Vector proof Quiz (Grade 8+)

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Score: 0 / 10

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Explanation:

To find a vector between two points, trace a known path along the diagram, reversing the sign of a vector when travelling against its arrow (e.g. AB = AO + OB = −a + b = b − a). If a point divides a line in the ratio m : n, use the fraction mm + n of the full vector.

To prove two vectors are parallel, factorise both so they are scalar multiples of the same bracketed vector (e.g. PQ = k × RS). If they also share a common point (such as P, Q, R where PQ = k × QR), then the three points are collinear (lie on a straight line).

Example 1: In ΔOAB, OA = a and OB = b. M is the midpoint of OA and N divides OB in the ratio 1 : 2. Find MN in terms of a and b.

Solution:

MO = −12a; ON = 13b (ratio 1 : 2)

MN = MO + ON ⇒ Answer: 13b − 12a

Example 2: Given PQ = 2a + 4b and QR = 3a + 6b, prove that P, Q, and R lie on a straight line.

Solution:

Factorise both vectors:

PQ = 2(a + 2b) and QR = 3(a + 2b)

Therefore QR = 32PQ, so PQ and QR are parallel

Since they also share the common point Q, P, Q, and R are collinear.

One Page Guide (PDF) Vector proof Worksheets

Question 1

[1 mark]

Given that AB→ = 3a + 3b and BC→ = 9a + 9b, express AC→ in terms of a and b and find the ratio AB : BC.

Worked Solution & Answer

BC→ = 3AB→, so A, B, C are collinear with AB : BC = 1 : 3. B1

Answer: AC→ = 12a + 12b, AB : BC = 1 : 3

Question 2

[1 mark]

Given that AB→ = 3a + 3b and BC→ = 6a + 6b, express AC→ in terms of a and b and find the ratio AB : BC.

Worked Solution & Answer

BC→ = 2AB→, so A, B, C are collinear with AB : BC = 1 : 2. B1

Answer: AC→ = 9a + 9b, AB : BC = 1 : 2

Question 3 AQA Nov 2022 2H Q25

[3 marks]

Two congruent parallelograms, PQRV and VRST, are joined.

QP→ = a PV→ = b
X is the midpoint of VT.
VW : WR = 1 : 2
Prove that Q, W and X lie on a straight line.

Worked Solution & Answer

VR = PQ = –a M1
QW = QP + PV + VW = a + b – 1/3 a = 2/3 a + b
QX = QP + PV + VX = a + b + 1/2 b = a + 3/2 b M1
QX = 3/2 (2/3 a + b) = 3/2 QW
Since QX is a multiple of QW and Q is a common point, Q, W and X are collinear. C1

Answer: QW = a + b – 1/3 a = 2/3 a + b.
QX = a + 3/2 b = 3/2 (2/3 a + b) = 3/2 QW.
Since QX is a scalar multiple of QW and they share point Q, Q, W and X lie on a straight line.

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Question 4 AQA Nov 2023 1H Q28

[3 marks]

PQRS is a quadrilateral.
PQ is not parallel to SR.
X is a point on QR.
QX : XR = 2 : 3
QX→ = 2a + 4b

Prove that PQRS is a trapezium.

Worked Solution & Answer

Since QX : XR = 2 : 3, QR = (5/2)QX = (5/2)(2a + 4b) = 5a + 10b M1
PS = PQ + QR + RS = –5a + (5a + 10b) + (a – 8b) = a + 2b M1
Since QR = 5(a + 2b) = 5PS, QR is parallel to PS, so PQRS has a pair of parallel sides and is a trapezium. C1

Answer: QR = 5/2(2a + 4b) = 5a + 10b
PS = –5a + (5a + 10b) + (a – 8b) = a + 2b
QR = 5PS so QR is parallel to PS, hence PQRS is a trapezium

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Question 5 AQA Nov 2019 3H Q27

[3 marks]

Here is quadrilateral PQRS.
PS = a, SR = b, RQ = c
X is a point on PS where PX : XS = 1 : 2
Y is a point on RQ where RY : YQ = 2 : 1

Is XY parallel to PQ?
Show working to support your answer.

Conclusion

Worked Solution & Answer

PQ = PS + SR + RQ = a + b + c M1
From PX : XS = 1 : 2, XS = 23a
From RY : YQ = 2 : 1, RY = 23c.
XY = XS + SR + RY = 23a + b + 23c
For XY to be parallel to PQ, XY must be a scalar multiple kPQ = k(a + b + c) M1
Comparing coefficients:
k = 23 for a and c, but k = 1 for b
Since the coefficients are not in the same ratio, XY is not a multiple of PQ
Hence XY is not parallel to PQ. A1

Answer: Conclusion No (vector XY is not a scalar multiple of vector PQ)

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Question 6 AQA Jun 2024 2H Q23

[4 marks]

Vector diagram with DC = 6a + b, CE = 2a – 5b, CF = 4a – 6b. Prove that DEF is a straight line. [4 marks]

Prove that DEF is a straight line.

