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1. (a) 2y + 8x = 4 –3x + 2y = –7 2x + 6 – 2x = 6 M1 A1 Note: Award M1 for attempt at components, A1 for two correct equations. No penalty for not checking the third equation. solving : x = 1, y = –2 (b) A1 4 4 │a + 2b│= 3 2 2 2 2 4 = 7 6 a 2b 4 2 (7) 2 6 2 = (M1) 101 A1 [5] 2. (a) (b) AB = b – a A1 CB = a + b A1 AB CB = (b – a)•(b + a) = │b│2–│a│2 = 0 since │b│=│a│ M1 A1 R1 Note: Only award the A1 and R1 if working indicates that they understand that they are working with vectors. so AB is perpendicular to CB i.e. AB̂C is a right angle AG [5] 3. (a) 4 Finding correct vectors AB 3 1 3 AC 1 1 Substituting correctly in scalar product AB AC = 4(–3) + 3(1) – 1(1) = –10 IB Questionbank Mathematics Higher Level 3rd edition A1A1 A1 AG N0 1 (b) AB 26 AC 11 Attempting to use scalar product formula, cos BÂC (A1)(A1) 10 M1 26 11 = –0.591 (to 3 s.f.) BÂC = 126° A1 A1 N3 [8] 4. (a) OP = i + 2j – k the coordinates of P are (1, 2, –1) (b) EITHER x = 1 + t, y = 2 – 2t, z = 3t – 1 y2 z 1 x – 1 = t, t, t 2 3 y 2 z 1 x–1= 2 3 (M1) A1 M1 A1 AG N0 OR x 1 1 y 2 t 2 z 1 3 x–1= (c) y 2 z 1 2 3 (i) 2(1 + t) + (2 – 2t) + (3t – 1) = 6 t = 1 (ii) coordinates are (2, 0, 2) M1A1 AG M1A1 N1 A1 Note: Award A0 for position vector. (iii) distance travelled is the distance between the two points (2 1) 2 (0 2) 2 (2 1) 2 14 (= 3.74) IB Questionbank Mathematics Higher Level 3rd edition (M1) (M1)A1 2 (d) (i) distance from Q to the origin is given by d(t) = t 4 (1 t ) 2 (1 t 2 ) 2 (or equivalent) 2 e.g. for labelled sketch of graph of d or d (ii) M1A1 (M1)(A1) the minimum value is obtained for t = 0.761 A1 the coordinates are (0.579, 0.239, 0.421) A1 N3 Note: Accept answers given as a position vector. (e) (i) (ii) 0 1 4 a 1 , b 0 and c 1 1 0 3 substituting in the equation a – b = k(b – c), we have 1 4 0 1 1 3 1 0 k 0 1 1 k 1 0 3 1 0 1 3 1 k 1 and k which is impossible 3 so there is no solution for k BA and CB are not parallel (hence A, B, and C cannot be collinear) (M1)A1 (M1) A1 R1 R2 Note: Only accept answers that follow from part (i). [23] IB Questionbank Mathematics Higher Level 3rd edition 3 5. (a) Use of cos = OA AB (M1) OA AB AB = i j + k A1 AB = 3 and OA = 3 2 A1 OA AB = 6 A1 substituting gives cos = (b) 6 or equivalent 6 3 2 M1 N1 L1: r = OA + s AB or equivalent L1: r = i j + 4k + s(i j + k) or equivalent (M1) A1 Note: Award (M1)A0 for omitting “r =” in the final answer. (c) Equating components and forming equations involving s and t 1 + s = 2 + 2t, 1 s = 4 + t, 4 + s = 7 + 3t Having two of the above three equations Attempting to solve for s or t Finding either s = 3 or t = 2 Explicitly showing that these values satisfy the third equation Point of intersection is (2, 2, 1) (M1) A1A1 (M1) A1 R1 A1 N1 Note: Position vector is not acceptable for final A1. (d) METHOD 1 1 2 3 r = 1 1 3 4 3 3 x = 1 + 2 3, y = 1 + + 3 and z = 4 + 3 3 Elimination of the parameters x + y = 3 so 4(x + y) = 12 and y + z = 4 + 3 so 3(y + z) = 