H2 Chapter 6B 3D Vector Geometry Planes Assignment Solutions 2024

Chapter 6 3D Vector Geometry TMJC 2024 Page 1 of 12 H2 Mathematics (9758) Chapter 6B 3D Vector Geometry Assignment 2 (Lines & Planes) Solutions 1 2018/MJC Promo/I/7 The line l and plane 1 have equations 2 1 1 3 , 1 1                       r  and 2 5 1x y z   respectively. (i) Find the acute angle between l and 1 . [2] (ii) Find the coordinates of P, the point of intersection of l and 1 . [3] (iii) Find the shortest distance from point  2,1, 1 to 1 . [3] The plane 2 has equation 2 1x ky z   , where k is a real constant. (iv) Give a reason why 1 and 2 intersect in a line. [1] (v) Given that L is the line where 1 and 2 meet, explain why point P lies on L. Hence or otherwise, find in terms of k, a vector equation of L. [4] (vi) The plane 3 has equation 2 3 4.x y z   Find the equation of the line of intersection between 1 and 3 . [1]

Chapter 6 3D Vector Geometry TMJC 2024 Page 2 of 12 1 Solution (i) Let  be the acute angle between l and 1 . 1 1 3 2 1 5 sin 1 1 3 2 1 5 12 11 30                                        41.344 41.3     (1 d.p.) or 0.722 rad (3 s.f) (ii) 1 1 : 2 1 5          r  2 1 2 1 3 1 3 1 1 1 OP                                        for some   . 1 2 1 5 OP            2 1 1 3 2 1 1 5                         2 2 6 5 5 1 12 4 1 3              7 3 0 2 3 7 2,0,3 3 OP P                     (iii) 7 1 2 13 3 10 1 1 3 32 1 1 1 3 3 AP                                                To find point of intersection between line and plane, substitute the equation of the line into the equation of plane

Chapter 6 3D Vector Geometry TMJC 2024 Page 3 of 12 Method 1: Shortest distance 1 1 1 1 3 23 301 5 12 3 30 4 30 2 30 or 0.730 15 AP                       n     Method 2: Shortest distance sinAP      Shortest distance sin 1 1 3 sin 41.344 3 1 1 11sin 41.344 3 0.730 AP              (iv) 1 2 1 2 1 1 2 , 5 2 k                         n n n n  Since the normal vectors of both planes are not parallel, they are not parallel planes or equal planes. Hence they must intersect in a line. (v) 2 1 : 1 2 k          r  7 1 1 3 0 2 2 2 3 7 4 3 3 1 OP k k                                 A P n

Chapter 6 3D Vector Geometry TMJC 2024 Page 4 of 12 Since P satisfies the equation of 2 , it lies on 2 . Since P lies on 1 and 2 , it will lie on L. 1 2 1 1 : 2 1 : 1 5 2 k                      r r  1 2 1 1 2 5 2 4 5 3 2 k k k                              n n 7 4 5 1: 0 3 , 3 2 2 k L k                        r  (vi) Using GC, the equation of the line of intersection between 1 and 3 is 3 2 5 1 4 2 4 , 0 1                     r  .

Chapter 6 3D Vector Geometry TMJC 2024 Page 5 of 12 2 2018/DHS Prelim/I/9 The line l1 passes through the point A with the position vector 1 3 2       and is parallel to 2 1 , 3 t t       while the cartesian equation of the plane p is given by 2 3,tx y z   where t is a real constant. It is known that l1 and p have no point in common. (i) Show that 1.t  [3] (ii) Find the distance between l1 and p. [2] (iii) The line l2 has the cartesian equation 2 , 3.y z x  Show that l2 lies on p. [2] (iv) Given that point B and point C lie on l1 and l2 respectively, find BC   such that it is perpendicular to both l1 and l2. [3] (v) Find the vector equation of the line of reflection of l1 in p. [3] 2 Solution (i) l1: 2 1 3 1 , 2 3 t t                      r  : 2 3 2 3 1 t p tx y z              r Since l1 and p have no point in common, the direction vector of l1 is perpendicular to normal vector of p and the point A does not lie on p. 2 2 21 2 0 2 2 3 0 1 or 1 3 1 t t t t t t                            1 3 2 3 6 2 3 1 2 1 t t t                            Therefore, 1t  (shown) Since l1 and p have no point in common, l1 is parallel to p

Chapter 6 3D Vector Geometry TMJC 2024 Page 6 of 12 (ii) The distance between l1 and p = shortest distance of A to p 1 : 2 3 2 3 1 p x y z                r Note that point  3,0,0M lies in p, since 3 1 0 2 3 0 1                    3 1 2 0 3 3 0 2 2 AM OM OA                                        Shortest distance from A to the plane p ˆ 2 1 3 2 2 1 2 6 2 2 6 units31 6 6 2 1 AM                               n     (iii) l2 : 3 0 2 , 3, 0 1 , 0 2 y z x                          r  Method 1 0 1 1 2 0 2 2 0 2 1                       Therefore, l2 is parallel to p. 3 1 0 2 3 0 1                    Therefore, l2 lies on p (shown). Method 2 A F p M Next, observe that finding the shortest distance from A to the plane p which is AF   is same as finding the length of projection of AM   onto n. For l2 lies on p, 1) Direction vector of l2 is perpendicular to normal vector of p 2) Given position vector of l2 lies on p

Chapter 6 3D Vector Geometry TMJC 2024 Page 7 of 12 l2 : 3 0 2 , 3, 0 1 , 0 2 y z x                          r  ---(1) 1 : 2 3 2 3 1 p x y z               r  --- (2) Substitute (1) into (2) 3 0 1 0 1 2 3 2 2 3 RHS of 0 2 1 y y p                                         Therefore, l2 lies on p (shown). (iv) Method 1 1 3 2 , for some 2 3 3 , for some 2 3 1 2 3 2 3 2 2 2 3 2 2 3 OB OC BC                                                                              Since BC is perpendicular to both l1 and l2, 1 2 0 and 0BC BC  d d               2 1 2 0 3 2 2 0 and 3 2 1 0 2 2 3 3 2 2 3 2 7 4 7 1 8 5 7 2 7 7Solve 1 and 2 : and 3 3                                                                          1031 1 7 53 2 332 3 5 OB                                      and 33 7 3 2 143 OC                           103 3 1 17 5 23 3 3 114 53 BC OC OB                                        

Chapter 6 3D Vector Geometry TMJC 2024 Page 8 of 12 Method 2 1 2 1 0 1 2 1 2 3 2 1 3 1 2 1 3 2 3 2 2 2 2 3 2 2 3 1 BC                                                                                    d d   Solving the equation, 1 7,3 3     Hence 1 1 23 1 BC           . (v) Method 1 Let l1’