Pure Mathematics
Vector – 2
⃗⃗⃗⃗⃗ = 5𝐚, 𝑂𝐵
1. Figure shows Δ𝑂𝐴𝐵 in which 𝑂𝐴 ⃗⃗⃗⃗⃗ = 5𝐛, and 𝑂𝐸
⃗⃗⃗⃗⃗ = 2𝐛. The point 𝐶 is on 𝑂𝐴 such that
𝑂𝐶 ∶ 𝑂𝐴 = 2 ∶ 5 and 𝐷 is the point on 𝐵𝐶 such that 7𝐵𝐷 = 5𝐵𝐶 .
(a) Express, in terms of 𝐚 and 𝐛, the vectors
⃗⃗⃗⃗⃗
(i) 𝐴𝐵 (ii) ⃗⃗⃗⃗⃗
𝐴𝐸 ⃗⃗⃗⃗⃗
(iii) 𝐵𝐶
(iv) ⃗⃗⃗⃗⃗
𝐴𝐷 ⃗⃗⃗⃗⃗
(v)𝐶𝐸
Hence prove that
(b) 𝐶𝐸 is parallel to 𝐴𝐵 .
(c) The points 𝐴, 𝐷 and 𝐸 are collinear.
(d) Find the ratio (area Δ𝑂𝐶𝐵 ∶ 𝑎𝑟𝑒𝑎 Δ𝐴𝐶𝐵).
(e) Find the ratio Δ𝑂𝐶𝐸: Δ𝑂𝐴𝐵 [Jan 02/P2/Q10]
25 10
Answers: [1] (a) (i) −5𝐚 + 5𝐛, (ii) −5𝐚 + 2𝐛, (iii) 2𝐚 − 5𝐛, (iv) − 7
𝐚 + 7
𝐛, (v) −2𝐚 + 2𝐛, (d) 4: 25,
(e) 2: 3
2. In fig, ⃗⃗⃗⃗⃗
𝑂𝐴 = 3𝐚, ⃗⃗⃗⃗⃗
𝑂𝐵 = 3𝐛 and ⃗⃗⃗⃗⃗
𝑂𝐶 = 2𝐚 + 2𝐛. The point 𝐷 is on 𝐴𝐶 such that 𝐴𝐷 ∶ 𝐷𝐶 = 2 ∶ 1 and 𝐸
is the point on 𝐵𝐶 produced such that ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ , where 𝑘 is a constant.
𝐵𝐸 = 𝑘𝐵𝐶
(a) Express, in terms of 𝐚 and 𝐛,
(i)⃗⃗⃗⃗⃗⃗
𝐵𝐶 (ii)⃗⃗⃗⃗⃗⃗
𝐴𝐶 (iii)⃗⃗⃗⃗⃗⃗⃗
𝑂𝐷
(b) Show that ⃗⃗⃗⃗⃗
𝑂𝐸 = 2𝑘𝐚 + (3 − 𝑘)𝐛.
Given also that 𝑂𝐷𝐸 is a straight line, find
(c) The value of 𝑘.
⃗⃗⃗⃗⃗ in terms of 𝐚 and 𝐛 only.
(d) 𝑂𝐸
(e) The ratio 𝑂𝐷: 𝐷𝐸.
(f) Find the ratio (area Δ𝑂𝐷𝐵 ∶ area Δ𝑂𝐸𝐵). [May 96/P1/Q14]
7 4 7 14 8
Answers: [2] (a) (i) 2𝐚 − 𝐛, (ii) −𝐚 + 2𝐛, (iii) 3 𝐚 + 3 𝐛, (c) 3, (d) 5
𝐚 + 5 𝐛, (e) 5: 1, (f) 5: 6
⃗⃗⃗⃗⃗ = 𝐚 and 𝑂𝐵
3. In Δ𝑂𝐴𝐵, 𝑂𝐴 ⃗⃗⃗⃗⃗ = 𝐛. The point 𝐶 divides 𝐴𝐵
in the ratio 2 : 3 and 𝐷 is the mid-point of 𝑂𝐵 as shown in figure.
⃗⃗⃗⃗⃗
(a) Find, in terms of 𝑎 and 𝑏, 𝑂𝐶 𝐾
FAISAL MIZAN 1
Pure Mathematics
𝑂𝐶 and 𝐴𝐷 meet at 𝐾.
