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    −    = 4 3 8 3 2 1 2 −    = 29 3 units2 or 1.80 units2

Chapter 5 Vectors TMJC 2024 Page 5 of 9 (iv) Since the two triangles share the same height, Area of triangle OPB : Area of triangle OPA = PB : PA 2 :1= Qn 6 2015/NYJC Promo/4 (i) ( )2 5 0+=a a b 2 2 5 0+=a a b 2 5 0+=ab since a is a unit vector 2 5=−ab Since angle between a and b is 2 3  , 2cos 3  = ab ab 21 5 2 (1) − −= b 4 5=b (shown) (ii) By Ratio theorem, ( )1OM = + −ba ( ) ( ) ( ) 1 11 ON MB   = = − + − = − − − b b a ba Area of triangle OAN ( ) ( ) 1 2 1 112 OA ON  = =  − − −a b a A B P O 1 2

Chapter 5 Vectors TMJC 2024 Page 6 of 9 ( ) ( ) ( ) ( ) ( ) ( ) 1 112 1 12 11 2 4 31 sin2 3 2 5 2 13 5     = −  − −  = −  −   = − =          −= a b a a ab ab Note that 2 4 3 2 3sin 3 5 2 5      = = =          a b a b Qn 7 2012/DHS/I/7 (i) (ii) , where q is a constant (iii) Shortest distance from the point E to OB (iv) It is the length of projection of a onto b. 32 5 pOD += ab 3 4OE += ab OD qOE= 3 2 3 54 3 3 4 5 4 5 2 1 1 5 4 2 p q qq p q p ++ =   =  =  =  = a b a b ( ) ( ) 3 45 1 320 3 ()20 OBOE OB= += =  +  =   = a b b a b b b a b b b 0 3 20k = ˆab b a

Chapter 5 Vectors TMJC 2024 Page 7 of 9 Qn 8 2008/HCI/I/12a (i) (− μ − 1) a + μb + c = 0 λ = − μ − 1 μ (b − a) + (c − a) = 0 0 1where AB AC AB k AC k   += = =−  A, B, C are collinear (ii) p = 4a - 3b  4a = p + 3b  a = (p + 3b) / 4 A divides PB in the ratio 3 : 1  P lies on BA produced with ratio : 3: 4PA PB = Alternatively 4 3 3 3PA= − = − + =− +a p a a b a b 4 3 4 4PB= − = − + =− +b p b a b a b  P lies on BA produced with ratio : 3: 4PA PB = Qn 9 2019/TJC/Prelim 9758/01/Q9 (a) Since +u v u is perpendicular to +u v v , we have (( ) ).(( ) ) 0 +  + =u v u u v v ( ).( ) ( ). ( ). . 0   +  +  + =u v u v u v v u v u u v 2 0 0 1 0  + + − =uv since () ⊥u v v and () ⊥u v u 1  =uv (shown) Let  be the angle between u and v. 1 cos 1 =−  =−u.v u v ---(1) 1 sin 1  =  =u v u v ---(2) (2) :(1) tan 1 =− ο135= (b) OF OA AF= + = − ab 1 2OX OB BX= + = − ba Method 1 Let : :1AY YC =− and : :1FY YX =− (1 ) (1 )OY OC OA OX OF    = + − = + − 1( ) (1 ) (1 )( ) 2     − + − = − + − −  b a a b a a b 1( 1 ) 1 1 2       − + − = − + − + − ba Since a and b are non-zero and non-parallel,

Chapter 5 Vectors TMJC 2024 Page 8 of 9 2 1 0 320 2   − + = −=  solving gives 34,55== 33: :1 3: 255AY YC = − = Method 2 Line AC: ( 2 )= + −r a b a ,  Line FX: 32 2= − + − r a b b a ,  When the lines intersect at Y, 3( 2 ) 2 2 + − = − + − a b a a b b a Since a and b are non-zero and non-parallel, 2 1 0 320 2   − + = −=  solving gives 34,55== 3 1 3( 2 )5 5 5OY = + − =− +a b a a b 63 55AY =− +ab and 42 55YC=− +ab : 3: 2AY YC=

Chapter 5 Vectors TMJC 2024 Page 9 of 9 Qn 10 2012/HCI/II/Q4 (i) ( ) ( ) 22 2 2 0 50 5 5 0 5 4 0 5 4 cos60 0 2 5 0 2 24 6 1 or 6 1 (rejected as 0)2 61 AB OP AB OP⊥   =  −  + =  +  −  −  = − −  = − − = + − = −= = − − −   = − b a a b b a b b a a a b b a a b a a b aa aa a (ii) 1 2OC= a By Ratio Theorem, 3 4 3 2 77 OB OCOE ++== ba Let : :1AD AB = By Ratio Theorem, ( ) ( )11OD OB OA    = + − = + − ba Since O, E, D are collinear, OE OD= for some \{0} . ( )32 17 32 177        + = + −    − = − −       ba ba ba Since a and b are non-zero and non-parallel, 3 0 --- (1)7 21 0 --- (2)7   −= − − = Using GC to solve (1) and (2), 3 5 = : 3: 5AD AB=