RI revision programme vectors (hard)
Uploaded by blahblahblah03 · 7 October 2025
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RAFFLES INSTITUTION H2 Mathematics 9758 2025 Year 6 Term 3 Revision 5 (Zeta) Topic(s): Vectors (Vectors II and III (Solutions)) ________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Vectors Page 1 of 19 Source of Question: JPJC Prelim 9758/2022/02/Q2 1 (a) With reference to the origin O , the points A , B and X are OA= a , OB= b and 13 88OX = +ab . The point Y lies on AB such that O, X and Y are collinear. Express OY in terms of a and b and find the ratio of :AY YB . [5] (b) The points P, Q and R have non zero position vectors p , q and r respectively. P and Q are fixed, and R varies. Describe geometrically the set of possible positions of the point R such that (i) ( )− ×=rp q 0 , [2] (ii) ( ) 0−=r p .q . [2 ] Solution: 1(a) Y lies on line AB ( ),OY β⇒= +− a ba for some .β∈ Since O, X and Y are collinear, 13() 88βα + −= + a ba a b for some .α∈ Since a and b are non-parallel, Comparing, 1 8 αβ−= ---(1) 3 8 αβ = ---(2) Solving (1) and (2), 32, 4αβ= = 13 132 88 44OY = += + ab ab 3 4AY AB= : 3:1AY YB = O A B •X •Y
Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Vectors Page 2 of 19 (b)(i) ( )− ×=rp q 0 We see that either ( ) //−r p q or ( )−=rp 0 , 0kk= +≠rp q or =rp Thus , kk= +∈rp q The set of all possible positions of the point R is the line that passes through the point P and parallel to the vector q. (b)(ii) ( ) 0 0 −= −= = r p .q r.q p.q r.q p.q The set of all possible positions of the point R is the plane that is perpendicular to q and containing the point P.
Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Vectors Page 3 of 19 Source of Question: HCI Prelim 9758/2023/02/Q5 2 The equations of two planes 1p and 2p are 2 3, 2 3 1, x sy z xyz ++= − −+= respectively, where s is a negative real constant. (a) Find a vector parallel to both 1p and 2p , giving your answer in terms of .s [2] (b) It is given that 1p and 2p intersect in a line l , and l meets the xz-plane at a point A . Find the coordinates of A . [3] (c) The shortest distance between the origin O and 1p is 3 6 . Find the value of s . [2] It is now given that 2,s=− with the normal vectors to 1p and 2p denoted as 2 2 1 = − 1n and 1 2 3 = − 2n respectively. It is given that 1p and 2p divide the real space into four regions 1R , 2R , 3R and 4R
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