RI revision programme vectors (hard)

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 as shown in the following two-dimensional diagram (not drawn to scale). (d) Relative to the origin O , the point B has position vector k . Determine whether B lies in 1R , 2R , 3R or 4R , justifying your answer. [2] Solution: 2(a) A vector parallel to both 1p and 2p is 2 1 32 25 13 4 s s s +     ×− = −   −−   (b) xz-plane 0y⇒= 2 3 ...(1) 3 1 ...(2) xz xz += − += 2 (2) (1):×− 55 1zz=⇒= 2x∴= −

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Vectors Page 4 of 19 Hence coordinates of A is ( 2,0,1 )− . (c) 1 2 :. 3 1 prs   =− 2 22 2 22 2 21 13. ( 3) 21 21 5 1 rs s ss  − = −=++ ++ + ∴ shortest distance between O and 1p is 2 2 2 2 2 2 33 65 33 65 6 15 6 15 65 1 1 (reject 1 since 0) s s s s s s s ss − = + = + =+ =+ = + = = ±= < 1s∴= − (d) Given 0 0 1 OB k →  = =  , 1 2 :. 2 3 1 pr  −= − , 2 1 :. 2 1 3 pr  −= Method 1: Consider the origin O : 1p : ( ) ( ) ( )20 20 0 0 3− + = >− ( ) ( ) ( )2 :0 20 30 0 1p −+= < Consider the point B : 1p : ( ) ( ) ( )20 20 1 1 3− + = >− ( ) ( ) ( )2 : 0 2 0 31 3 1p −+= > Hence, O and B are on the same side of 1p but O and B are on opposite side of 2p . ∴ B is in 2R . Method 2: Using angles 1 2 :. 2 3 1 pr  −= − ( )0, 0, 3Q − is a point that lies in 1p . Since 1 30OQ⋅ = −<n   , then the angle between OQ and 1n is obtuse. O B Q

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Vectors Page 5 of 19 1 02 0. 2 40 41 BQ    ⋅ = − = −< −  n   , then the angle between BQ and 1n is obtuse. ∴ O and B are on same side of 1p . ( )1−−− 2 1 :.2 1 3 pr  −= ( )1 ,0 ,0T is a point that lies in 2p . Since 2 10OT⋅= >n   , then the angle between OT and 2n is acute. 2 11 0 2 20 13 BT      ⋅ = ⋅− = −<  −  n   , then the angle between BT and 2n is obtuse. ∴ O and B are on opposite side of 2p . ( )2−−− Combining (1) and (2) with reference to given diagram, B∴ lies in 2R . Method 3: (using shortest distance of planes to origin O) Let 3p be plane containing B and parallel to 1p . Equation of 3p is 2 02 .2 0 .2 1 1 11 r        −= −=       2 22 2 22 211 .2 2 ( 2) 1 2 ( 2) 1 1 211i.e. . 2 033 1 r r  −=+− + +− +  −=>   Since equation of 1p 213. 2 1033 1 r  −− = = −< , 1p∴ and 3p are on opposite sides of origin O. …(1) Let 4p be plane containing B and parallel to 2p . Equation of 4p is 1 01 .2 0 .2 3 3 13 r        −= −=       O B T

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ________________________________________ Y6 H2 Math Term 3 Revision Lecture 5: Vectors Page 6 of 19 2 22 2 22 11 33.2 0 141 ( 2) 3 1 ( 2) 3 3 r  −= = >+− + +− + i.e. 113.2 0 14 143 r  −= > Since equation of 2p 111.2 0 14 143 r  −= > , 2p∴ and 4p are on same side of origin O. …(2) Combining (1) and (2) with reference to given diagram, B∴ lies in 2R . Source of Question: NYJC Prelim 9758/2022/01/Q10 3 Two charged particles, U and V, are confined to the planes F1 and F2 with position vectors given by (3 6 ) (1 4 ) (6 2 4 )p pq p q+ + +++ +−ij k and ( 9 3 ) (1 2 ) (3 8 )p pq pq−+ + + − + − +ijk respectively, where ,.pq ∈  (i) Obtain the equation of F 1 in scalar product form. [2] (ii) Find the acute angle between F1 and F2. [3] The forces of the two particles U and V allow another charged particle W to remain suspended between them, such that 2UW = WV. (iii) As the positions of U and V vary, show that the set of points described by the path of W is a line l, whose vector equation is to be determined. [3] (iv) An uncharged particle A is fired along a path described by a line r = sk, where s∈  and crosses F1 at some instant in time. Find the shortest distance of A from l at this instant. [4] Solution: 3(i) 1F : 36 3 6 0 14 1 4 1 62 4 6 2 4 p pq p q pq +         =++= + +        +− −     r 1 6 0 18 3 4 1 24 6 4 24 6 1 −−         = ×= =        −     n

Raffles Institution H2 Mathematics 2025 Year 6 _______________________________________________________________________________________________ ________________________________________ Y6 H2 Math Term 3 Revision Lec