Worked Solution & Answer

DE = DC + CE = (6a + b) + (2a – 5b) = 8a – 4b = 4(2a – b) M2
EF = EC + CF = –(2a – 5b) + (4a – 6b) = 2a – b M1
Since DE = 4EF, DE and EF are parallel and share common point E, so DEF is a straight line. C1

Answer: DE = 8a – 4b = 4(2a – b) and EF = 2a – b, so DE = 4EF; parallel and share point E, so DEF is a straight line

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Question 7 AQA Jun 2019 1H Q22

[4 marks]

ABC and ACD are triangles.
k is a constant.

(a) Show that CD = 6a + 4.5b [1 mark]

(b) BCD is a straight line.
Work out the value of k.
You must show your working. [3 marks]

Answer

Worked Solution & Answer

(a)

CD = CA + AD = -3b + (6a + 7.5b) = 6a + 4.5b C1

Answer (a): Answer 6a + 4.5b

(b)

BC = BA + AC = ka + 3b M1
Since BCD is a straight line, BC is a scalar multiple of CD = 6a + 4.5b M1
Comparing the b coefficients: the scalar factor is 34.5 = 23
Therefore, ka = 23(6a) = 4a ⇒ k = 4. A1

Answer (b): Answer 4

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Question 8

[5 marks]

A, B and C are three points such that
[INDENT]AB→ = 3a + 4b
[INDENT]AC→ = 15a + 20b

(a) Prove that A, B and C lie on a straight line. [2 marks]

(b) Find the ratio
[INDENT]length of DF : length of DE [3 marks]

Worked Solution & Answer

(a)

AC = 5(3a + 4b) = 5AB M1 C1

Answer (a): AC→ = 5(3a + 4b) = 5AB→, so AC→ is a scalar multiple of AB→ (parallel) and they share point A

(b)

DF = DE + EF = (3 – 10.5)e + (6 – 21)f = –7.5e – 15f = –2.5(3e + 6f) = –2.5 DE M2
⇒ |DF| : |DE| = 2.5 : 1 = 5 : 2 A1

Answer (b): 5 : 2

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Question 9

[5 marks]

The diagram shows quadrilateral OACB.
M is the midpoint of OA.
N is the point on BC such that BN : NC = 4 : 5
OA→ = a OB→ = b AC→ = kb where k is a positive integer.

(a) Express MN→ in terms of k, a and b.
Give your answer in its simplest form. [4 marks]

(b) Is MN parallel to OB?
Give a reason for your answer. [1 mark]

Worked Solution & Answer

(a)

BC = BO + OA + AC = –b + a + kb M2
BN = (4/9)BC = (4/9)(–b + a + kb) M1
MN = MO + OB + BN = –½a + b + (4/9)(a – b + kb) = –(1/18)a + (5/9)b + (4/9)kb = (1/18)(8kb + 10b – a) A1

Answer (a): 118(8kb + 10b - a)

(b)

No, because MN contains a non-zero multiple of a so it is not a scalar multiple of b

Answer (b): No, because MN contains a non-zero multiple of a so it is not a scalar multiple of b

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Question 10 AQA Nov 2021 3H Q27

[6 marks]

In the diagram
DE = a
DH = b
HG = 8b
EX : XH = 3 : 1
EF : FG = 1 : 3

(a) Show that DX = 14a + 34b [2 marks]

(b) Is DXF a straight line?
Show working to support your answer. [4 marks]

Worked Solution & Answer

(a)

EH = ED + DH = -a + b M1
Since EX : XH = 3 : 1, EX = 34EH = 34(-a + b).
DX = DE + EX = a + 34(-a + b) = 14a + 34b. C1

(b)

DG = DH + HG = b + 8b = 9b. M1
EG = ED + DG = -a + 9b M1
Since EF : FG = 1 : 3, EF = 14EG = 14(-a + 9b).
DF = DE + EF = a + 14(-a + 9b) = 34a + 94b M1
Notice that DF = 3(14a + 34b) = 3DX
Since DF is a scalar multiple of DX and they share the common point D, DXF is a straight line. A1

Answer (b): Yes, DF = 3 DX, so DX and DF are parallel and share point D, meaning DXF is a straight line

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Score: 0 / 10 (0 / 35 marks)

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Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum