RI 5 Vectors Soln

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 6 ____________________________________ Y6 H2 Math Term 3 Revision Session 5: Vectors Page 1 of 9 Term 3 Revision Session 5: Vectors Questions 1 MI Prelim 9758/2021/01/Q4 With respect to the origin O, the position vectors of the points A , B and C are a, b and c respectively. Point C lies on AB such that : 1:2AC CB = . It is given that a is a unit vector and the length of OB is 2 units. (i) Give a geometrical interpretation of ac . [1] (ii) It is given that the angle AOB is 60°. By considering ( ) ( )22−−ab ab , find 2 −ab . [3] (iii) Find c in terms of a and b. [1] (iv) Hence by considering cosine of angle AOC and cosine of angle COB, determine if the line segment OC bisects the angle AOB. [3] Suggested Solution (i) Since a is a unit vector, ac is the length of projection of c onto a. (ii) ( ) ( ) 22 2 o2 2 2 4( ) 2( ) 2( ) ( ) 4 4( ) ( ) 4(1) 4 cos 60 (2) 14 4(1)(2) 4 4.2 −− =−−+ = −+ = = −+ =− +=  a b a b aa ab ba bb a ab b ba ab ab        ( ) ( )22− −=ab ab 2 2 4 2 2.− =⇒ −=ab ab (iii) By Ratio Theorem, 2 3 += bac . (iv) 2 2 cos 2 3 [from ] 12 3 12 [since 1 ]3 AOC∠= +=  +=   += = = a c ac baa ( iii)ac ab aa ac ab aa ac     

Raffles Institution H2 Mathematics 2024 Year 6 _________________________________________________________________________________________ ____________________________________ Y6 H2 Math Term 3 Revision Session 5: Vectors Page 2 of 9 2 2 2 cos 2 3 [from ] 1 2( )= 3 1 2 2( ) [since 2 ]32 12 3 COB∠= +=  +  += = =  +=  bc bc bab (iii)bc bb ab bc ab bb bc ab c       Since cos COB∠ cos AOC= ∠ , line OC bisects the angle AOB. 2 DHS Prelim 9758/2021/02/Q5 The points A, B and R have position vectors ,a b and r respectively. (a) The point C has position vector 2 77 3−ab and the point D is such that the origin O is the midpoint of the line segment CD. The point R lies on BD extended such that the ratio of BD to BR is 4 : 7. Show that the points A , O and R are collinear and state the ratio of OA to OR. [4] (b) It is given that the point R has position vector , x y z  = r and that 1 3, 2  = a and 1 5. 3 − = b (i) Determine the exact area of the triangle AOB. [2] (ii) Give the geometrical interpretation of the point ,R given that ( ) 0.⋅×=rab [2] (iii) Find the shortest distance between the point ( 8, 2, 9)−− and the collection of all points R satisfying ( ) 0.⋅×=rab [2] Suggested Solution (a) ( ) 23 77 1 327 OD OC  →  → =− = −− = − d ab ba

Raffles Institution H2 Mathematics 2024 Year 6 _________________________________________________________________________________________ ____________________________________ Y6 H2 Math Term 3 Revision Session 5: Vectors Page 3 of 9 By ratio theorem, Since is parallel to and O is a common point, the points A , O and R are collinear. (shown) The ratio of AO to OR is 2:1 (b) (i) Area of triangle AOB 1 2 11 1 352 23 1 1 52 8 1 9023 102 OA OB= × −   = ×    − = − = =     (ii) ( )0⋅×=rab refers to the collection of all points on the plane that is perpendicular to ×ab and containing the origin. (iii) ( )0⋅×=rab 1 50 8 − ⋅− = r Shortest distance from point to plane 1 5 8 901 90 5 8 8 2 9 90 − = ⋅− = =− − − −  43 7 473 13273477 11 22 OR OBOD OR OD OB OR OA  →  →  →  →  →  →  →  → += = −  = −−   = −= − ba b a OR  → OA  → R B D 4 3 O Perpendicular distance P

