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 6 2 9 3 3 3 3 1                       3d d , 3d // 4d , thus 3 4 and l l are parallel lines. If 3 4 and l l are the same line, then there exist   such that 1 2 2 1 2 3 3 3 0 0 0 1 0                                           Since the values of  are inconsistent for all 3 equations,  1,3,0 does not lie on 3l . Thus 3 4 and l l are distinct parallel lines.

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 4 of 19 3 The lines 1 2 and l l have equations 3 1 1 2 0 4 s                   r and 1 2 1 1 1 1 t                     r respectively, where s and t are real parameters. (i) Show that 1l passes through the point  2, 1, 4A   , but that 2l does not. (ii) Find the acute angle between 2l and the line joining  2, 1, 4A   and  1, 1,1B  . Q3 Solution (i) 1 3 1 : 1 2 , 0 4 l s s                    r  2 1 2 : 1 1 , 1 1 l t t                      r  When 1s ,   3 1 2 1 1 2 1 0 4 4                                 r . Therefore, 1l passes through  2, 1, 4  . Alternatively, Let 3 1 2 1 2 1 0 4 4 s                               for some s  3 2 1 1 2 1 1 4 4 1 s s s s s s              = The value of s are consistent for all 3 equations, so 1l passes through  2, 1, 4  . Let 1 2 2 1 1 1 1 1 4 t                                for some t  11 2 2 2 1 1 0 1 4 5 t t t t t t                 = The values of t are inconsistent for all 3 equations, so 2l does not pass through  2, 1, 4  . (ii) 1 2 1 1 1 0 1 4 5 AB                                  The line joining  2, 1, 4A   and  1, 1,1B  has direction vector 1 0 5         . Let  be the required acute angle.   2 1 1 0 1 5 7cos 55.9 1 d.p. 6 26 6 26                         

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 5 of 19 4 Given that point A has position vector 2 0 1         and point B has position vector 1 0 2         , find the (i) length of the projection of AB   onto the z-axis, (ii) projection vector of AB   onto the z-axis. Q4 Solution (i) 1 2 1 0 0 0 2 1 1 AB OB OA                                       Length of projection of AB   onto z-axis 1 0 0 0 0 0 11 1 1 1) 1                          unit (ii) Projection vector of AB   onto z-axis 1 0 0 0 0 01 1 0 01) 1 1                                       

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 6 of 19 5 Find the coordinates of the foot of perpendicular from the point P(7, –2, 4) to the li ne kjir  )23()2(  , where  is a real parameter. Q5 Solution                       0 2 1 1 3 2 r , λ  Let F be the foot of perpendicular from P to the line. Since F lies on the line, OF                1 23 2   for some λ  . PF                                         3 25 5 4 2 7 1 23 2     Since PF is perpendicular to the line, PF   1 2 0 0          . 0)25(2)5(    3 Substitute  = 3 into OF   : The position vector of F is OF               1 3 5 and the coordinates of F is  5, 3,1

Chapter 6A 3D Vector Geometry (Lines) TMJC 2024 Page 7 of 19 Level 2 6 The equation of a straight line l is 1 1 2 1 3 1 t                   r , where t is a parameter. The point A on l is given by 0t , and the origin of the position vectors is O. (a) Calculate the acute angle between OA and l, giving your answer correct to the nearest degree. (b) Find the position vector of the point P on l such that OP is perpendicular to l. (c) A point Q on l is such that the length of OQ is 5 units. Find the two possible position vectors of Q. (d) The points R and S on l are given by t  and 2t  respectively. Show that there is no value of  for which OR and OS are perpendicular. Q6 Solution (a) 1 1 : 2 1 , 3 1 l t t                    r  At 0t , 1 2 3 OA          Let  be the required (acute) angle   1 1 2 1 3 1 4cos 52 nearest degree 14 3 14 3 .                         (b) Since P is on l, 1 2 3 t OP t t           for some t  . 1 1 1 41 0 2 1 0 31 3 1 t OP l OP t t t                                           4 3 4 3 4 3 1 7 12 2