RI Vectors C2A Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ______________________________ Additional Practice C2AVectors I Page 1 of 21 Additional Practice Questions for Chapter 2A: Vectors I Vector Algebra, Ratio Theorem, Scalar and Vector Products 1 Referred to an origin O, points A and B have position vectors given respectively by and22 236 .OA OB    ij k ij k    The point P on AB is such that :: 1 .AP PB  Show that ( 1 ) ( 25 ) ( 28 ) .OP      ij k  (i) Find the value of  for which OP is perpendicular to AB. (ii) Find the value of  for which angles AOP and POB are equal. 9205/1984/02/Q5 Solution:       By Ratio Theorem, 1 1 121 2 12 3 2 13 262 1 6 1 2 5 , that is, (1 ) (2 5 ) ( 2 8 ) (shown) 28 OA OBOP OP                                      ij k      B P A O

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 2 of 21 (i)   21 1 32 5 62 8 0 11 2 5 5 0 28 8 1 10 16 1 25 64 0 AB OB OA OP AB OP AB                                                 25 5 90 18  (ii)       cos cos 11 12 11 25 2 25 337 28 2 28 6 11 1 4 4 1 10 16 2 6 12 2 15 4837 AOP POB AOP POB OP OA OP OB OP OA OP OB                                                                      79 2 5 3 1 6 6 5 3 10        OR : Use the fact that in a rhombus, the diagonal is an angle bisector. We can form a rhombus with the vectors andab ab which has diagonal ab ab . So OP // ab ab ie. OP k   ab ab  for some .k  By Ratio Theorem, since P on AB is such that :: 1 .AP PB       1 1 Since and are non-zero and non parallel vectors, 1( 1 ) a n d ( 2 ) 33(2) (1) 17 1 0 OP k kk                      ab abab ab ab ab a b 

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 3 of 21 2 Referring to an origin O, the position vectors of points A, B and C are given by 7OA i  , 4OB ik   and 4OC ij   respectively. A parallelepiped has OA, OB, OC as three edges, and the remaining vertices are X, Y, Z and D as shown in the diagram. (i) Write down the position vector of Z in terms of i, j and k. [1] (ii) The point P divides CZ such that CP PZ  . Given that OP  is perpendicular to CZ  , find the value of  and evaluate OP  . [6] IJCPrelim9233/2006/01/Q14(i), (ii) Solution: (i) 8 4 OZ OA AZ OA OB    ik          (ii) Since P divides CZ such that CP PZ  , by Ratio Theorem, 1 18 1 4 01 04 OC OZOP                    A O B Z Y D X C Z P C O

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 4 of 21 Since OP CZ , 0OP CZ    where 7 4 4 CZ OZ OC              18 7 1 404 01 04 4 71 6 5 61 6 0 1 8                       1 8 11 414 48 91 11/2 OP              and 22 24 481 49OP      Alternative Method From (i), 87 0. S o 4 44 OZ CZ OZ OC                        CZ = 2272 ( 4 ) 9 Since 1 ,8  1 1a n d 88 CP CP PZPZ    17 1 4OP    1 8 C Z P O

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 5 of 21 3 The position vectors of A and B with respect to the origin O are a and b respectively. M is the mid-point of OA and C is the point on MB such that 2CB MC      . Given that 6a , 5b and the angle between a and b is 45 , evaluate the exact value of the scalar product of b and c . MI Prelim 9233/2006/01/Q1 Solution:  1Given that is the mid-point of , 2 Given 2 2 1 : 2 :1 By Ratio Theorem, 21 33 MO A O M CB MC CB MC CB MC OM OB      a c= a b                 2 2 1 3 1 3 1 cos 453 12 5 6 532 1 15 2 253 5 5 3 23                 bc b a b ba bb ba b  M B C 1 2 O

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 6 of 21 4 Referred to the origin O, the points A and B have position vectors a and b such that aij k and 22 bi j k . (i) Find the size of angle OAB. [2] The point C has position vector c given by  cab , where λ and µ are positive constants. Given that the area of triangle OAC is twice that of triangle OBC, (ii) find µ in terms of , [3] (iii) hence, if OC = 118, find the position vector c. [4] JPJC Prelim 9758/2021/01/Q2 Solution: (i) 11 0 12 1 12 1 BA                     10 11 11 2cos 32 6 144.7 (1 d.p) OA BAOAB OA BA OAB                   (ii)   1Area of since 0 22 1Area of since 0 22 OAC a a b a b OBC b a b b a                     Given area of OAC is twice that of OBC, 222ab ba    Since , 2ab ba    

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 7 of 21 (iii)  222 222 2 118 118 21 1 8 s i n c e 2 2 41 1 8 4 355 1 1 8 92 52 51 1 8 59 118 2 Since 0, 32 2, 5 2 52 OC ab ab c                                   OR: 222 3 5 118 5 3 5 118 since 0 5 3 5 5 118 118 2 59 32 52 52 c                   

Raffles Institution H2 Mathematics 2025 Year 5 _____________________________________________________________________________________________ ___________________________ Additional Practice C2AVectors I Page 8 of 21 5 With respect to the origin O, the position vectors of the points A, B and C are a , b and c respectively. Point C lies on AB such that :1 : 2AC CB  . It is given that a is a unit vector and the length of OB is 2 units. (i) Give a geometrical interpretation of ac . [1] (ii) It is given that the angle AOB is 60 . By considering    22 ab ab , find 2 ab . [3] (iii) Find c in terms of a and b. [1] (iv) Hence by considering cosine of angle AOC and cosine of angle COB, determine if the line segment OC bisects the angle AOB. [3] MI Prelim 9758/2021/01/Q4 Solution: (i) ac is the length of projection of c onto a .