RI Vectors C2B Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ___________________ Tutorial 2B: Vectors II Page 1 of 4 Tutorial 2B: Vectors II – Equations of Straight Lines Section A (Basic Questions) 1 For each of the following, write down a vector equation of the line l and convert it to Cartesian form. (a) l passes through the point with position vector 2ij k  and is parallel to the vector ij . (b) l passes through the points (1, 1, 3)P  and (2,1 , 2)Q  . (c) l passes through the origin O and is parallel to the line 11 :1 2 , 33 m              r  . (d) l passes through the point (1, 0,1)C and is parallel to the y axis. Solution (a) 11 :2 1 , 10 rl              Let x y z   r . Then 11 21 10 x y z             r 11x x      2+ 2yy     1z Cartesian equation is 1 2, 1x yz  (b) 1 1 3 OP    , 2 1 2 OQ     . 1 2 5 PQ OQ OP          11 :1 2 , 35 l           r  Then 11x x    112 2 yy      335 5 zz      Cartesian equation is 131 25 yzx   

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ___________________ Tutorial 2B: Vectors II Page 2 of 4 (c) 1 2, 3     r  Cartesian equation is 23 yzx (d) A vector parallel to the y-axis is 0 1 0    . 10 :0 1 , 10 rl          1 , 1 , xz y   Cartesian equation is 1 , 1 .xz

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ___________________ Tutorial 2B: Vectors II Page 3 of 4 2 For each of the following, find the acute angle between the lines 1l and 2l . Determine if 1l and 2l are parallel, intersecting or skew. In the case of intersecting lines, find the position vector of the point of intersection. (a) 1 :1 2lx y z  and 2 21 3: 22 2 x yzl    (b) 1 14 :0 2 , 03 l               r  and 2 03 :1 0 8 , 11 l             r  (c) 1 :( 5 )( ) , l      rik i j k  and 2 :( )( 5 4 ) , l     ri j k ij k  [(a) 0 , parallel (b) 81.3 , skew (c) 44.5, intersecting, 6 5 0   ] Solution (a) 1 :1 2lx y z  and 2 21 3: 22 2 x yzl    1 11 :0 1 , 21 l              r  2 22 :1 2 , 32 l           r  Since 12 112 212        , 1 1 1   is parallel to 2 2 2   . Hence 1l and 2l are parallel and the angle between 1l and 2l is 0  . [Note that 1l and 2l are actually the same line] (b) 1 14 :0 2 , 03 l               r  and 2 03 :1 0 8 , 11 l             r  Let  be the angle between 1l and 2l . 43 28 31 12 16 3 7cos 16 4 9 9 64 1 29 74 2146                  98.691  Hence required acute angle between 1l and 2l is 180 98.691 81.3  (1 d.p.) Since 43 28 31 m           for any m  , 1l and 2l are not parallel. [Note that the above step of checki ng whether the direction vectors of 1l and 2l may no longer be necessary since the angle between these 2 lines is non-zero.]

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ___________________ Tutorial 2B: Vectors II Page 4 of 4 Let 14 03 02 1 0 8 03 1 1                      14 3 4 3 1 2 1 08 2 8 1 0 31 3 1                Solving using GC, there is no solution for  and  . Hence 1l and 2l do not intersect. Read Appendix A on how to solve simultaneous equations with/without a GC Since 1l and 2l are non-parallel and non-intersecting, they are skew lines. (c) 1 :( 5 )( ) , l      rik i j k  and 2 :( )( 5 4 ) , l     ri j k ij k  1 11 :0 1 , 51 l          r  and 2 15 :1 4 , 11 l            r  Let be the angle between 1l and 2l . 15 14 11 541 8cos 32 51 61 34 2 1 2 6             44.5  (to 1d.p) Since 15 14 11 k        for any k  , 1l and 2l are not parallel. [Note that the above step of checki ng whether the direction vectors of 1l and 2l may no longer be necessary since the angle between these 2 lines is non-zero. (similar to Qn 2(b)] To check if the lines intersect: Let 11 15 01 1 4 51 1 1               50 41 6         Solving using GC, we obtain a unique solution 5  and 1  . Hence 1l and 2l are intersecting lines. Read Appendix A on how to solve simultaneous equations with/without a GC Position vector of point of intersection 15 6 14 5 11 0          (We can of course also use 5  to obtain the same answer)