P1 Vector Past Paper Questions

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1 P1 Vector Past Paper Questions Doublestruck & CIE - Licensed to Brillantmont International School 1

2 1. Relative to an origin O, the position vectors of the points A and B are given by OA = 2i + 3j k and OB = 4i 3j + 2k. (i) Use a scalar product to find angle AOB, correct to the nearest degree. (ii) Find the unit vector in the direction of AB. (iii) The point C is such that OC = 6j + pk, where p is a constant. Given that the lengths of AB and AC are equal, find the possible values of p. [3] Doublestruck & CIE - Licensed to Brillantmont International School 2

3 2. Relative to an origin O, the position vectors of the points A and B are given by OA = 2i 8j + 4k and OB = 7i + 2j k. (i) Find the value of OA. OB and hence state whether angle AOB is acute, obtuse or a right angle. [3] 2 (ii) The point X is such that AX = AB. Find the unit vector in the direction of OX. 5 Doublestruck & CIE - Licensed to Brillantmont International School 3

4 3. Relative to an origin O, the position vectors of points P and Q are given by OP 2 2 = 3 and OQ = 1, 1 q where q is a constant. (i) In the case where q = 3, use a scalar product to show that cos POQ = (ii) Find the values of q for which the length of PQ is 6 units [3] Doublestruck & CIE - Licensed to Brillantmont International School 4

5 4. D 6 m E B 8 m C j k O i 14 m A The diagram shows the roof of a house. The base of the roof, OABC, is rectangular and horizontal with OA = CB = 14 m and OC = AB = 8 m. The top of the roof DE is 5 m above the base and DE = 6 m. The sloping edges OD, CD, AE and BE are all equal in length. Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. (i) Express the vector OD in terms of i, j and k, and find its magnitude. (ii) Use a scalar product to find angle DOB. Doublestruck & CIE - Licensed to Brillantmont International School 5

6 The position vectors of points A and B are 6 and respectively, relative to an origin O. (i) Calculate angle AOB. [3] (ii) The point C is such that AC = 3 AB. Find the unit vector in the direction of OC. Doublestruck & CIE - Licensed to Brillantmont International School 6

7 6. Relative to an origin O, the position vectors of the points A and B are given by OA 4 3 = 1 and OB = (i) Given that C is the point such that AC = 2 AB, find the unit vector in the direction of OC. 1 The position vector of the point D is given by OD = 4, where k is a constant, and it is given k that OD moa nob, where m and n are constants. (ii) Find the values of m, n and k. Doublestruck & CIE - Licensed to Brillantmont International School 7

8 7. The diagram shows a cube OABCDEFG in which the length of each side is 4 units. The unit OA, OC OD vectors i, j and k are parallel to and respectively. The mid-points of OA and DG are P and Q respectively and R is the centre of the square face ABFE. (i) Express each of the vectors PR and PQ in terms of i, j and k. [3] (ii) (iii) Use a scalar product to find angle QPR. Find the perimeter of triangle PQR, giving your answer correct to 1 decimal place. [3] Doublestruck & CIE - Licensed to Brillantmont International School 8

9 8. Relative to an origin O, the position vectors of points A and B are 2i + j + 2k and 3i 2j + pk respectively. (i) (ii) Find the value of p for which OA and OB are perpendicular. In the case where p = 6, use a scalar product to find angle AOB, correct to the nearest degree. [2] [3] (iii) Express the vector AB is terms of p and hence find the values of p for which the length of AB is 3.5 units. Doublestruck & CIE - Licensed to Brillantmont International School 9

10 9. F E 8 cm P B M C k A 6 cm O j i D N 20 cm The diagram shows a semicircular prism with a horizontal rectangular base ABCD. The vertical ends AED and BFC are semicircles of radius 6 cm. The length of the prism is 20 cm. The mid-point of AD is the origin O, the mid-point of BC is M and the mid-point of DC is N. The points E and F are the highest points of the semicircular ends of the prism. The point P lies on EF such that EP = 8 cm. Unit vectors i, j and k are parallel to OD, OM and OE respectively. (i) Express each of the vectors PA and PN in terms of i, j and k. [3] (ii) Use a scalar product to calculate angle APN. Doublestruck & CIE - Licensed to Brillantmont International School 10

11 Solutions Doublestruck & CIE - Licensed to Brillantmont International School 11

12 1. OA = 2i + 3j k OB = 4i 3j + 2k (i) OA. OB = = 3 Correct use of a 1 a 2 + b 1 b 2 + c 1 c 2 OA. OB = cos AOB Modulus. Correct use of abcos θ AOB = 99 o CAO A14 (ii) AB = b a = 2i 6j + 3k B1 CAO Magnitude of AB = 49 = 7 Use of Pythagoras + division. 1 Unit vector = (2i 6j + 3k) A1 3 7 CAO (use of BA for AB has max 2/3). (iii) AC = 2i + 3j +(p + 1)k B1 CAO - condone a c here (p + 1) 2 = 49 A1 Correct method for forming an equation p = 5 or 7 CAO A14 [11] Doublestruck & CIE - Licensed to Brillantmont International School 12