𝐾 divides 𝐴𝐷 in the ratio 𝜆 ∶ 3
(b) Find ⃗⃗⃗⃗⃗⃗
𝑂𝐾, hence find the value of 𝜆.
(c) Find the ratio ∆𝑂𝐶𝐵: ∆𝐴𝐷𝐵 [Jan 11/P1/Q4]
3 2
Answers: [3] (a) 5 𝐚 + 5 𝐛, (b) 𝜆 = 4, (c) 6: 5
2
4. In figure, ⃗⃗⃗⃗⃗
𝑂𝐴 = 𝐚, ⃗⃗⃗⃗⃗
𝑂𝐵 = 𝐛 and ⃗⃗⃗⃗⃗⃗
𝑂𝐷 = 𝐛. The points 𝐸 divides 𝐴𝐷 in the ratio 2:3
3
(a) Find as simplified expressions in terms of 𝑎 and 𝑏
(i) ⃗⃗⃗⃗⃗
𝐴𝐷 (ii) ⃗⃗⃗⃗⃗
𝑂𝐸 (iii) ⃗⃗⃗⃗⃗
𝐵𝐸
⃗⃗⃗⃗⃗ = 𝜆𝑂𝐴
The point 𝐹 lies of 𝑂𝐴 such that 𝑂𝐹 ⃗⃗⃗⃗⃗
and 𝐹, 𝐸 and 𝐵 are collinear.
(b) Find the value of 𝜆.
The area of triangle 𝑂𝐹𝐵 is 5 square units.
(c) Find the area of triangle 𝑂𝐴𝐷.
𝑝
Give your answer in the form 𝑞 , where 𝑝 and 𝑞 are integers. [May 16/P1/Q13]
2 3 4 3 11 9 110
Answers: [4] (a) (i) 𝐛 − 𝐚, (ii) 𝐚 + 𝐛, (iii) 𝐚 − 𝐛, (b) 𝜆 = , (c) .
3 5 15 5 15 11 27
2
5. In figure, ⃗⃗⃗⃗⃗
𝑂𝐴 = 𝐚, ⃗⃗⃗⃗⃗
𝑂𝐵 = b and ⃗⃗⃗⃗⃗⃗
𝑂𝐷 = 𝐛. The point of 𝐴𝐷 is 𝐸.
5
Find, in terms of 𝑎 and 𝑏,
⃗⃗⃗⃗⃗
(i) 𝐴𝐷 ⃗⃗⃗⃗⃗
(ii) 𝑂𝐸 (iii) ⃗⃗⃗⃗⃗
𝐵𝐸
The point 𝐹 is such that 𝑂𝐹 = 𝜆𝑂𝐴
Given that 𝐹, 𝐸 and 𝐵 are collinear,
(b) find the value of 𝜆.
Given that area Δ𝑂𝐴𝐷 = 𝜇 × 𝑎𝑟𝑒𝑎 ∆𝑂𝐹𝐵,
(c) Find the value of 𝜇. [May 05/P1/Q17]
2 1 1 1 4 5 16
Answers: [5] (a) (i) −𝐚 + 5 𝐛, (ii) 2 𝐚 + 5 𝐛, (iii) 2 𝐚 − 5 𝐛, (b) 8, (c) 25.
FAISAL MIZAN 2
Pure Mathematics
6. In fig, ⃗⃗⃗⃗⃗
𝑂𝐴 = 𝐚 and ⃗⃗⃗⃗⃗
𝑂𝐵 = 𝐛. Given that 𝑂𝐵: 𝐵𝐶 = 2: 1 and the point 𝐷
Divides 𝐴𝐵 internally in the ratio 1 : 4.
(a) Express, in terms of 𝐚 and 𝐛,
(i) ⃗⃗⃗⃗⃗⃗
𝑂𝐷 (ii) ⃗⃗⃗⃗⃗
𝑂𝐶 (iii) ⃗⃗⃗⃗⃗
𝐴𝐶
The point 𝐸 divides 𝐴𝐶 internally in the ratio 1 : 𝜆.
Given that 𝑂, 𝐷 and 𝐸 are collinear,
(b) Find the value of 𝜆.
