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   −       − • − =      −+    ( ) ( ) ( )0 8 16 42 9 0+ − + + − + = 2= ( ) 1 5 ans 8 OQ  =− (iii) 10 1 4 1 1 49 0 16 9 cos 73         − • − = + + + +          25cos 45.6 (to 1 d.p.) 5 51 =  =

Chapter 6 3D Vector Geometry TMJC 2024 Page 4 of 10 Qn 5 2008/HCI/I/12 ( )1 1Required Distance 11 1 13 427 2 2 21 21 21 2110 4 −= = −        = − = =      −−    a r n rnan n n n Alternatively Take a point in 1 , say C ( 13,0,0) which satisfies 1 2 13 4   =− r 12 7 10 AC OC OA  = − = −  Length of projection of AC onto 1 2 4   − = 12 1 17 2 2 21 2110 4 AC        = − =       −    n n Alternatively: Finding foot of perpendicular first (Long Method) Vector equation of a line through A and parallel to 1 2 4   − : 11 7 2 , 10 4         = +       −−    r Let C be the foot of the perpendicular from A to 1 Then 1 72 10 4 OC    + =+−− C lies on 1 : 11 7 2 2 13 10 4 4    +       +=      − − −     λ = 2− 1 72 10 4 OC    + =+−− = 1 3 2 −  − ; 11 7 2 7 10 4 10 AC    +− = + −− − +

Chapter 6 3D Vector Geometry TMJC 2024 Page 5 of 10  Required distance 84 2 21AC= = = (ii) Since d is positive, the angle between ( )1−ar & n is acute ( ) 1 1 3 12 7 2 2 21 2 1 2110 4 6 OB OA d −           = − = − = −           −−      n n Alternatively: Finding foot of perpendicular first (Long Method) Vector equation of a line through A and parallel to 1 2 4   − : 11 7 2 , 10 4         = +       −−    r Let C be the foot of the perpendicular from A to 1 Then 1 72 10 4 OC    + =+−− C lies on 1 : 11 7 2 2 13 10 4 4    +       +=      − − −     λ = 2− 1 72 10 4 OC    + =+−− = 1 3 2 −  − By ratio theorem : ( ) 1 22OC OA OB OB OC OA= +  = − 3 1 6 OB −  = − (iii) 1 : x + 2y − 4z = 13 ……. (1) 2 : x + 3y + 3z = −8 ……. (2) By G.C. solve equations (1) & (2) The vector equation of the line of intersection is l : 55 18 21 7 01         = − + −          r where  or 1 18 07 31         = + −      −    r etc (iv) Since B and l lie on the image plane of 2 so the equation of image plane is 3 18 55 3 1 7 21 1 6 1 0 6  −  −                = − + − + − − −                         r 3 18 29 1 7 10 6 1 3  −           = − + − + −           −      r where  and 

Chapter 6 3D Vector Geometry TMJC 2024 Page 6 of 10 Qn 6 2010 DHS Prelim/P2/Q4 (i) Let  be acute angle between the 2 planes. 12 12 21 43 11 15cos | || | 21 11 21 11 9.3 .                    = = = = nn nn (ii) 2 4 10 38 x y z x y z + + = + + = Using GC, 1 Let , 1 , 3 ,22 11 : 3 1 , 202 zt ttxy tl  =  =− + = − −        = + − =           r (iii) Since the point with co-ordinates (6,m.5) lies on the first plane, 11 62 4 10 51 12 4 5 10 D m m =        =           + + = ad 7 4m =− (iv) 2 22 0 , . 71 lm           = +         −    22: r a + d = 12 1 0 2 2 0 ( ) 21 independent of thevalueof m        = − = − =      −    12d d Therefore lines l1 and l2 are perpendicular for all real values of m.

Chapter 6 3D Vector Geometry TMJC 2024 Page 7 of 10 Qn 7 2009/CJC/I/11 (i) Let n1 and n2 be the normals of p1 and p2 respectively. n1 = 10 01 11        =      −    1 1 1  − n2 = 2 1 3 3 0 2 0 1 3             = −          −      1 3 5 1 2 6 1 3 1            −  − =          −      Therefore l is parallel to 5i + 6j + k (ii) ( ) 1 12 13 12 13Acute angle bet. and cos 3 22 75.7 1 d.p. pp −        −−      −   = = (iii) 4 2 3 2 4 2 2 6 3 Perpendicular distance 22 23 22 43 22 22 22 22            −−           −     =        −−      −−   = −== (iv) p1: 1 . -1 4 1   = r and p3: 2 . -2 2 b   = r Distance between p1 and p3 = 1 3 41 3 2 3 3 b−= 41 3 2 3 3 b−= or 41 3 2 3 3 b −−= b = 6 or 10

Chapter 6 3D Vector Geometry TMJC 2024 Page 8 of 10 Qn 8 2009/MJC/I/9 (i) Sub ( ), ,0 into  1 and  2. 1 2 : 1( ) 3( ) (0) 8 : 3( ) 1( ) (0) 0 a b       + + = +