13 2. (i) OA.OB = = 6 A1 Must be scalar from correct method. (ii) This is ve Obtuse angle. co. Correct deduction from his scalar. AB = 5i + 10j 5k 2 AX = (AB) 5 OX = OA + AX Needs AB and OX attempting. OX = 4i 4j + 2k co Divides by the modulus Must finish with a vector, not a scalar. B1 3 A1 1 Unit vector = (4i 4j + 2k) A1 4 6 Correct for his OX. [7] 2 3. OP 3 and 1 2 OQ 1 q 2 2 (i) 3. 1 with q = 3, = =2 1 q Use of a 1 a 2 +b 1 b 2 +c 1 c 2. 1 = cos θ = 2, cos θ = 7 Dot product linked with moduli and cos. co A13 4 (ii) PQ = q p = 2 q 1 Allow for p q or q p (q 1) 2 = 36 A1 Use of modulus and Pythagoras q = 5 or q = 3 Co (for both) A14 [7] Doublestruck & CIE - Licensed to Brillantmont International School 13

14 Doublestruck & CIE - Licensed to Brillantmont International School 14

15 4. (i) Vector OD = 4i + 4j + 5k B2, 1 One off for each error. Column vectors ok Magnitude = = 57 Correct use of Pythagoras. Magnitude = 7.55 m Accept 57. A14 (ii) Vector OB = 14i + 8j B1 co OD.OB = = 88 Use of x 1 x 2 + y 1 y 2 + z 1 z 2 for his vectors OD.OB = cosθ Used correctly Angle DOB = 43.7 co A14 [8] Doublestruck & CIE - Licensed to Brillantmont International School 15

16 a = 6 b = Ok to work throughout with column vectors or with i.j.k (i) a,b = = 27 Use of x 1 x 2 + y 1 y 2 +z 1 z 2 a.b = 54 21cosθ Use of cos θ θ = 36.7 or radians In either degrees or in radians. A13 (ii) Vector AB = b α 2 = 4 1 For use of b a Vector OC = = For a + 3(b a) or equivalent Unit vector = Vector OC + 9. For division by Modulus of OC. = 1 i 2 j + 2 k A Co. [7] Doublestruck & CIE - Licensed to Brillantmont International School 16

17 6. (i) AB 1 2 = 1 and AC = For ± (b a) (not b + a) OC = OA + AC 2 = 3 6 A1 Co 2 1 Unit vector = Division by the modulus for his OC A1 4 (ii) m 1 + n 2 = k 4m + 3n = 1 and m + 2n = 4 Forming 2 simultaneous equations m = 2 and n = 3 A1 co k = 8 A1 4 Equation for k in terms of m and n. co [8] Doublestruck & CIE - Licensed to Brillantmont International School 17

18 (i) PR 2 PQ = 2 B1 2 4 PR All elements of any notation ok. Loses one mark for each error in PQ B2, 13 (ii) PQ. PR = = 8 Must be scalar PQ = 24 PR = 12 As long as this is used with dot product PQ PR = cos QPR Everything linked ( QP. PR used still gains all M marks) Angle QPR = 61.9 or 1.08 rad Co A14 4 (iii) QR = 0 QR = 20 2 For correct QR cosine rule ok. Perimeter = = 12.8 cm Adds three roots. Co beware fortuitous answers from incorrect sign in vectors. A13 [10] Doublestruck & CIE - Licensed to Brillantmont International School 18

19 8. OA = 2i + j + 2k, OB = 3i 2j + pk (i) (2i + j + 2k). (3i 2j + pk) = 0 For x 1 x 2 + y 1 y 2 + z 1 z 2 (in (i) or (ii)) p = 0 p = 2 A12 co (ii) (2i + j + 2k). (3i 2j + 6k) nb Part (ii) gains 4 marks if (i) missing allow for ± this co ( here if (i) not done) A1 = 9 49 cos θ All connected correctly θ = 40 A13 co (iii) AB = i 3j + (p 2)k B1 Must be for AB, not BA (p 2) 2 = Pythagoras (allow if wrong once) Method of solution. p = 0.5 or 3.5 co (use of BA can score the last 3 marks) D A14 [9] 9. (i) PA = 6i 8j 6k B1 Co column vectors ok PN = 6i + 2j 6k B2, 1 3 One off for each error (all incorrect sign just one error) Doublestruck & CIE - Licensed to Brillantmont International School 19

20 (ii) PA.PN = = 16 Use of x 1 x 2 + y 1 y 2 + z 1 z 2 16 cos APN = Modulus worked correctly for either one Division of 16 by product of moduli APN = 99 Allow more accuracy A14 [7] Doublestruck & CIE - Licensed to Brillantmont International School 20

PAST QUESTIONS ON VECTORS P1

PAST QUESTIONS ON VECTORS P1 1. Diagram shows a solid cylinder standing on a horizontal circular base, centre O and radius 4 units. The line BA is a diameter and the radius OC is at 90 o to OA. Points