(c) Find the ratio Δ𝑂𝐴𝐵: Δ𝐴𝐵𝐶 [Jan 04/P1/Q6]
4 1 3 3
Answers: [6] (a) (i) 𝐚 + 𝐛, (ii) 𝐛, (iii) −𝐚 + 𝐛, (b) 6, (c) 2: 1
5 5 2 2
7. In fig,⃗⃗⃗⃗⃗⃗
𝑂𝐴 = 3𝐚,⃗⃗⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ = 2𝑂𝐷
𝑂𝐵 = 2𝐛, 𝑂𝐵 ⃗⃗⃗⃗⃗⃗ , 𝑂𝐴
⃗⃗⃗⃗⃗ = 3𝑂𝐶
⃗⃗⃗⃗⃗ , 𝐷𝐸 ⃗⃗⃗⃗⃗ ,⃗⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = 𝑝𝐷𝐴 ⃗⃗⃗⃗⃗ and 𝐵𝐹
𝐶𝐸 = 𝑞𝐶𝐵 ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = 𝑟𝐵𝐴
where 𝑝, 𝑞 and 𝑟 are positive constants.
(a) Find, in terms of 𝑎 and 𝑏 express for
(i) ⃗⃗⃗⃗⃗
𝐷𝐴 (ii) ⃗⃗⃗⃗⃗
𝐶𝐵
(b) Show that ⃗⃗⃗⃗⃗
𝑂𝐸 = 3𝑝𝐚 + (1 − 𝑝)𝐛.
(c) Find in terms of 𝐚, 𝐛 and 𝐪, another expression for ⃗⃗⃗⃗⃗
𝑂𝐸
(d) Using our answer to (b) and (c), find the value of 𝑝 and 𝑞.
(e) Write down, in terms of 𝑎 and 𝑏 only an expression for ⃗⃗⃗⃗⃗
𝑂𝐸 .
(f) Find the value of 𝑟. [May 97/P1/Q13]
1 2 3 4 1
Answers: [7] (a) (i) 3𝐚 − 𝐛, (ii) −𝐚 + 2𝐛, (c) (i) (1 − 𝑞)𝐚 + 2𝑞𝐛, (d) 𝑝 = 5, 𝑞 = 5, (e) 5 𝐚 + 5 𝐛, (f) 3.
8.
2
In Figure ⃗⃗⃗⃗⃗
𝑂𝐴 = 𝐚, ⃗⃗⃗⃗⃗
𝑂𝐵 = 𝐛 and ⃗⃗⃗⃗⃗⃗
𝑂𝐷 = 3 𝐛. The point 𝐸 divides 𝐴𝐷 in the ratio 2: 3.
FAISAL MIZAN 3
Pure Mathematics
(a) Find as simplified expression in terms of 𝐚 and 𝐛:
⃗⃗⃗⃗⃗ ,
(i) 𝐴𝐷 ⃗⃗⃗⃗⃗ ,
(ii) 𝑂𝐸 (iii) ⃗⃗⃗⃗⃗
𝐵𝐸 .
The point 𝐹 lies on 𝑂𝐴 such that ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗ and 𝐹, 𝐸 and 𝐵 are collinear.
𝑂𝐹 = 𝜆𝑂𝐴
(b) Find the value of 𝜆.
The area of triangle 𝑂𝐹𝐵 is 5 square units.
𝑝
(c) Find the area of triangle 𝑂𝐴𝐷. Give your answer in the form 𝑞 , where 𝑝 and 𝑞 are integers.
[May 16/P1/Q8]
2 3 4 3 11 9 110
Answers: [8] (a) (i) 𝐛 − 𝐚, (ii) 𝐚 + 𝐛, (iii) 𝐚 − 𝐛, (b) , (c) ,
3 5 15 5 15 11 27
9.
⃗⃗⃗⃗⃗ = 𝐚 and 𝑂𝐵
In Figure, 𝑂𝐴 ⃗⃗⃗⃗⃗ = 𝐛. The point 𝐶 divides 𝑂𝐵 in the ratio 1: 3.
The point 𝐷 is the midpoint of 𝐴𝐶.
(a) Find, as a simplified expression in terms of 𝐚 and 𝐛
⃗⃗⃗⃗⃗ ,
(i) 𝐴𝐶 ⃗⃗⃗⃗⃗⃗ ,
(ii) 𝑂𝐷 ⃗⃗⃗⃗⃗⃗ .
