These twelve questions cover the five lessons in two-dimensional vector proofs. Attempt each one on paper first, draw a sketch, then open the answer. All questions are original.
Unless stated otherwise, a and b are non-parallel base vectors and O is the origin.
Questions 1 to 4: displacements
Q1. →OA = 3a and →OB = a + 2b. Find →AB.
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End minus start: →AB = →OB − →OA = (a + 2b) − 3a = a − 3a + 2b.
→AB = −2a + 2b
Q2. OABC is a parallelogram with →OA = 2p and →OC = 3q. Find →OB, →AC and →BC.
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→OB = →OA + →AB = 2p + 3q (since →AB = →OC). →AC = −→OA + →OC = −2p + 3q. →BC = −→OA = −2p (opposite sides of a parallelogram).
→OB = 2p + 3q, →AC = 3q − 2p, →BC = −2p
Q3. In triangle OAB, M is the midpoint of AB. Find →OM and →BM in terms of a and b.
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→AB = b − a. →OM = a + ½(b − a) = ½a + ½b. →BM = −½→AB = −½(b − a) = ½a − ½b.
→OM = ½a + ½b, →BM = ½a − ½b
Q4. →OP = 2a − b and →OQ = a + 3b. Find →QP, and state the vector that is half of →QP.
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→QP = →OP − →OQ = (2a − b) − (a + 3b) = a − 4b. Half of it is ½a − 2b.
→QP = a − 4b, half is ½a − 2b
Questions 5 to 7: dividing points
Q5. P is on AB with AP : PB = 1 : 4, where →OA = a and →OB = b. Find →OP.
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Five parts in all, so →AP = 1/5(b − a). →OP = a + 1/5b − 1/5a = 4/5a + 1/5b.
→OP = 4/5 a + 1/5 b. Check: the coefficients sum to 1, and a has the larger share because P is nearer A.
Q6. A is (−2, 3) and B is (8, −7). P is on AB with AP : PB = 2 : 3. Find the coordinates of P.
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→AB = (8 − (−2), −7 − 3) = (10, −10). →AP = 2/5 × (10, −10) = (4, −4). →OP = (−2, 3) + (4, −4) = (2, −1).
P = (2, −1). Check with 3/5a + 2/5b: x = −6/5 + 16/5 = 2, y = 9/5 − 14/5 = −1.
Q7. P is on AB and →OP = 1/3 a + 2/3 b. Find AP : PB.
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→OP = a + k(b − a) has b coefficient k, so k = 2/3, meaning →AP = 2/3 →AB. AP is 2 parts and PB is 1 part.
AP : PB = 2 : 1. Check: P is nearer B and has more b, as expected.
Questions 8 and 9: collinearity
Q8. →OA = a + 2b, →OB = 3a + b, →OC = 7a − b. Show that A, B, C are collinear and find AB : BC.
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→AB = (3a + b) − (a + 2b) = 2a − b. →AC = (7a − b) − (a + 2b) = 6a − 3b = 3(2a − b) = 3→AB. They are parallel and share the point A, so A, B and C are collinear. →BC = →AC − →AB = 2→AB.
Collinear, AB : BC = 1 : 2
Q9. →PQ = 2a − 3b and →PR = 6a + kb. Given that P, Q and R are collinear, find k.
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→PR must be a multiple of →PQ. The a coefficient is multiplied by 3 (2 to 6), so →PR = 3→PQ = 6a − 9b.
k = −9
Questions 10 and 11: intersections
Q10. →OA = a, →OB = b. M is the midpoint of OB and N is on OA with ON = 2/3 a. Lines AM and BN meet at X. Find →OX.
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AM: →OX = a + λ(½b − a) = (1 − λ)a + (λ/2)b. BN: →OX = b + μ(2/3a − b) = (2μ/3)a + (1 − μ)b.
Equate: a: 1 − λ = 2μ/3. b: λ/2 = 1 − μ, so λ = 2 − 2μ. Then 1 − 2 + 2μ = 2μ/3, so 4μ/3 = 1, μ = 3/4 and λ = 1/2.
→OX = ½a + ¼b. Check with μ = 3/4 in BN: (2 × 3/4 ÷ 3)a + (1/4)b = ½a + ¼b.
Q11. Line 1: r = (0, 1) + λ(2, 1). Line 2: r = (10, −2) + μ(−2, 1). Find the point of intersection.
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x: 2λ = 10 − 2μ. y: 1 + λ = −2 + μ, so μ = 3 + λ. Substitute: 2λ = 10 − 6 − 2λ, so 4λ = 4, λ = 1 and μ = 4. Line 1 at λ = 1: (2, 2). Line 2 at μ = 4: (10 − 8, −2 + 4) = (2, 2).
(2, 2)
Question 12: direction
Q12. Line l: r = (−1, 4) + λ(6, −9). (a) Give a simpler direction vector. (b) Is (5, −5) on l? (c) Is (2, 0) on l?
Show answer
(a) (6, −9) = 3(2, −3), so a simpler direction is (2, −3).
(b) x: −1 + 6λ = 5, so λ = 1. y: 4 − 9(1) = −5. It matches, so (5, −5) is on l.
(c) x: −1 + 6λ = 2, so λ = 1/2. y: 4 − 9/2 = −1/2, not 0. (2, 0) is not on l.
If you got these wrong
| What went wrong | Where to revise |
|---|---|
| Sign errors, →AB written as a − b, wrong route | Express a displacement using base vectors |
| Weights on the wrong vector in a ratio question | Find a dividing point on a segment |
| Only one coefficient compared, or no shared point or conclusion | Show points are collinear using a scalar relation |
| One parameter used for two lines, or slips solving the pair of equations | Find an intersection of vector-defined lines |
| A point tested on one coordinate only, or position and direction confused | Explain direction in a vector equation |
Record each slip with the mistake log and retest queue and retest it in a few days. Fraction slips are worth practising with the non-calculator working trainer.
If the same type of question keeps going wrong after revision, see how our teachers work in online one-to-one Additional Mathematics tuition.