These eight original questions follow the four lessons in order. Try each one with full working before you open the answer.
Questions use a and b for OA and OB, and these vectors are not parallel. Start with the lessons if you need them: finding a ratio from two expressions and parallel versus collinear.
Questions
Question 1. Given p = (k + 2)i + 9j and q = 2i + 3j, find k so that p is parallel to q.
Answer
Write p = cq. The j parts give 9 = 3c, so c = 3. The i parts give k + 2 = 2c = 6, so k = 4.
Check: p = 6i + 9j = 3(2i + 3j).
Question 2. The points are A(0, 1), B(2, 4) and C(6, h). Find h if A, B and C are collinear.
Answer
AB = (2, 3) and BC = (4, h − 4). Write BC = cAB: 4 = 2c, so c = 2. Then h − 4 = 2 × 3 = 6, so h = 10.
Check: AC = (6, 9) = 3 × (2, 3), so C lies on the line AB. B is the shared point.
Question 3. In a figure, OA = 3a, OC = b and CB = 6a. A student writes “CB is parallel to OA, so O, A and B are collinear.” Is the conclusion correct?
Answer
CB = 6a and OA = 3a, so CB = 2OA. The vectors are parallel, which is correct.
But AB = AO + OC + CB = −3a + b + 6a = 3a + b. This is not a multiple of OA = 3a, because of the b. So O, A and B are not collinear.
Question 4. P lies on AB such that AP:PB = 3:2. Express OP in terms of a and b.
Answer
The total is 3 + 2 = 5, so AP = (3/5)AB. OP = a + (3/5)(b − a) = (2/5)a + (3/5)b.
Check: the coefficients add up to 1, and P is nearer B, which matches the larger weight on b.
Question 5. M lies on AB and OM = μ(2a + b). Find AM:MB.
Answer
Route 1: OM = (1 − λ)a + λb. Route 2: OM = 2μa + μb.
Equate: 1 − λ = 2μ and λ = μ. So 1 − λ = 2λ, giving λ = 1/3. AM = (1/3)AB, so AM:MB = 1:2.
Check: OM = (2/3)a + (1/3)b, and the coefficients add up to 1.
Question 6. In triangle OAB, M lies on OA with OM = (2/3)a, and N is the midpoint of OB. Lines AN and BM meet at P. Find OP.
Answer
Along AN: AN = (1/2)b − a, so OP = (1 − λ)a + (λ/2)b. Along BM: BM = (2/3)a − b, so OP = (2μ/3)a + (1 − μ)b.
Equate: 1 − λ = 2μ/3 and λ/2 = 1 − μ. From the second, λ = 2 − 2μ. Then 1 − 2 + 2μ = 2μ/3, so (4/3)μ = 1 and μ = 3/4. Then λ = 1/2.
OP = (1/2)a + (1/4)b.
Check on AN: (1 − 1/2)a + (1/4)b = (1/2)a + (1/4)b. It matches.
Question 7. R lies on AB produced beyond B such that AR:RB = 5:2. Express OR in terms of a and b.
Answer
Let AR = 5t and RB = 2t in length. Since R is beyond B, AB = 5t − 2t = 3t. So AR = (5/3)AB.
OR = a + (5/3)(b − a) = −(2/3)a + (5/3)b.
Check: the coefficients add up to 1. RB = b − r = (2/3)a − (2/3)b = −(2/3)AB, so lengths are 5/3 : 2/3 = 5:2. The negative coefficient of a puts R beyond B.
Question 8. OA = 2a + b, OB = 4a + 3b and OC = 7a + 6b. Show that A, B and C are collinear, and find AB:BC.
Answer
AB = OB − OA = 2a + 2b. BC = OC − OB = 3a + 3b.
So BC = (3/2)AB. The vectors are parallel and B is a common point, so A, B and C are collinear.
AB = 2(a + b) and BC = 3(a + b), so AB:BC = 2:3.
If you got some wrong
Questions 1 to 3 test parallel vectors and collinearity. Questions 4 and 7 use ratios: see solving geometric vector ratios and checking the sign of an external division ratio. Question 6 is covered in finding an intersection using two vector paths.
Log each slip in the mistake log and paper-error review tool, and use the timed original practice session builder to repeat the set under time. A teacher in online one-to-one Additional Mathematics tuition can give you fresh questions on whichever type slips most.