(iii) 𝐵𝐷
⃗⃗⃗⃗⃗ = 𝜆𝑂𝐴
The point 𝐸 is such that 𝑂𝐸 ⃗⃗⃗⃗⃗ . Given that 𝐸, 𝐷 and 𝐵 are collinear.
(b) find the value of 𝜆.
area ∆OAC
Given that area ∆𝑂𝐸𝐵 = 𝜇.
(c) Find the value of 𝜇. [Nov 20/P1/Q11]
1 1 1 1 7 4 7
Answers: [9] (a) (i) −𝐚 + 4 𝐛, (ii) 2 𝐚 + 8 𝐛, (iii) 2 𝐚 − 8 𝐛, (b) 7, (c) 16,
FAISAL MIZAN 4
Pure Mathematics
10. Figure shows the triangle 𝑂𝐴𝐵 with ⃗⃗⃗⃗⃗
𝑂𝐴 = 2𝐚 and ⃗⃗⃗⃗⃗
𝑂𝐵 = 3𝐛.
1 𝐴
⃗⃗⃗⃗⃗ = 𝑂𝐴
The point 𝐶 lies on 𝑂𝐴 such that 𝑂𝐶 ⃗⃗⃗⃗⃗ .
3
2
⃗⃗⃗⃗⃗⃗ = 𝑂𝐵
The point 𝐷 lies on 𝑂𝐵 such that 𝑂𝐷 ⃗⃗⃗⃗⃗ .
3
⃗⃗⃗⃗⃗ in terms of 𝐚 and 𝐛.
(a) Find 𝐶𝐷 𝑂 𝑃
𝐵
The point 𝑃 is such that 𝑂𝐷𝐵𝑃 is a straight line and 𝐴𝑃 is parallel to 𝐶𝐷.
(b) Find ⃗⃗⃗⃗⃗
𝑂𝑃 in term of 𝐛
2
Answers: [10] (a) − 𝐚 + 𝐛, (b) 6𝐛,
3
11. In Figure, ⃗⃗⃗⃗⃗
𝑂𝐴 = 𝐚, ⃗⃗⃗⃗⃗
𝑂𝐵 = 𝐛 and 𝑀 is the mid-point
of 𝐴𝐵. The point 𝑃 is on 𝑂𝐴 such that 𝑂𝑃: 𝑃𝐴 = 3: 2.
The point 𝑋 lies on 𝑂𝐵 produced.
(a) Find, as simplified expressions in terms of a and b,
(i) ⃗⃗⃗⃗⃗
𝐴𝐵 , (ii) ⃗⃗⃗⃗⃗⃗
𝑂𝑀, (i) ⃗⃗⃗⃗⃗⃗
𝑃𝑀,
Given that 𝑃, 𝑀 and 𝑋 are collinear
⃗⃗⃗⃗⃗ ,
(b) find, in terms of 𝐛, 𝑂𝑋
(c) Find the ratio (area ∆𝑂𝐴𝑀) : (area ∆𝑂𝐴𝑋). [Jan 13/P2/Q8]
1 1 1
Answers: [11] (a) (i) 𝐛 − 𝐚, (ii) 2 (𝐚 + 𝐛), (iii) 2 𝐛 − 10 𝐚, (b) 3𝐛, (c) 1: 6,
⃗⃗⃗⃗⃗ = 𝐚 and 𝑂𝐵
12. Figure shows the triangle 𝑂𝐴𝐵 with 𝑂𝐴 ⃗⃗⃗⃗⃗ = 𝐛.
(a) Find ⃗⃗⃗⃗⃗
𝐴𝐵 in terms of 𝐚 and 𝐛.
3
⃗⃗⃗⃗⃗ = 𝑂𝐴
The point 𝑃 is such that 𝑂𝑃 ⃗⃗⃗⃗⃗ , and the point 𝑄 is
4 𝑃
the midpoint of 𝐴𝐵. 𝑄
(b) Find ⃗⃗⃗⃗⃗
𝑃𝑄 as a simplified expression in terms of 𝐚 and 𝐛.
The point R is such that PQR and OBR are straight lines
⃗⃗⃗⃗⃗
(c) Find 𝑂𝑅 𝑅
(d) Find the ratio (area ∆𝐴𝑃𝑄) : (area ∆𝑂𝐴𝑅) [Jan 18/P2/Q6]
1
Answers: [12] (a) – 𝐚 + 𝐛, (b) 4 (2𝐛 − 𝐚),
FAISAL MIZAN 5
Pure Mathematics
13.
