes in the one equation left over, which is the check as well as the answer.

9. Directions (2, 1, −1) and (1, −1, 2) are not multiples, so the lines are not parallel [1]. x: 1 + 2s = 2 + t; y: s = 2 − t [1]. Substituting: 1 + 4 − 2t = 2 + t, so t = 1 and s = 1 [1]. z for line 1: 2 − 1 = 1; z for line 2: 0 + 2 = 2 [1]. The values differ, so the lines never meet: they are skew [1] Examiner insight: The two statements “not parallel” and “no common point” are both needed for skew. Give the two z values side by side.

10. (a) (2, −2, 4) = −2(−1, 1, −2), so the directions are parallel [1]. Point (4, −1, 9) on l₁: 1 + 2s = 4 gives s = 3/2 [1]; y = 2 − 3 = −1 and z = 3 + 6 = 9 both match, so l₂ passes through a point of l₁ and the lines are the same line [1] (b) (1, −1, 2) = ½(2, −2, 4), so l₃ is parallel to l₁ [1]. If (0, 1, 1) were on l₁, then x gives s = −1/2 and y = 2 + 1 = 3, not 1, so it is not on l₁ and the lines are distinct [1] Examiner insight: Parallel directions alone do not decide between “same” and “distinct”. Test a point from one line on the other, and check all three coordinates.

11. (a) In a parallelogram AB = DC, so OD = OA + OC − OB [1] = (1 + 5 − 4, 0 + 5 − 2, 2 + 7 − 3) = (2, 3, 6) [1] (b) Midpoint of AC = ½(6, 5, 9) = (3, 5/2, 9/2) [1]; midpoint of BD = ½(6, 5, 9) = (3, 5/2, 9/2). They are equal, so the diagonals bisect each other [1] (c) AB = (3, 2, 1), |AB| = √14 [1]; BC = (1, 3, 4), |BC| = √26 [1]; perimeter = 2√14 + 2√26 [1] Examiner insight: A parallelogram has two pairs of equal opposite sides, so the perimeter is twice the sum of two adjacent lengths. Keep surds exact because √14 and √26 do not combine.

12. (a) x: 1 + 2s = 4 + t; y: −2 + s = 1 − t [1]. From the y equation, t = 3 − s, so 1 + 2s = 7 − s, giving s = 2 and t = 1 [2]. z: 0 + 2(2) = 4 and 2 + 2(1) = 4, which agree, so the lines intersect [1]. P = (1 + 4, −2 + 2, 0 + 4) = (5, 0, 4) [1] (b) |(2, 1, 2)| = √(4 + 1 + 4) = 3, so 9 units is 3 times the direction vector [1]. The points are P ± 3(2, 1, 2) [1]: (11, 3, 10) and (−1, −3, −2) [1] Examiner insight: “Two points” means both signs. Check (11, 3, 10) and (−1, −3, −2) each differ from P by a vector of length 9.

Where marks are usually lost

• Using OA − OB for AB, which reverses every sign.
• Calling vectors parallel after matching only one pair of components.
• Using one letter for both parameters in an intersection question.
• Concluding that 3D lines intersect after solving only two equations.
• Writing “skew” without showing the directions are not parallel.
• Giving only the value of t when the question asks for coordinates or a position vector.
• Dropping a sign when subtracting a negative component, such as −4 − (−2).
• Giving only one of the two points when “two points” are requested.
• Converting a surd to a rounded decimal when the question asks for an exact answer.

Next steps

• Revisit the revision notes for the method boxes.
• Reread the study guide for any worked example you found hard.
• Return to the course hub and the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA. This page covers section P2.10, first part (Unit P2, International A-level only).