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 normal vector to the plane.  vector equation of plane in scalar product form is 1 3 1 1 1 1 4 2 3 2                               r   , i.e. 1 . 1 4 2          r . From 1 . 1 4 2          r , let x y z        r . 1 . 1 4 2 2 4 x y z x y z                      Cartesian equation of plane is 2 4x y z   . For point of intersection between 1 1 2 5 1 3                     r — (1) and 1 1 4 2         r  — (2) Substitute (1) into (2) 1 1 12 5 . 1 4 101 3 2                          . Substitute 1 10 into (1)

Chapter 6B 3D Vector Geometry TMJC 2024 Page 4 of 32 1 111 10 10 1 52 5 10 2 1311 3 1010                                           r Thus, point of intersection is 11 5 13, ,10 2 10      . Alternative: Using GC to find the point of intersection between line and plane: Using the Cartesian equation of the line and plane, we have 2 4 1 2 5 1 3 x y z x y z            Rearranging  2 4 1 5 2 3 1 x y z x y z           Using GC, 11 5 13 1, , ,10 2 10 10x y z    Thus, point of intersection is 11 5 13, ,10 2 10      .

Chapter 6B 3D Vector Geometry TMJC 2024 Page 5 of 32 (c) A direction vector parallel to plane 1 2 1 2 5 7 4 6 2                                 . Vector equation of plane is 1 2 1 2 3 7 , , 4 6 2                                   r  . In parametric form, 1 2 , 2 3 7 , 4 6 2 , where ,x y z              1 2 36 7 3 2 2 6 11                               is a normal vector to the plane.  vector equation of plane in scalar product form is 36 1 36 2 2 2 4 11 4 11                                 r   i.e 36 2 4 11         r  From 36 . 2 4 11         r , let x y z        r . 36 2 4 11         r  36 2 11 4x y z   Cartesian equation of plane is 36 2 11 4.x y z   For point of intersection between 1 1 2 5 1 3                     r — (1) and 36 2 4 11          r  — (2) Substitute (1) into (2) 1 36 552 5 . 2 4 791 3 11                          . Substitute 55 79 into (1)

Chapter 6B 3D Vector Geometry TMJC 2024 Page 6 of 32 55 241 79 79 55 1172 5 79 79 86551 3 7979                                            r Thus, point of intersection is 24 117 86, ,79 79 79      . Alternative: Using GC to find the point of intersection between line and plane: Using the Cartesian equation of the line and plane, we have 36 2 11 4 1 2 5 1 3 x y z x y z            Rearranging  36 2 11 4 1 5 2 3 1 x y z x y z         Using GC, 24 117 86 55, , , 79 79 79 79 x y z    Thus, point of intersection is 24 117 86, ,79 79 79      .

Chapter 6B 3D Vector Geometry TMJC 2024 Page 7 of 32 2 Find the coordinates of the point where the line ( )r i i k    intersects the plane with equation 2x – 3y + z = 1. Q2 Solution Line: 1 1 ( ) 0 0 0 1 r i i k                          , Plane: 2x – 3y + z = 1  2 3 1 1         r Since line intersects the plane, 1 1 2 0 0 3 1 0 1 1                                      2 2 1     1   The position vector of the point of intersection is 0 0 1         and the coordinates are (0, 0, 1).

Chapter 6B 3D Vector Geometry TMJC 2024 Page 8 of 32 3 Find the equation of the line of intersection between the two planes with equations 3 2 10 1         r and 5 1 6 2         r respectively. Q3 Solution The cartesian equations of the two planes are 3 2 10x y z   ---------- (1) 5 2 6x y z   ---------- (2) Using the G.C, an equation of the line of intersection is 2213 3 32 1113 130 r                     ,   . Learning point: Although the GC gives 22 313 13 32 1113 13 0 1                            r ,  , it is helpful to factorise out 1 13 from the direction vector so that the remaining numbers are nicer to work with.

Chapter 6B 3D Vector Geometry TMJC 2024 Page 9 of 32 4 Find the acute angle between (a) the line 1 3 2 5 1 2 ,           