Pearson Edexcel International Advanced Level Mathematics: Vectors, lines and the scalar product (YMA01) – 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 cover Unit P4 section 7 (Vectors, topics 7.1 to 7.7) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification (YMA01), Issue 3, April 2019. P4 is compulsory for IAL Mathematics and Pure Mathematics and has IAS weighting N/A. Learn the content in the study guide and the revision notes. The course hub is /boards/edexcel/a-level/mathematics/ and the checklist is /checklists/edexcel/a-level/mathematics/. Marks in square brackets show the credit for each step.

Questions

1. The vector p = 6i − 2j + 9k.

(a) Find |p|. [2] (b) Write down a unit vector in the direction of p. [1] (c) Find a vector of magnitude 33 in the direction of p. [1]

2. The points A and B have coordinates (−2, 3, 1) and (2, 1, 7).

(a) Find the vector AB. [1] (b) Find the distance AB, giving your answer as 2√14. [2] (c) Find the position vector of the midpoint of AB. [1]

3. The points A and B have position vectors a = 2i + j − 3k and b = 10i + 9j − 11k. The point C lies on AB with AC : CB = 3 : 5. Find the position vector of C, and the distance OC, where O is the origin. [5]

4. The points A, B and C have coordinates (2, 0, 1), (5, 1, −1) and (7, 5, 2).

(a) The quadrilateral ABCD is a parallelogram. Find the coordinates of D. [3] (b) Show that ABCD is not a rhombus. [2]

5. The line l passes through P(4, 0, −1) and Q(6, 3, −7).

(a) Write down a vector equation of l in the form r = c + t(d − c). [2] (b) Determine whether R(10, 9, −13) lies on l. [2]

6. Two lines have equations

l₁: r = (i + 2k) + λ(i + 2j − k) l₂: r = (3i + j) + μ(2i − j + k)

Show that l₁ and l₂ are skew. [6]

7. Two lines have equations

l₁: r = (2i − j + 4k) + λ(3i − k) l₂: r = (3i − 5j − k) + μ(i + 2j + 2k)

(a) Show that l₁ and l₂ intersect, and find the coordinates of the point of intersection. [4] (b) Find the acute angle between l₁ and l₂, to 1 decimal place. [3]

8. (a) The vectors 2i + pj − 3k and 4i − j + 5k are perpendicular. Find p. [2] (b) The points A, B and C have coordinates (1, 2, 3), (3, −1, 5) and (0, 3, 4). Find angle ABC to 1 decimal place. [4]

9. The line l has equation r = (−i + 2j) + λ(i − 2j + 2k). The point P has coordinates (4, −2, 7). The foot of the perpendicular from P to l is F.

(a) Find the coordinates of F. [5] (b) Find the shortest distance from P to l. [1] (c) Find the position vector of the point P′, the reflection of P in l. [2]

10. In the quadrilateral OABC, OA = a = 2i + 2j + k and OC = c = 2i − j + 2k, and B is such that OABC is a parallelogram.

(a) Show that |a| = |c|, and state the name of the shape OABC. [2] (b) Show that the diagonals OB and AC are perpendicular. [3] (c) Find angle AOC to 1 decimal place. [3] (d) Show that the midpoint of OB is also the midpoint of AC. [2]

11. The lines l₁ and l₂ have direction vectors d₁ = pi + pj + 2k and d₂ = pi − 5j + 3k, where p is a constant.

(a) Find the two values of p for which l₁ and l₂ are perpendicular. [3] (b) For p = 1, find the acute angle between the lines to 1 decimal place. [3]

Answers

1. (a) |p|² = 36 + 4 + 81 = 121 [1], so |p| = 11 [1] (b) (6i − 2j + 9k)/11 [1] (c) 3 × the unit vector: 18i − 6j + 27k [1] Examiner insight: “A unit vector” needs every component divided by 11. Leaving a component undivided means the vector is not a unit vector, and part (c) builds on it, so check that the length is 1.

2. (a) AB = (2 − (−2))i + (1 − 3)j + (7 − 1)k = 4i − 2j + 6k [1] (b) |AB|² = 16 + 4 + 36 = 56 [1], so |AB| = √56 = 2√14 [1] (c) (a + b)/2 = 2j + 4k [1] Examiner insight: When the form is given, showing the line √56 = √(4 × 14) is how you evidence the simplification. A decimal alone does not show the surd form was reached.

3. AB = b − a = 8i + 8j − 8k [1]. AC = (3/8)AB = 3i + 3j − 3k [1]. c = a + AC [1] = 5i + 4j − 6k [1]. OC = √(25 + 16 + 36) = √77 [1] Examiner insight: Take the fraction 3/8, not 3/5, because the parts add to 8 and AC is measured along the whole of AB. A sketch of the ratio check is worth doing.

4. (a) In a parallelogram ABCD, AD = BC [1]. BC = 2i + 4j + 3k [1], so d = a + BC, giving D = (4, 4, 4) [1] (b) |AB| = √(9 + 1 + 4) = √14 [1]. |BC| = √(4 + 16 + 9) = √29. These differ, so adjacent sides are unequal and ABCD is not a rhombus [1] Examiner insight: A “show that” statement should end with a conclusion that refers to the property tested. Two numerical magnitudes with no comment leave the argument unfinished.

