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HIGHER

5.21H Vectors: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 65 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Write down the vector from P to Q given P(2, -1) and Q( -3, 4).

(Total for Question 1 is 2 marks)

Given vectors a = (3, 1) and b = ( -2, 4), calculate 2a - 3b and give your answer as a column vector.

(Total for Question 2 is 3 marks)

Vectors p and q are shown on a grid. p = (4, 0) and q = (1, 3).
(a) Write down p + q.
(b) Find a scalar k such that kq = (4, 12).

(a)Write down p + q.(2)

(b)Find a scalar k such that kq = (4, 12).(2)

(Total for Question 3 is 4 marks)

On a diagram, A, B and C are points with position vectors relative to origin O: OA = (2, 1), OB = ( -1, 3), OC = (5, 4).
(a) Find the vector AB.
(b) Show that the three points are not collinear by using vectors.

(a)Find the vector AB.(2)

(b)Show that A, B and C are not collinear by using vectors.(3)

(Total for Question 4 is 5 marks)

Vectors u and v satisfy u = (t, 2) and v = (1, k). Given that u + 2v = (5, 8), find t and k.

(Total for Question 5 is 5 marks)

Points A and B have position vectors OA = (2, -1) and OB = (8, 5). A point X lies on AB such that AX : XB = 2 : 1.
(a) Find the position vector of X.
(b) Verify by vector calculation that X divides AB in the ratio 2 : 1.

(a)Find the position vector of X.(3)

(b)Verify by vector calculation that X divides AB in the ratio 2 : 1.(3)

(Total for Question 6 is 6 marks)

Vectors a and b satisfy 3a - 2b = (7, 1) and a + b = (3, 2).
(a) Solve for a and b.
(b) Hence find 2a - b.

(a)Solve for a and b.(4)

(b)Hence find 2a - b.(3)

(Total for Question 7 is 7 marks)

Show that the vectors (2, -1), (5, 2) and (7, 1) are coplanar in R² and then determine whether they are linearly dependent by finding constants x, y, z, not all zero, such that x(2, -1) + y(5, 2) + z(7, 1) = (0, 0). Give one nontrivial solution.

(Total for Question 8 is 8 marks)

Mark scheme · 5.21H Vectors: Higher Tier Practice

Question 1

M1 Subtract coordinates: Q - P to form vector PQ
A1 PQ = ( -5, 5) cao
Answer: (-5, 5)

Question 2

M1 Multiply vectors and form 2a and 3b components: 2a = (6,2), 3b = ( -6,12)
M1 Subtract components: (6 - ( -6), 2 - 12) or show (6+6, -10)
A1 (12, -10) cao as a column vector
Answer: [12; -10]

Question 3

(a) M1 Add components: (4+1, 0+3)
(a) A1 (5, 3) cao
(a) Answer: (5, 3)
(b) M1 Set k*(1,3) = (4,12) and compare components
(b) A1 k = 4 cao
(b) Answer: 4

Question 4

(a) M1 Use AB = OB - OA
(a) A1 AB = ( -3, 2) cao
(a) Answer: (-3, 2)
(b) M1 Compute AC = OC - OA = (3, 3) and note AB = ( -3, 2)
(b) M1 Attempt to find scalar t with AB = t*AC or AC = t*AB and show no single t fits both components
(b) A1 Conclude not collinear since components would need same ratio but -3/3 ≠ 2/3
(b) Answer: AC = (3, 3), AB = ( -3, 2); no scalar t satisfies AB = t AC since -3/3 ≠ 2/3, so points are not collinear.

Question 5

M1 Write u + 2v = (t,2) + 2(1,k) = (t + 2, 2 + 2k)
M1 Equate components to (5,8): t + 2 = 5 and 2 + 2k = 8
A1 t = 3 cao
M1 Solve 2 + 2k = 8 => 2k = 6
A1 k = 3 cao
Answer: t = 3, k = 3

Question 6

(a) M1 Use section formula: X = OA + (2/3)AB since AX is 2/3 of AB
(a) M1 Compute AB = OB - OA = (6,6)
(a) A1 OX = (2, -1) + (2/3)(6,6) = (2, -1) + (4,4) = (6,3) cao
(a) Answer: (6, 3)
(b) M1 Compute AX = OX - OA = (6,3) - (2,-1) = (4,4) and XB = OB - OX = (8,5) - (6,3) = (2,2)
(b) M1 Compare components to show AX = 2*XB in each component
(b) A1 Conclude AX : XB = 2 : 1 cao
(b) Answer: AX = (4,4), XB = (2,2); AX = 2 XB so ratio 2 : 1.

Question 7

(a) M1 Use system: from a + b = (3,2) get a = (3,2) - b and substitute into 3a - 2b
(a) M1 Substitute: 3[(3,2) - b] - 2b = (7,1) => (9,6) - 3b - 2b = (7,1)
(a) M1 Combine -5b = (7,1) - (9,6) = ( -2, -5) so b = ( ( -2)/ -5, ( -5)/ -5 )
(a) A1 b = (2/5, 1) and then a = (3,2) - b = (13/5,1) cao
(a) Answer: a = (13/5, 1), b = (2/5, 1)
(b) M1 Substitute a and b: 2a - b = 2(13/5,1) - (2/5,1)
(b) M1 Compute components: (26/5 - 2/5, 2 - 1) = (24/5,1)
(b) A1 2a - b = (24/5, 1) cao
(b) Answer: (24/5, 1)

Question 8

M1 Note that in R² any three vectors are coplanar; state this or show they lie in R²
M1 Set up equations for linear dependence: 2x + 5y + 7z = 0 and -x + 2y + z = 0
M1 Solve system: eliminate x using x = 2y + z from second equation or solve by elimination
M1 Substitute x = 2y + z into first: 2(2y+z) + 5y + 7z = 0 => 4y + 2z + 5y + 7z = 0
M1 Combine to get 9y + 9z = 0 so y = -z
M1 Choose z = 1 then y = -1 and x = 2y + z = 2( -1) + 1 = -1
A1 Provide nontrivial solution x = -1, y = -1, z = 1 and conclude vectors are linearly dependent
A1 State that the vectors are coplanar (in R²) and linearly dependent since a nonzero solution exists
Answer: Coplanar in R². One nontrivial solution is x = -1, y = -1, z = 1 since -1*(2,-1) + -1*(5,2) + 1*(7,1) = (0,0).

Vectors: Higher Tier Practice - Worksheets, Questions and Revision