RI Vectors C2B Add Prac (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 _____________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 1 of 23 Additional Practice Quest ions for Chapter 2B: Vectors II – Equation of Straight Lines (Solutions) 1 Referred to the origin O, the position vector of the point A is 226ij k and the cartesian equation of the line l is 12 6x yz  . Find (i) the position vector of the foot of the perpendicular from A to l, [3] (ii) the perpendicular distance from A to l, [2] TPJC Prelim 9740/2013/01/Q10 (modified) (i) 11 :2 1 , 61 l           r  Let B be the foot of perpendicular from A to l. 11 2 6 1 .1 0 1 4 11 533 17 OB AB AB OB                           (ii) 6Perpendicular distance from to 2 11s 3i1 1 3AlA B    

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 2 of 23 2 The line 1l has equation 12 12 , 3      r where  is a parameter. (i) Find the exact shortest distance between the origin O and 1.l [3] Another line 2l has equation 13 .229 x yz  (ii) Write down a vector equation of 2.l [1] (iii) Show that 1l and 2l are skew lines. [3] (iv) Find a vector that is perpendicular to both 1l and 2.l [1] The points P and Q lie on 1l and 2l respectively, such that the line PQ is perpendicular to both 1l and 2.l (v) Find a vector equation of the line PQ. [3] ASRJC Promo 9758/2021/Q8 (i) 1Shortest distance from to 12 12 03 449 3 3 4 449 34 17 2 units Ol                (ii) 2 2 13 1 3 0: 229 2 29 12 :3 2 , 09 xy z x y zl l                      r 

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 3 of 23 (iii) 1 2 12 :1 2 , 03 12 :3 2 , 09 l l                                 r r   Clearly, 2 2 3   and 2 2 9   are not scalar multiples of each other, hence both lines are not parallel. 12 12 12 32 03 09 1 2 1 2 --- (1) 1 2 3 2 --- (2) 3 9 --- (3)                                  From (3), we have 3 --- (4)  Substituting (4) into (2), we get  12 3 32 16 32 42 1 2               Hence 133. 22   Substituting 1 2  and 3 2   into (1), LHS of (1) 312 2 2     RHS of (1) 112 0 L H S o f ( 1 )2    Hence both lines do not intersect. Therefore, both lines are skew. (shown)

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 4 of 23 (iv)  22 22 39 18 6 18 6 44 12 1 12 12 1 00                         Thus, 1 1 0    is a vector that is perpendicular to both 1l and 2.l (v) Since 1 1 0    is parallel to ,PQ  1 1 for some 0 PQ k k     Since P lies on 1,l 12 1 2 for some 03 OP                   Since Q lies on 2 ,l 12 3 2 for some . 09 OQ                   Thus, 1212 32 12 09 03 12 12 32 12 09 03 22 22 2 93 PQ OQ OP                                                                          

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 5 of 23 Comparing, 22 1 22 2 1 93 0 22 0 22 2 930 k k k                            Using GC, we obtain 13, , 144 k   Therefore, 12 0 . 5 312 2 . 5 .40 3 2.25 OP                     Hence an equation of the desired line is 0.5 1 2.5 1 , 2.25 0 tt              r 

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 6 of 23 3 The line 1l has equation 55 3 x z m   , 2y and the line 2l has equation 5 x y , 0z  , where m is a constant. It is given that 1l and 2l intersect at point A. (i) Find the value of m, and the coordinates of A. [4] (ii) Find the position vector of the point P on 1l such that OP is perpendicular to 1l , where O is the origin. [3] (iii) Find a vector equation of the line which is a reflection of 2l in 1l . [3] PJC JC2CT2 9740/2014/02/Q5 (i) Since y = 2, from 5 x y we have x = –10 and substituting into 55 3 x z m   (together with z = 0) we have 15 5 3 m   and thus m = 1. We also have therefore A(–10, 2, 0). Alternatively, (i) Let 55 3 xz m  53 5 x zm     1 53 :2 0 5 l m         r ,   Let 5 x y  5 0 x y z      2 5 :1 0 l     r ,   Given that 1l and 2l intersect, 53 5 20 1 50 m                   2 53 5 5 50 1mm            

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 7 of 23 Coordinates of A is  10, 2,0 (ii) 3 00 1 OP    53 3 20 0 0 51 1 15 9 5 0 2                     531 22 0 2 51 3 OP               (iii) Observe that O is on l2. Let P’ be the point of reflection of O in l1. Then the line of reflection of l2 in l1 passes through A and P’. 2 '2 4 6 OP OP       84 '' 2 2 1 63 AP OP OA                Line of reflection: 10 4 21 03 r         ,   

Raffles Institution H2 Mathematics 2025 Year 5 ____________________________________ _____ ______________________________________________ Additional Practice Questions for Chapter 2B: Vectors II Page 8 of 23 4 The points A and B have position vectors 8 3 2     and 2 3 4     respectively. (i) Show that 22 6AB . [1] (ii) Find the cartesian equation for the line AB . [2] (iii) The line l has equation r 22 36 45 t        . Find the length of the projection of AB onto l . [2] (iv) Calculate the acute angle between AB and l , giving your answer correct to the nearest degree. [2] (v) Find the position vector of the foot N of the perpendicular from A to l . Hence find the position vector of the image of A in the line l