5. (a) d − c = (6 − 4)i + (3 − 0)j + (−7 + 1)k = 2i + 3j − 6k [1]. So r = (4i − k) + t(2i + 3j − 6k) [1] (b) i: 4 + 2t = 10, so t = 3 [1]. j: 0 + 9 = 9 matches, but k: −1 − 18 = −19, not −13, so R is not on l [1] Examiner insight: One matching component proves nothing. Check all three with the same t, and state the mismatch explicitly.

6. The directions (1, 2, −1) and (2, −1, 1) are not multiples of each other, so the lines are not parallel [1]. Equate components: i: 1 + λ = 3 + 2μ; j: 2λ = 1 − μ; k: 2 − λ = μ [1] From k, μ = 2 − λ. In j: 2λ = 1 − 2 + λ, so λ = −1 and μ = 3 [2] Check i: the left side is 1 − 1 = 0 and the right side is 3 + 6 = 9. They differ [1], so the lines do not meet. Not parallel and no common point, so they are skew [1] Examiner insight: The conclusion should state both facts: not parallel, and no common point. Answers that show only the failed equation can leave the first fact unstated.

7. (a) j: −1 = −5 + 2μ, so μ = 2 [1]. k: 4 − λ = −1 + 4 = 3, so λ = 1 [1]. Check i: 2 + 3 = 5 and 3 + 2 = 5, so they match [1]. The lines meet at (5, −1, 3) [1] (b) d₁ = 3i − k, d₂ = i + 2j + 2k, so d₁·d₂ = 3 − 2 = 1 [1]. |d₁| = √10 and |d₂| = 3 [1]. cos θ = 1/(3√10), so θ = 83.9° [1] Examiner insight: Choose the two equations that are quickest to solve (here j then k) and use the third as a check. Using the position vectors, rather than the direction vectors, in part (b) is a typical way to go wrong.

8. (a) (2)(4) + p(−1) + (−3)(5) = 8 − p − 15 = 0 [1], so p = −7 [1] (b) BA = (−2, 3, −2) and BC = (−3, 4, −1) [1]. BA·BC = 6 + 12 + 2 = 20 [1]. |BA| = √17 and |BC| = √26 [1]. cos ∠ABC = 20/√442, so ∠ABC = 18.0° [1] Examiner insight: Both vectors must start at B. Mixing BA with CB reverses one sign and gives the supplementary angle, 162.0°, so check that both vectors leave B.

9. (a) A general point on l is F = (λ − 1, 2 − 2λ, 2λ) [1]. PF = F − P = (λ − 5)i + (4 − 2λ)j + (2λ − 7)k [1]. For a perpendicular, PF·(i − 2j + 2k) = 0: (λ − 5) − 2(4 − 2λ) + 2(2λ − 7) = 9λ − 27 = 0 [1], so λ = 3 [1]. F = (2, −4, 6) [1] (b) At λ = 3, PF = −2i − 2j − k, so the distance is √(4 + 4 + 1) = 3 [1] (c) F is the midpoint of PP′, so p′ = 2f − p [1] = (4, −8, 12) − (4, −2, 7) = −6j + 5k [1] Examiner insight: Substitute λ back into PF as a check: PF·d should be exactly 0. Keep the sign of each component when writing PF, since F − P and P − F lead to the same λ but different vectors.

10. (a) |a| = √(4 + 4 + 1) = 3 and |c| = √(4 + 1 + 4) = 3, so they are equal [1]. A parallelogram with two equal adjacent sides is a rhombus [1] (b) OB = a + c = 4i + j + 3k [1]. AC = c − a = −3j + k [1]. OB·AC = 0 − 3 + 3 = 0, so OB and AC are perpendicular [1] (c) a·c = 4 − 2 + 2 = 4 [1]. cos ∠AOC = 4/(3 × 3) = 4/9 [1]. ∠AOC = 63.6° [1] (d) The midpoint of OB is b/2 = 2i + 0.5j + 1.5k [1]. The midpoint of AC is (a + c)/2 = 2i + 0.5j + 1.5k, which is the same point [1] Examiner insight: For a perpendicularity proof, write the scalar product line in full and then the word “perpendicular”. A zero stated without the working is easy to dispute.

11. (a) d₁·d₂ = p² − 5p + 6 [1]. Setting this to 0 gives (p − 2)(p − 3) = 0 [1], so p = 2 or 3 [1] (b) For p = 1, d₁ = i + j + 2k and d₂ = i − 5j + 3k. d₁·d₂ = 1 − 5 + 6 = 2 [1]. |d₁| = √6 and |d₂| = √35 [1]. cos θ = 2/√210, so θ = 82.1° [1] Examiner insight: Two perpendicular values means a quadratic, so give both. Offering only one root is a common way to lose the final credit in part (a).

Where marks are usually lost

• Reversing a vector: AB = a − b instead of b − a.
• Dividing only some components by the magnitude when asked for a unit vector.
• Using one parameter for both lines, so that an intersection cannot be found.
• Announcing an intersection after solving only two of the three component equations.
• Calling lines skew without saying they are not parallel.
• Using position vectors instead of direction vectors for the angle between lines.
• Quoting an obtuse angle when the question asks for the angle between two lines.
• Rounding too early and drifting in the last decimal place.
• Stating a·b = 0 without the arithmetic when a question says “show that”.

Next steps

• Review the revision notes and the study guide.
• Go to the Edexcel A Level Mathematics hub and tick off topics on the checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited. Unit P4, section 7 (Vectors).