RI 2024 RTT2 Vectors(Solutions)

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 5 ________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 1 of 21 Term 4 RTT2: Post-Promo Revision Lesson Worksheet Session 3: C4A to C4C: Vectors 1 Referred to the origin O, the points A and B have position vectors a and b respectively. The vectors a and b are given by 263a n d 2 ,ppp   aij k b i j k where p is a constant. Find ab a and give a geometrical interpretation of ab a . [2] Find ab and give a geometrical interpretation of ab . [2] Given that a is a unit vector, find the possible value(s) of p . [2]  222 21 61 32 266 10 72 263 6 3 p p p p p p                  ab a   What does the phrase “geometrical interpretation” mean? Draw a picture to help determine for yourself. Let  be the angle AOB. cos  ab ba = length of projection of b onto a . cos cos cos     abab ab ab ab b a 

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 2 of 21 [ REMARKS ] What is the geometrical interpretation of ak ? Answer : ak represents the length of projection of r onto the z-axis. 21 2 19 61 6 1 7 32 3 2 8                                    ab = p pp p p which is a vector.  sin sinn   ab a b ab a b  Area of triangle OAB = 11 sin22  ab a b sin ab a b = area of parallelogram formed with adjacent side OA and OB Since a is a unit vector, 22 16 6 1 33 p pp p           a   22226 3 1p    71p  1 7p [ REMARKS ] Be careful. It is quite co mmon for students to miss out 1 7p as one of the answers.

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 3 of 21 2 DHS Prelim 9758/2020/01/Q6b modified With reference to the origin O, the points A and B are such that OA   a and OB   b. It is given that 2 3OX   a, 3 4OY   b and the line ON bisects the line XY at the point M. (i) By considering the ratio :XM MY , find the vector OM  in terms of a and b. [1] (ii) Given that :: 1AN NB  and :: 1ON OM k  where  and k are real constants, find the ratio AN : NB. [4] (i) 2 3OX   a and 3 4OY   b Since line ON bisects the line XY at the point M, :1 : 1XM MY  11 3 23 8OM OX OY         ab Recall Ratio Theorem : A B O X N Y M P a B O A b r

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 4 of 21 Let P be a point which divides AB in the ratio :, then      abp (MF27). (ii) Since :: 1AN NB  for some   , (1 )ON      ab Moreover, since :: 1ON OM k  , 13 38ON kOM k k  ab     Hence, 13 (1 )38kk   ab a b . Vectors anda b are not parallel and non-zero: 3 8 k   8 3k  18() 133   9 17  :9 : 8AN NB A   N B   1   Let and be non-zero and non-parallel vectors: If for some then .

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 5 of 21 3 (a) The non-zero vectors a, b and c are such that ab ca . Given that bc , find a linear relationship between a, b and c. [3] (b) The variable vector vij kabc satisfies the equation  3.vi k j  Find the set of vectors v and describe this set geometrically. [3] (a)          ab ca ab 0 ab 0 ab c 0 ca ac Given that a0 and bc which implies b + c  0, a is parallel to bc  ,0  ab c . (b) 13 03 3 ab ba c cb              30 31 0 3 1 0 b ac b a n dac b                 Now a can still vary, but b = 0, and value of c depends on a. 01 00 0 0 13 13 1 3 01 00 , . 13 aa a a bb a ca c a                                      v  v describe the set of position vectors of all points on the line which passes through the point (0, 0, 1) and is parallel to 3ik . Important property of cross product :  uv vu or orab 0 a = 0 b = 0 ab  Since cross product satisfies the equation, the values of a, b and c are restricted by RHS of the equation. This equation is in the form of ‘equation of a line’ r = a b . Hence describe it in terms of a line.

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 6 of 21 4 ACJC Promo 9758/2020/Q8 The lines l and m are defined by the equations :( 2 6 3 ) , 13:. 44 l xa y zm a     ri k i j k (i) Given that the lines intersect, show that 6a . [2] (ii) Find the position vector of N, the foot of perpendicular from the point (5, 0,1)A to the line l. [3] (iii) Find the position vector of the two points on l that are 5 units from A. [3] Solution : (i) 12 :0 6 13 l          r 14 : 34 ma a          r For intersection, 12 1 4 06 133 4 12 14 2 4 0 ( 1 ) 66 ( 2 ) 13 34 3 4 2 ( 3 ) aa aa aa                                        Solving the first and third equation gives 2, 1   Because the lines intersect, therefore 2, 1   satisfies the second equation 6( 2) ( 1) 2 12 6aa a a   (shown) 13: 44 x ay zm a  Let 13 44 xa y z a  14 , , 34 14 1 4 34 3 4 x ya a z x y aa a a z                                        r

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 7 of 21 (ii) 12 5 :0 6 0 13 1 lO A                     r  Since N is on l, 12 6f o r s o m e 13 ON         42 6 23 AN          AN  is perpendicular to l, therefore 42 2 6. 6 0 23 3 ( 8 6) (4 36 9) 0 2 7                         12 7 4 1 1 21 1060 1 2 1 277 713 7 6 1 ON                                           . Remarks : If the question ask for coordinates of N (instead of position vector of N), you have to conclude your answer as 11 12 1,,77 7   . (iii) Let B be a point on l such that 5AB . Then 12 6 13 OB       for some   12 5 42 60 6 13 1 23 AB                           N A(5,0,1) l

Raffles Institution H2 Mathematics 2024 Year 5 __________________________________________ Y5 H2 Math RTT2 Focus Lesson on Vectors Page 8 of 21    22 2 2 2 42 65 23 (2 4) ( 6 ) (3 2) 5 4 36 9 16 12 16 4 25 49 28 5 0 (7 5)(7 1) 0 51 or77 AB                                  12 12 5106 o r 067713 13 OB OB             