Figure shows triangle 𝑂𝐴𝐵 and triangle 𝑂𝐶𝐷.
⃗⃗⃗⃗⃗ = 5𝐩
𝑂𝐴 ⃗⃗⃗⃗⃗ = 3𝐪
𝐴𝐵 ⃗⃗⃗⃗⃗ = 3 𝑂𝐵
𝑂𝐶 ⃗⃗⃗⃗⃗ ⃗⃗⃗⃗⃗⃗ = 3 𝑂𝐴
𝑂𝐷 ⃗⃗⃗⃗⃗
2 5
⃗⃗⃗⃗⃗ as a simplified expression in terms of 𝐩 and 𝐪.
(a) Find 𝐷𝐶
The line 𝐷𝐶 meets the line AB at 𝐹.
⃗⃗⃗⃗⃗ as a simplified expression in terms of 𝐩 and 𝐪.
(b) Using a vector method, find 𝑂𝐹 [May 21/P1/Q10]
9 9
Answers: [13] (a) 𝐩 + 𝐪, (b) 5𝐩 + 2𝐪,
2 2
14. Referred to a fixed origin 𝑂, the position vectors of the points 𝑃 and 𝑄 are (10𝑖 − 3𝑗) and (4𝑖 + 6𝑗)
respectively. The point 𝑅 divides 𝑃𝑄 internally in the ratio 2 ∶ 1.
(a) Find the position vector of 𝑅.
The point 𝑆 divides 𝑂𝑄 internally in the ratio 5 : 4 and area ∆𝑂𝑃𝑄 = 𝜆 area Δ𝑆𝑅𝑄.
(b) Find the exact value of 𝜆. [Jan 12/P1/Q1]
27
Answers: [14] (a) 6𝑖 + 3𝑗, (b) 4
15. Referred to an origin 𝑂,the points 𝐴,𝐵 and 𝐶 have position vectors 2i + 7j, 8i + 9j
and 11i + 10j respectively.
(a) Find in terms of i and j: (i) ⃗⃗⃗⃗⃗
𝐴𝐵 (ii) ⃗⃗⃗⃗⃗
𝐵𝐶
(b) Show that 𝐴, 𝐵 and 𝐶 are collinear. The point 𝐷 lies on 𝑂𝐴 and has position vector 𝜆(2i +7j)
⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ = 𝐵𝐶
The point 𝐸 lies on 𝑂𝐵 and has position vector 𝜇(8𝑖 + 9𝑗). Given that 𝐷𝐸
(c) Find the value of 𝜆 and the value of 𝜇.
(d) Write down the ratio 𝐷𝐸 ∶ 𝐴𝐵.
(e) Find the ratio (area Δ𝑂𝐷𝐸) : (area Δ𝑂𝐴𝐶). [Jan 01/P2/Q9]
FAISAL MIZAN 6
Pure Mathematics
1 1
Answers: [15] (a) (i) 6𝑖 + 2𝑗, (ii) 3𝑖 + 𝑗, (c) 𝜆 = 2, 𝜇 = 2, (d) 1: 2, (e) 1: 6
16. The points 𝐴, 𝐵, 𝐶 and 𝐷 are the vertices of a quadrilateral 𝐴𝐵𝐶𝐷 such that ⃗⃗⃗⃗⃗
𝐴𝐵 = 7𝐢 + 𝑝𝐣,
⃗⃗⃗⃗⃗⃗
𝐴𝐶 = 11𝐢 − 𝑝𝐣, ⃗⃗⃗⃗⃗
𝐴𝐷 = 4𝐢 − 2𝑝𝐣.
(a) Show that, for all values of 𝑝, 𝐴𝐵𝐶𝐷 is a parallelogram.
⃗⃗⃗⃗⃗⃗ | = 3√ 10.
Given that|𝐵𝐷
(b) Find the possible values of 𝑝.
Given that 𝑝 > 0.
⃗⃗⃗⃗⃗⃗ .
(c) Find a unit vector which is parallel to 𝐵𝐷 [Jan 20/P2/Q4]
(−3i+9𝐣)
Answers: [16] (b) ±3, (c) ± 3√10
,
FAISAL MIZAN 7
