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.