H2MA Remedial _ Vectors I, II_Solution (RVHS)

Teacher’s copy 1 RVHS H2 Mathematics Remedial Programme Topic: Vectors I, II Basic Mastery Questions 1. ACJC Promo 9758/2021/Q10(i) Referred to the origin O, the points A, B and C have position vectors 42ij− , 2 −+i j k and 7 − − +i j k respectively, where  and  are constants. Given that A, B and C are collinear, show that  =5, and find the value of  . [3] Answer: 10 =− 41 2 1 7 02 , and are collinear 1 4 5 4 5 5 1 5 1 5 5 2 2 10 OA OB OC A B C AB k AC kk kk k        −           = − = − = −                =  =−− − − =−        = −  =−  =        = =−       2. JPJC Prelim 9758/2021/01/Q2 Referred to the origin O, the points A and B have position vectors a and b such that =++a i j k and 22= + +b i j k . (i) Find the size of angle OAB. [2] The point C has position vector c given by =+c a b , 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]

Teacher’s copy 2 Answer: (i) 144.7 (ii) 2= (iii) 32 52 52 c   =  (i) 1 1 0 1 2 1 1 2 1 BA            = − = −          −      10 11 11 2cos 3 2 6 144.7 OA BAOAB OA BA OAB        −      − −   = = = = (ii) ( ) ( ) 1Area of since 022 1Area of since 022 OAC a a b a b OBC b a b b a       =  + =   =  + =   Given area of triangle OAC is twice that of triangle OBC, 222a b b a  =  since 2 a b b a   =  =

Teacher’s copy 3 (iii) ( ) ( ) ( ) 222 222 2 118 118 2 118 since 2 2 4 118 4 3 5 5 118 9 25 25 118 59 118 2 Since 0 32 2, 5 2 52 OC ab ab c               = += + = = +  += + + + = + + = = =    ==   3. RI Prelim 9758/2021/02/Q4(a)(i) Referred to the origin O, points A, B and C have position vectors a, b and c respectively. The three points lie on a circle with centre O and diameter AB (see diagram). Using a suitable scalar product, show that the angle ACB is 90 . [4] Or 3 5 118 5 3 5 118 since 0 5 59 118 2 32 52 52 c     