12 + 9 (A1) M1A1 M1 3(y + z) = 4(x + y) + 9 A1 Cartesian equation of plane is 4x + y 3z = 9 (or equivalent) A1 IB Questionbank Mathematics Higher Level 3rd edition N1 4 METHOD 2 EITHER The point (2, 4, 7) lies on the plane. The vector joining (2, 4, 7) and (1, 1, 4) and 2i + j + 3k are parallel to the plane. So they are perpendicular to the normal to the plane. (i j + 4k) (2i + 4j + 7k) = i 5j 3k i j k n 1 5 3 2 (A1) 1 M1 3 = 12i 3j + 9k or equivalent parallel vector A1 OR L1 and L2 intersect at D (2, 2,1) AD = (2i + 2j + k) (i j + 4k) = 3i + 3j 3k i j k n 2 1 3 (A1) M1 3 3 3 = 12i 3j + 9k or equivalent parallel vector A1 THEN r • n = (i j + 4k) • (12i 3j + 9k) = 27 Cartesian equation of plane is 4x + y 3z = 9 (or equivalent) M1 A1 A1 N1 [20] IB Questionbank Mathematics Higher Level 3rd edition 5 6. METHOD 1 for finding two of the following three vectors (or their negatives) 0 2 2 AB 2 , AC 2 , BC 0 1 2 1 and calculating (A1)(A1) EITHER 2 AB AC 0 2 1 2 2 2 2 4 i j k 1 AB AC 2 area ∆ABC = M1A1 M1 OR 2 BA BC 0 2 1 2 2 0 1 4 i area ∆ABC = j k 1 BA BC 2 M1A1 M1 OR 2 CA CB 2 2 2 2 2 0 1 4 i area ∆ABC = j k 1 CA CB 2 M1A1 M1 THEN area ∆ABC = = 24 2 6 IB Questionbank Mathematics Higher Level 3rd edition A1 AG N0 6 METHOD 2 for finding two of the following three vectors (or their negatives) 0 2 2 AB 2 , AC 2 , BC 0 1 2 1 (A1)(A1) EITHER cos A = AB AC M1 AB AC = 6 5 12 6 3 or 60 15 2 5 sin A = A1 1 AB AC sin A 2 1 2 = 5 12 2 5 1 = 24 2 = 6 area ∆ABC = M1 A1 AG N0 OR cos B = BA BC M1 BA BC 1 = 5 5 1 5 24 24 or 25 5 1 area ∆ABC = BA BC sin B 2 1 24 = 5 5 2 25 1 = 24 2 = 6 sin B = IB Questionbank Mathematics Higher Level 3rd edition A1 M1 A1 AG N0 7 OR cos C = CA CB M1 CA CB 6 = 12 5 sin C = 6 3 or 60 15 2 5 1 CA CB sin C 2 1 2 = 12 5 2 5 1 = 24 2 = 6 area ∆ABC = A1 M1 A1 AG N0 METHOD 3 for finding two of the following three vectors (or their negatives) 0 2 2 AB 2 , AC 2 , BC 0 1 2 1 AB = 5 c, AC 12 2 3 b, BC 5 a 52 3 5 3 5 2 area ∆ABC = s( s a)(s b)(s c) s= (A1)(A1) M1A1 M1 = ( 3 5 )( 3 )( 5 3 )( 3 ) = 3(5 3) A1 = 6 AG IB Questionbank Mathematics Higher Level 3rd edition N0 8 METHOD 4 for finding two of the following three vectors (or their negatives) 0 2 2 AB 2 , AC 2 , BC 0 1 2 1 AB = BC = 5 and AC = ∆ABC is isosceles 12 2 3 let M be the midpoint of [AC], the height BM = area ∆ABC = = 2 3 2 2 6 (A1)(A1) M1A1 53 2 M1 A1 AG N0 [6] IB Questionbank Mathematics Higher Level 3rd edition 9 7. (a) EITHER normal to plane given by i j k 2 3 2 6 3 2 = 12i + 8j – 24k equation of π is 3x + 2y – 6z = d as goes through (–2, 3, –2) so d = 12 π :3x + 2y – 6z = 12 M1A1 A1 (M1) M1A1 AG OR x = –2 + 2λ + 6μ y = 3 + 3λ – 3μ z = –2 + 2λ + 2μ (b) eliminating μ x + 2y = 4 + 8λ 2y + 3z = 12λ eliminating λ 3(x + 2y) – 2(2y + 3z) = 12 π : 3x + 2y – 6z = 12 M1A1A1 therefore A(4, 0, 0), B(0, 6, 0) and C(0, 0, –2) A1A1A1 M1A1A1 AG Note: Award A1A1A0 if position vectors given instead of coordinates. (c) area of base OAB = V= (d) (e) 1 4 6 = 12 2 1 12 2 = 8 3 3 1 2 0 = 3 = 7 × 1 × cos 6 0 3 arccos 7 3 so θ = 90 – arccos = 25.4° (accept 0.443 radians) 7 d = 4 sin θ = 12 (= 1.71) 7 IB Questionbank Mathematics Higher Level 3rd edition M1 M1A1 M1A1 M1A1 (M1)A1 10 (f) 8= 1 12 area area = 14 3 7 M1A1 Note: If answer to part (f) is found in an earlier part, award M1A1, regardless of the fact that it has not come from their answers to part (c) and part (e). [20] 8. EITHER l goes through the point (1, 3, 6), and the plane contains A(4, –2, 5) the vector containing these two points is on the plane, i.e. 1 4 3 3 2 5 6 5 1 i j k 1 3 2 5 1 2 1 = 7i + 4j + k 1 1 3 5 1 4 7 2 4 25 5 1 hence, Cartesian equation of the plane is 7x + 4y + z = 25 (M1)A1 M1A1 (M1) A1 OR finding a third point e.g. (0, 5, 5) three points are (1, 3, 6), (4, –2, 5), (0, 5, 5) equation is ax + by + cz = 1 system of equations a + 3b + 6c = 1 4a – 2b + 5c = 1 5b + 5c = 1 7 4 1 a= , from GDC ,b ,c 25 25 25 7 4 1 so x y z 1 25 25 25 or 7x + 4y + z = 25 M1 A1 M1 M1A1 A1 [6] IB Questionbank Mathematics Higher Level 3rd edition 11 9. (a) (i) METHOD 1 1 1 0 AB = b a = 2 1 1 3 2 1 (A1) 3 1 2 AC = c a = 0 1 1 1 2 1 (A1) i j k 1 1 2 1 1 AB AC = 0 M1 = i (1 + 1) j(0 2) + k (0 2) (A1) = 2j 2k Area of triangle ABC = A1 1 1 2 j 2k 8 2 sq. units 2 2 M1A1 Note: Allow FT on final A1. METHOD 2 AB 2 , BC 12 , AC 6 A1A1A1 Using cosine rule, e.g. on Ĉ 6 12 2 cos C = 2 72 Area ABC M1 2 2 3 A1 1 absin C 2 = 1 12 6 sin 2 M1 arccos 2 2 3 2 2 2 = 3 2 sin arccos 3 A1 Note: Allow FT on final A1. (ii) AB = 2 = 2 1 1 AB h 2 h , h equals the shortest distance 2 2 h=2 IB Questionbank Mathematics Higher Level 3rd edition A1 (M1) A1 12 (iii) METHOD 1 0 has form r • 2 d 2 (M1) Since (1, 1, 2) is on the plane 1 1 d = 1 • 2 24 2 2 2 M1A1 0 Hence r • 2 = 2 2 2y 2z = 2 (or y z = 1) A1 METHOD 2 1 0 2 r = 1 λ 1 μ 1 2 1 1 x = 1 + 2 (i) y=1+ (ii) z=2+ (iii) (M1) A1 Note: Award A1 for all three correct, A0 otherwise. From (i) = x 1 2 x 1 2 substitute in (ii) y = 1 + x 1 2 =y1+ substitute and in (iii) M1 x 1 x 1 2 2 z=2+y1+ y z = 1 IB Questionbank Mathematics Higher Level 3rd edition A1 13 (b) (i) The equation of OD is 0 r = 2 , 2 0 or r λ 1 1 M1 This meets where 2 + 2 = 1 = (M1) 1 4 A1 1 1 Coordinates of D are 0 , , 2 2 (ii) 2 2 1 1 1 OD 0 2 2 2 A1 (M1)A1 [20] 10. consider a vector parallel to each line, 4 3 e.g. u = 2 and v 3 1 1 let θ be the angle between the lines u v 12 6 1 cos θ = uv 21 19 7 = = 0.350... 21 19 7 so 0 = 69.5° or 1.21 rad or arccos 21 19 A1A1 M1A1 (A1) A1 N4 Note: Allow FT from incorrect reasonable vectors. [6] IB Questionbank Mathematics Higher Level 3rd edition 14 11. METHOD 1 (from GDC) 1 1 1 0 6 12 2 1 0 1 3 6 0 0 0 0 (M1) 1 1 x λ 6 12 A1 2 1 y λ 3 6 A1 1 1 r = i 6 12 2 1 j i j k 3 6 A1A1A1 N3 M1A1 (M1)A1 M1 A1 N3 METHOD 2 (Elimination method either for equations or row reduction of matrix) Eliminating one of the variables Finding a point on the line Finding the direction of the line The vector equation of the line [6] IB Questionbank Mathematics Higher Level 3rd edition 15