Raffles Institution H2 Mathematics 2024 Year 6 _________________________________________________________________________________________ ____________________________________ Y6 H2 Math Term 3 Revision Session 5: Vectors Page 4 of 9 3 MI Prelim 9758/2021/02/Q4 The plane p passes through the points with coordinates ( ) , 2, 5k− , ( )0, 2, 1− and 1, 3, 12 −− , and the line l has equation 24 23 xz y k +− =−=− , where k is a constant. (i) Show that the cartesian equation of the plane is 63 6x y kz k++= − . [2] (ii) Show that line l cannot be perpendicular to p. [2] For the rest of this question, let 2k =− . (iii) Given that l meets p at point N, find the coordinates of N. [3] (iv) Another plane π is parallel to the plane p. Given that the distance between p and π is 11 units, find the possible points of intersection between l and π. [3] Suggested Solution (i) Plane p is parallel to 0 22 0 15 6 kk−       −=      −−    and 11 22 0 321 110 −−     −=    −−   . Normal of p is parallel to 1 2 6 013 60 k k −     ×=    −   . Hence equation of p is 6 06 3 2 3 6. 1 k kk      = = −    −   r Cartesian equation is 63 6.x y kz k++= − (shown) (ii) 24: 2 3 xzly k +− =−=− 24Let 23 23 2 4 23 : 2 1 , 4 xz y k x y zk l k λ λ λ λ λλ +−= =−=− = −− = + = + −−   = +∈    r  ( ) , 2,5k− ( )1 2 , 3, 1− −

Raffles Institution H2 Mathematics 2024 Year 6 _________________________________________________________________________________________ ____________________________________ Y6 H2 Math Term 3 Revision Session 5: Vectors Page 5 of 9 To show that l cannot be perpendicular to p : Method 1: Show l i s not perpendicular to a direction parallel to p. Suppose l is perpendicular to p, then l is perpendicular to 1 2 1 0 −   . [from (i)] Since 3 0.5 1 1 1.5 1 2.5 0, 0k −−     • = += ≠       l is not perpendicular to 1 2 1 0 −   ⇒ l cannot be perpendicular to p. (shown) Method 2: Show l is not parallel to normal of p. Suppose l is perpendicular to p, then l is parallel to the normal of p, i.e. 36 13 t kk −     =       , for some t∈  . 0.536 113 3 1 tt tt k tk t =−−= ⇒=⇒=  = = Since there is no unique value of t, l is not parallel to the normal of p, i.e. l cannot be perpendicular to p. (shown) (iii) 23 : 2 1 , 42 l λλ −−   = +∈   −  r  6 : 3 8 2 p   =− r  When l intersects p, 2 36 2 1 3 8 4 22 λ − −      +=     −−    ( )12 6 8 3 18 4 8 11 22 2 λ λ λ − +−+ − + = −= =− 4 position vector of = 0 8 N    ∴Coordinates of N are ( )4 , 0, 8 . p

Raffles Institution H2 Mathematics 2024 Year 6 _________________________________________________________________________________________ ____________________________________ Y6 H2 Math Term 3 Revision Session 5: Vectors Page 6 of 9 (iv) Method 1: Let the point of intersection of l and πbe M. Since M lies on l, 23 2 1 42 OM λ −−   = +   −    for some λ∈  4 2 36 3 02 1 2 1 84 2 4 2 MN λλ − −  −       = − + = −−       −−      Distance from M to p = 11 ( ) ( ) 222 63 6 12 1 3 11 63 242 2 1 36 6 8 18 3 4 117 22 11 77 22 11 77 or 77 11 55 or 99 λ λ λ λ λ  −     −− =  + +−   −−   −−+ −− = += += − = −  5 or 9λ = − 17 25 7 or 7 6 22 OM −     = −    −     Hence, possible points of intersections between l and π are (−17, 7, −6) and (25, −7, 22). Method 2: Distance between two planes 6 : 3 8 2 p   =− r Let the equation of π be 6 3 2 a   =− r Distance between p and π ( ) 222 8 63 2 a−= + +− 8 7 a−= 8 117 8 77 8 77 or 77 69 or 8