TMJC Chp 5 Vectors Extra Practice Solutions

Chapter 5 Vectors TMJC 2024 Page 1 of 9 H2 Mathematics (9758) Chapter 5 Vectors Extra Practice Solutions Qn 1 2014/IJC Promo/4 (i) 1 3 2 AO − =− and 7 1 5 AB OB OA − = − = − 17 31 25 7 3 10 6cos 14 75 14 75 1050 OAB −−       −•      − − − −   = = = 1o 6cos 100.7 (1d.p) 1050 OAB − −==  (ii) Vector perpendicular to both OA and OB OA OB= 16 34 27 −       =      −−    13 19 22 − =  Required unit vector 13 19 22 13 19 22 −  = −   13 1 19 1014 22 − =  (alternatively, ......OA BO OA BO ==  13 1 19 1014 22  =− − )

Chapter 5 Vectors TMJC 2024 Page 2 of 9 Qn 2 2016/MI Prelim/I/Q3 Using ratio theorem, 2 3 1 3 3 3 2 3 5 2 1 17 4 2 8 OA OCOB OC OB OA += −           = − = − − = −                ( equal vectors ) 33 1 17 28 0 16 10 OD OA AD OA OC OC AD =+ = + = −       = + −           =− Area of OADC ( ) ( ) 222 3 3 42 7 1 17 30 6 5 2 8 48 8 6 7 5 8 6 138 OA OC= −               =  − = − = −              −−        = + − + − = Qn 3 2017/CJC Prelim/II/Q2 (i) Length of projection of a on to b ( b is a unit vector) (ii) sin π(2)(1)sin4 2 = = = a b a b (iii) ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) ( )( )   ( )( ) 2 2 2 3 2 3 3 3 3 5 6 2 3 5 6 26               + +  +       = +  +  + + +  + +  = − + + +   =  = = + +  p q = a b a+ b a a a b b a b b b a a a 0 and b b 0 ba A C B | | | 1 2 O O C D A

Chapter 5 Vectors TMJC 2024 Page 3 of 9 (iv) Area OPQ = ( ) ( ) 21 262 + +  ba ( ) 21 2 6 22 = + + ( ) 22 152 = + + Smallest Area OPQ = 252 unit2= Qn 4 2014/TPJC Promo/4 (i) ( )1By Ratio Theorem: 3 4OC=+ ab (ii) 5 5 10 50= = =ba 2 2 2 2 2 2 3 5 50 34 50 34 50 16 4 ( is a positive constant) k k k k kk  = + + =  + =  + = = = b (iii) 3 233 17 9542 04 1 OC            = + =             1Area of triangle 2 3 21 17 322 0 1 3 1 12 1 11 2 OAC OA OC=      =      =− − = Minimum value of ( ) 2 15++ is 5. (Minimum point of quadratic equation)

Chapter 5 Vectors TMJC 2024 Page 4 of 9 (iv) Length of projection of OC onto the line OB OBOC OB = 3 2 3 17 5252 41    =    1 9 35 42252 = + + 26 13 2 552 == Qn 5 2015/DHS Promo/2 (i) ( ) 24 16 10 AP AB OP OA OB OA OP    = − = − −       = − +          (ii) 0OP AB OP AB⊥  = 2 4 4 1 6 6 0 10 8 16 6 36 0 52 14 7 26      −−       − + =          − + − + = = = (iii) Area of triangle OPA = 1 2 OP OA = 2 3 2 1 112 11    −