TMJC H2 Chp 5 Vectors Discussion Solutions 2024

Chapter 5 Vectors TMJC 2024 Page 1 of 25 H2 Mathematics (9758) Chapter 5 Vectors Discussion Solutions Level 1 1 (a) OEFG is a parallelogram as shown in the diagram. (i) Find OF and EG . (ii) Find the size of the angle GEO. (b) The points P, Q and R have coordinates ( )4, 1, 1 , ( )8, 5, 15−− and ( )7, 0, 5 respectively. Show that P, Q, and R are collinear. (c) Given =+−a i j k and 3=−b j k , find a vector perpendicular to both a and b. Q1 Solution (a)(i) 32 01 45 1 1 9 OF OE EF OE OG =+ =+ −       =+           = 23 10 54 5 1 1 EG OG OE=− −       =−          − = (a)(ii) 53 1 and 0 14 EG EO OE −−       = =− =       −    ( ) 53 1.0 14 11cos 27 25 15 3 65.0 1 d.p EG EOGEO EG EO GEO −−              −    = = =  =   To find GEO , use vectors that are both either pointing ‘outwards’ from point E (vectors EG and EO ) or pointing ‘inwards’ from point E (vectors GE and OE ) E (3, 0, 4) O G (−2, 1, 5) F E G O Using the parallelogram, OF OE EF=+ and EF OG= Recall: EG OG OE=− Use position vectors OG and OE to find EG

Chapter 5 Vectors TMJC 2024 Page 2 of 25 (b) 8 4 12 5 1 4 15 1 16 PQ OQ OP −−           = − = − =           −−      7 4 3 0 1 1 5 1 4 PR OR OP            = − = − = −                12 3 4 4 1 4 16 4 PQ PR −       = =− − =−      −    Since 4,PQ PR=− P, Q, R are collinear. (Other answers such as 4 5PQ QR=− is also accepted) (c) A vector perpendicular to both a and b : 1 0 2 1 1 3 1 3 1 −            =  =          −−      ab 2 3 is also accepted 1     = −  − ba Recommended to work out the constant like this to ensure that you have the correct relationship between the vectors

Chapter 5 Vectors TMJC 2024 Page 3 of 25 2 2010(9740)/I/1 The position vectors a and b are given by 2 3 6p p p= + +a i j k and 22= − +b i j k , where 0p . It is given that =ab . (i) Find the exact value of p. [2] (ii) Show that ( ) ( ) 0+  − =a b a b . [3] Q2 Suggested Solutions (i) Since =ab , ( ) ( ) ( ) 2 2 2 222 2 2 2 2 2 3 6 1 2 2 4 9 36 9 9 49 3 7 p p p p p p p p + + = + + + + = = = Since 0p , 3 7p= (ii) Method 1: ( ) ( ) ( ) 22 22 0 +  − = −  +  − = − = = a b a b a b a b a b a a b a Method 2: ( ) ( ) 2 1 2 1 33 3 2 3 277 6 2 6 2 13 / 7 1/ 7 5 / 7 23 / 7 32 / 7 4 / 7 13 115 128 049 49 49                      +  − = + − − −                               −       =−          =− − + = a b a b Note that there are 2 possible values of p when you solve . Always reject with reason Make sure the