T2W10+Vectors+1+%28Soln%29
Uploaded by blahblahblah03 · 18 October 2025
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RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 6 _____________________________________ Y6 H2 Math T2W10 Math Focus: Vectors 1 Page 1 of 4 Term 2 Week 10 Math Focus Topic: Vectors 1 1 9740/2016/01/Q5 The vectors u and v are given by u = 2i – j + 2k and v = ai + bk, where a and b are constants. (i) Find (u + v) × (u – v) in terms of a and b. [2] (ii) Given that the i- and k-components of the answer to part (i) are equal, express (u + v) × (u – v) in terms of a only. Hence find, in an exact form, the possible values of a for which (u + v) × (u – v) is a unit vector. [4] (iii) Given instead that (u + v).(u – v) = 0, find the numerical value of |v|. [2] Qn 1 Solution (i) ( ) ( ) ( ) 2 (Since , and ) 2 2 0 1 2 2 2 2 2 4 4 2 a b b b a b b a a a +− = + − − = = = =− = − = − = + − − − u v u v u u v u u v v v v u u u 0 v v 0 u v v u i j k (ii) ( ) ( ) ( ) ( ) Given that 1 2 4 2 4 1 111 2 18 1 2 18 6 2 ba a aa a aa =− −− + − = − = − −− + − = = = = u v u v u v u v (iii) ( ) ( ) ( ) 22 222 . 0 . . 0 0 2 1 2 3 + − = − = − = = = + − + = u v u v u u v v uv vu
Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________ _____________________________________ Y6 H2 Math T2W10 Math Focus: Vectors 1 Page 2 of 4 2 9758/2018/01/Q6 Vectors a, b and c are such that a 0 and 3 2 .a b = a c (i) Show that 3 2 , −=b c a where is a constant. [2] (ii) It is now given that a and c are unit vectors, that the modulus of b is 4 and that the angle between b and c is 60 . Using a suitable scalar product, find exactly the two possible values of . [5] Qn 2 Solution (i) ( ) ( ) 3 2 3 2 22 / / 3 2 or 3 2 Thus, 3 2 , where is a constant. − = − = − = − − = −= a b c a b a c a c a c 0 a b c b c 0 b c a (ii) ( ) ( ) 2 2 2 2 2 . cos 60 2 3 2 . 3 2 . 9 12 . 4 144 24 4 124 2 31 == − − = − + = − + = = = b c b c b c b c a a b b c c a
Raffles Institution H2 Mathematics 2025 Year 6 _________________________________________________________________________________________ _____________________________________ Y6 H2 Math T2W10 Math Focus: Vectors 1 Page 3 of 4 3 9758/2019/02/Q5 With reference to the origin O. the points A, B, C and D are such that OA= a , OB= b , 24OC=+ ab and 5OD=+ba . The lines BD and AC cross at X (see diagram). (i) Express OX in terms of a and b. [4] The point Y lies on CD and is such that the points O. X and Y are collinear. (ii) Express OY in terms of a and b and find the ratio OX : OY. [6] Qn 3 Solution (i) ( ) ( ) ( ) ( ) ( ) ( ) 5 2 4 5 14 5 1 4 Since , and is not parallel to , 141 4 51 OX OB BX OX OA AX BD AC OD OB OC OA = +
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