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         =   = = =   =  O A B C

Teacher’s copy 4 ( ) 2 2 2 2 ( ) ( ) ( ) ( ) ( ) ( ) (since ) (since , ) 0 since radius AC BC OC OA OC OB = −  − = −  − = −  + =− =  −  +  −  =  −  = −  =  = = = = c a c b c a c a b a c c a c c a a a c c a a c a c c c a a a ca

Teacher’s copy 5 Standard Questions 1. MI Promo 9758/2021/PU2/02/Q4 Referred to the origin O, the points A and B are such that OA= a and OB = b . The mid-point of OA is P and the point M on PB is such that : 2 : 3PM MB = . By finding OM , show that the area of triangle OMP can be written as k ab where k is a constant to be found. [5] Given that 2, 2==ab and the angle AOB is radians4  , show that PM is perpendicular to OA. [4] Answer: 1 10k = (i) 1 2 23 55 2 3 1 5 5 2 23 5 10 OP OM OB OP = =+ =+  =+ a ba ba Area of triangle OMP 1 2 OM OP= 1 2 3 1 2 5 10 2 1 2 1 3 1 2 5 2 10 2 1 1 3 2 5 20 1 since 10 1 10 = +  =  +  =  +  =   = = b a a b a a a b a a a b a a a 0 ab where 1 10k = . P O A B M 1 1 2 3

Teacher’s copy 6 (ii) 2 3 1 5 10 2 21 55 PM OM OP=− = + − =− b a a ba ( )( ) ( ) 2 2 21 55 21 55 21 cos55 21 2 2 cos 25 4 5 4 2 2 4 5 2 5 0 PM OA AOB  =− =− =− =−  =−  = b a a b a a a b a a Since 0, is perpendicular to .PM OA PM OA= Alternative method: 1 2PB OB OP= − = − ba ( )( ) ( ) 2 2 1 2 1 2 1cos 2 12 2 cos 2 42 22 2 2 2 0 PB OA AOB  =− =− =− =−  =−  = b a a b a a a b a a Since 0, is perpendicular to . 2Since , is perpendicular to .5 PB OA PB OA PM PB PM OA = = Hence PM is perpendicular to OA.

Teacher’s copy 7 2. MI Promo 9758/2020/PU2/P1/Q7 In the parallelogram OABC, aOA= and cOC= . The point M on OA is such that OM : MA = 2 : 1 and the point N on AB is such that AN : NB = 1 : 2. It is given that the lines CM and ON intersect at point R. (i) Find OM and ON , giving your answers in terms of a and c . [2] (ii) Show that 62 11 11OR=+ ac . [4] (iii) Hence find the ratio CR : RM . [1] (iv) State, with a reason, whether the points O, B and R are collinear. [2] (i) 2 3OM = a 1 3ON OA AN= + = + ac (ii) Method 1 Let : :1CR RM =− and : :1OR RN =− . Using CM , by ratio theorem, ( ) ( ) 21 23 113OR    −+ = = + − ac ac Using ON , 33OR ON    = = + = + ca a c ( )2 133   + − = +a c a c Comparing coefficient of a : 2 3  = Comparing coefficient of c : 1 3 −= Solving simultaneously, 69,11 11== . 2 9 9 6 2 13 11 11 11 11OR     = + − = +       a c a c . ================================================= Method 2

Teacher’s copy 8 1: , 3 ONl = +  r a c 2 3CM OM OC= − = − ac 2: , 3 CMl = + −  r c a c Since the lines intersect, 12 33   + = + −       a c c a c . ( )2 133 + = + −a c a c Comparing coefficient of a : 2 3  = Comparing coefficient of c : 13  =− Solving simultaneously, 69,11 11== . 2 9 9 6 2 13 11 11 11 11OR     = + − = +       a c a c . (iii) From above, 9 11CR CM= . : 9 : 2.CR RM= (iv) 62 11 11OR=+ ac OB=+ac Since OR kOB for any real constant k, O, B and R are not collinear.