RI Yr 5 TP H2 Math Topical Revision (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 _____________________ 2025 Y5 TP Topical Revision Page 1 of 77 Year 5 H2 Math Timed Practice Topical Revision: Questions and Solutions You should check these solutions only after you have attempted the questions in this exercise. C2: Vectors 1 ACJC Promo 9758/2022/Q7 modified (a) Referred to the origin O, points A and B have position vectors a and b respectively, where a and b are non-zero and non parallel vectors . Point C lies on OA, between O and A, such that OC : CA = 2 : 1. Point D lies on OB produced such that 5BD OB . Find, in terms of a and b, the position vector of the point E where the lines AB and CD meet. [3] (b) The position vectors of the points D and E relative to the origin O are d and e respectively, where d and e are non-zero and non parallel vectors. It is given that the length OE is 2 units and 3 d e . The point F, with position vector f, is the reflection of the point D in the line OE. (i) Express f in terms of vectors d and e. [3] (ii) Show that the area of triangle ODF can be expressed as k d e , where k is a constant to be determined. [2] [(a) 5 3+8 8a b (bi) 3 2 e d (bii) 3 4 ] Solutions (a) OC : CA = 2 : 1 2 3OC  a   5BD OB 6OD  b       on line for some 1 E AB OE OA AB                a b a a b       

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________ 2025 Y5 TP Topical Revision Page 2 of 77   on line for some 2 2 6 3 3 2 1 63 E CD OE OC CD                   a b a a b           21 1 6 3       a b a b Since a and b are non-zero and non-parallel,    21 1 and 63 6 1,16 16 6 6 5 31 +16 16 8 8OE                     a b a b   (bi) Let the foot of perpendicular from point D to line OE be N, with position vector n. N is on line OE for some    n e  DN is perpendicular to OE 0DN  . e    0  e d . e 0  e. e d. e 2 3 4  d. e e 3 4 n e  1 2 3 32 2 4 2         n d f f n d e d e d (bii) Area of triangle ODF = 1 3 2 2 3 4 3 since 4               d e d d e d d d e d d 0 OR   1 1 32 22 2 2 2 4ON DN               d e d e d e

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________ 2025 Y5 TP Topical Revision Page 3 of 77 2 CJC Promo 9758/2022/Q9 Relative to the origin O, the position vectors of points A, B and C are a , b and c respectively, where b is a unit vector. It is given that b and b a are perpendicular and C lies on AB such that : 3 :1AC CB  . (i) Show that seca , where  is the angle between a and b . [3] By expressing c in terms of a and b, (ii) find the value of c b and state the geometrical interpretation of c b , [4] (iii) find the value of   b c a b . [2] [(ii) 1 (iii) 1 4 ] Solutions (i) Method : Let  be the angle between a and b . Since b and b a  are perpendicular,   2 b b a 0 b b b a 0 b b a a b                        Since b is a unit vector, a b 1   a b cos 1   a cos 1 1a seccos        Method : cos 1 since is a unit vector 1 sec (shown)cos OB OA b a ba a            O B A θ

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________ 2025 Y5 TP Topical Revision Page 4 of 77 (ii) By Ratio Theorem, 1 3c a b4 4    2 1 3c b a b b4 4 1 3a b b b4 4 1 3= b b b b4 4 b 1                              since a b b b      c b  is the length of projection of c on b . (iii) Method :   1 3b a bb c 4 4 a b a b 1 b a4 , where b b 0a b a b1= , where b a a b4 a b 1 4                                 Method : 1 b cb c 2 1a b a b2 area of area of 1 2 1 2 1 4 OBC OAB BC h AB h               O B A θ C 3 1 O B A θ C 1 3

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________ 2025 Y5 TP Topical Revision Page 5 of 77 3 EJC Promo 9758/2022/Q7 Referred to the origin O, the points A and B have position vectors a and b respectively, where a and b are non-parallel vectors. The lines AB and OC intersect at point X such that : 3:1AX XB  and : 3 : 5OX XC  (see diagram). (a) Express O C   in terms of a and b. [2] (b) Given that OC is perpendicular to AB and : 3 : 2OA OB  , show that 23. 8a b b . [3] (c) Hence, by considering 2 OC    , find OC : OB. [3] [(a)  2 33 a b (c) 6 :1 ] Solutions (a) By ratio theorem, 3 4OX  a b  . Since 8 3 OC OX  ,  8 3 2 33 4 3OC       a b a b   . (b) Since OC is perpendicular to AB,    2 3 03    a b b a . 2 2 2 2 . 3 3 . 0 3 2 . 0         a b a b a b a b a b . Since 3 2a b , 2 2 9 4a b We have 2 29 3 2 . 04   b b a b Rearranging, 23. 8a b b . (shown) A X O B C

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________ 2025 Y5 TP Topical Revision Page 6 of 77 (c) Observe that       2 2 2 2 2 2 2 2 2 2 3 . 33 3 4 6 . 99 4 9 3 6 99 4 8 4 27 9 2 6 OC                        a b a b a a b b b b b b b   So OC:OB = 6 :1 4 NYJC Promo 9758/2022/Q9 The line l has equation 1 , 1 2 zx y    , where  is a constant. (i) Given that the point of intersection between the line l and the yz-plane is  0,1,5 , show that 3  . [2] Referred to the origin, the points A and B have position vectors 3j k  and 12 5i j k  respectively. (ii) Find the cartesian equation of the plane 1 which contains the point A and the line l. [3] (iii) Find the acute angle between the line AB and 1 . [3] The plane 2 has cartesian equation 2 5 x z . (iv) Find the cartesian equations of the planes such that the perpendicular distance from each plane to 2 is 2 5 . [4] [(ii) 2 4x y z   (iii) 24.1 (iv) 2 15 or 2 5x z x z     ]

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ______________________ 2025 Y5 TP Topical Revision Page 7 of 77 Solutions (i) Method 1 (using vector equation of l): l : 1 1 1 0 , . 2 r                      0 1 1 1 1 0 5 2                           1 0 1 5 2           3  Method 2 (using cartesian equation of l): Since the point  0,1,5 is on the line l, 51 0 , 2   3  Method 3: Equation of yz- plane is 1 0 0. 0         r . [Note: yz-plane contains the y and z axes and the origin, with the x axis perpendicular to the plane] When  meets the yz-plane, 1 1 1 0 0 3 2 0 1 0 1                         .  0 1 1 1 1 5 2 3                      (ii) l : 1 1 1 0 , . 3 2                    r  Let C be the point (1, 1, 3) 1 0 1 1 1 2 3 3 0 AC