TMJC H2 Chapter 6B 3D Vector Geometry (Lines & Planes) Discussion Questions Solutions 2024

Chapter 6B 3D Vector Geometry TMJC 2024 Page 1 of 32 H2 Mathematics (9758) Chapter 6B 3D Vector Geometry (Lines & Planes) Discussion Questions Solutions Level 1 1 Find the equation of the following planes in parametric form, scalar product form and Cartesian form. (a) The plane passing through the points A(0, 1, 1), B(1, −3, 2) and C(1, 0, 1). (b) The plane containing the lines 32 1,3  zyx and 3 2 1 4 3 3 ,                     r  . (c) The plane that includes the line  2 4 2 3 6      r i j k i j k and the point with position vector 2 5 6 . i j k In each case, find the coordinates of the point of intersection of the plane and the line 1 1 2 5 1 3 ,                      r  . Q1 Solution (a) The plane is parallel to 1 1 4 and 1 1 0 AB AC                         . Vector equation of plane is 0 1 1 1 4 1 , , 1 1 0                                  r  . In parametric form, , 1 4 , 1 , where ,x y z         A vector normal to the plane is 1 1 1 4 1 1 1 0 3                               . r  n 0 1 1 1 0 1 3 4 1 3                       .

Chapter 6B 3D Vector Geometry TMJC 2024 Page 2 of 32 In scalar product form, the equation of the plane is r 1 1 4 3          . r 1 1 4 3           1 1 4 3 x y z                    The cartesian equation of the plane is 43  zyx 1 1 2 5 1 3 ,                      r  - (1) r 1 1 4 3          -(2) Substitute (1) into (2) 1 2 5 1 3 1 1 4 3 1 2 5 3 9 4 2 5                                    Substitute 2 5 into (1) 1 2 5 1 3 2 5 7 2 1 205 5 11 2 5                                       r Thus, point of intersection is 7 11, 4,5 5      . Alternative: Using GC to find the point of intersection between line and plane: Using the Cartesian equation of the line and plane, we have 3 4 1 2 5 1 3 x y z x y z            Rearranging  3 4 1 5 2 3 1 x y z x y z            Using GC, 7 11 2, 4, ,5 5 5x y z    Thus, point of intersection is 7 11, 4,5 5      .

Chapter 6B 3D Vector Geometry TMJC 2024 Page 3 of 32 (b) 32 1,3  zyx 3 0 1 2 3 1 ,                     r  Vector equation of plane is 3 0 2 1 2 4 , , 3 1 3                                 r  . In parametric form, 3 2 , 1 2 4 , 3 3 , where ,x y z             0 2 2 1 2 4 2 2 1 1 3 4 2                                         is a n