TMJC H2 Chapter 6A 3D Vector Geometry (Lines) Discussion Questions Solutions 2024

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 1 of 19 H2 Mathematics (9758) Chapter 6A 3D Vector Geometry (Lines) Discussion Questions Solutions Level 1 1 Find a vector equation , a cartesian equation and a set of parametric equations of the following lines: (a) passing through the point with position vector 7 2 4 i j k and parallel to 3 , i j k (b) passing through the points  1, 2,1 and  0, 4,9 , (c) passing through the point  3,0,2 and parallel to the line 4 , 1.3 yx z   Q1 Vector Equation Cartesian Equation Parametric Equation (a) 1 7 1 2 3 , 4 1 l                      : r  27 4 3 yx z     OR 27 4 3 yx z     7x   2 3y   4z   ,   (b) 1 1 2 6 ,2 1 8 l                        : r  2 11 6 8 y zx      OR 2 11 6 8 y zx     1x   2 6y   1 8z   ,   (c) 3 1 0 3 ,5 2 0 l                     : r  3 , 23 yx z   3x   3y  2z ,   

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 2 of 19 2 For the following pairs of lines, determine whether they are parallel lines, intersecting lines or skew lines. Find the coordinates of the point of intersection for intersecting lines. (a) 1 1 13 2 x y z    ,  2 3 2 3     r i j i j k where  is a real parameter. (b)  2 3 2 3     r i j i j k ,      1 6 3 9 3       r i j k , where  and  are real parameters. Q2 Solution (a) 1 1 3 1 1: 1 1 2 ,3 2 1 1 x yl z                             r  2 2 2 : 3 3 , 0 1 l                       r  3 2 1        is not parallel to 2 3 1         , therefore 1 2 and l l are not parallel lines. Let       1 3 2 2 11 3 2 2 1 2 3 3 1 2 3 3 2 1 1 0 1 1 3                                                                 3 2 : 2 2 2 (4)         2 4 : 3 3 0       From (3), 1  Put 1  , 0  in equation (1): LHS 1 3 2   RHS 2 LHS  Thus (1) holds. 1 2 and l l are intersecting lines and the coordinates of their point of intersection are  2,3,0 . Alternative Re-arranging, we get       2 3 3 5 3 2 2 6 1 7                   Using GC, to solve equations (5), (6) and (7), 0  , 1  . 1 2 and l l are intersecting lines and the coordinates of their point of intersection are  2,3,0 .

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 3 of 19 (b)  3 : 2 3 2 3l      r i j i j k      4 : 1 6 3 9 3l        r i j k 3 2 2 : 3 3 , 0 1 l                       r  4 1 6 : 3 9 , 0 3 l                        r  Since 4