TMJC Chapter 6 3D Vector Geometry Extra Practice Solutions

Chapter 6 3D Vector Geometry TMJC 2024 Page 1 of 10 H2 Mathematics (9758) Chapter 6 3D Vector Geometry Extra Practice Questions Solutions Qn 1 21 : 2 1 , 34 l        = − + −           r , and 1 : . 5 5 1  =  r 11 1 5 1 5 4 0 41        − = − + =           l and  are parallel. ( )2, 2,3P − is a point on l . ( )0,1,0A is a point on  . Shortest distance between l and  0 2 1 2 1 1 1 10 101 2 5 3 5 3 927 27 270 3 1 3 1 PA   −                   = = − − = = =                   −          n units Qn 2 2007/IJC/I/5 (i) 1 4 7 1 , 3 , 2 1 2 1 OA OB OC −           = = = −           −      Find any 2 of the 3 vectors: 3 11 8 2 , 5 , 3 132 AB BC AC −−           = = − = −           −−      Since AB not parallel to BC (or equivalent), therefore A, B & C not collinear. (ii) 3 11 3 8 2 5 or 2 3 1 3 1 2                                      or equivalent, Vector ⊥ to plane ABC = 1 2 7   − (or 1 2 7  −− ) P A

Chapter 6 3D Vector Geometry TMJC 2024 Page 2 of 10 (iii) 2 4 7 OP  = and 4 4 6 OQ  =  2 0 1 PQ  =− Method 1: 2 1 2 0 2 13 1 7 4Length of projection 1 4 49 54 4 169 16 7 14 2254 PQ n n or                  −−     = = = ++ ++== Method 2: Length of projection of PQ onto n = 21 02 17 9 1 4 49 54 PQ n n           −−   == ++ Using Pythagoras’ Theorem, length of projection of PQ onto plane = 81 7 145 54 2 2 or−= Qn 3 2007/PJC/I/6 (i) 1 1 : 3 1 , 70 p l         = +           r Given 1q= and 4p= , Since C lies on 1l , 41 31 70 OC         =+          ; 41 01 40 AC OC OA         = − = +           2AC l⊥ 41 00 42   +        • =          12 =− 8 9 7OC=− − +i j k (ii) 2AB q=+ ik Given acute angle between 1l and 2l is 60 , 2 1 10 02cos 60 24 q q        •         = +

Chapter 6 3D Vector Geometry TMJC 2024 Page 3 of 10 22 4 2qq+= 2q = Qn 4 2008/JJC/I/3 (i) 1 11 : 3 1 , 27 l         = + −           r since P lies on 1l , 11 1 3 1 for some suitable 16 2 7 a             = + −                 31  − = 2= ( )1 3 provena  = + = (ii) since Q lies on 2l , 10 3 4 for some suitable 23 OQ         = + −           2 0 2 2 4 2 4 14 3 14 3 PQ OQ OP   −−           = − = + − = −          − − +      since 2PQ l⊥ 20 2 4 4 0 14 3 3   −   