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

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