SRJC_H2_MATHS_P2_Soln

1 [Turn Over SERANGOON JUNIOR COLLEGE 2014 JC2 PRELIMINARY EXAMINATION MATHEMATICS Higher 2 9740/2 Tuesday 26 Aug 2014 Additional materials: Writing paper List of Formulae (MF15) TIME : 3 hours READ THESE INSTRUCTIONS FIRST Write your name and class on the cover page and on all the work you hand in. Write in dark or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps u sing mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. The number of marks is given in brackets [ ] at the end of each question or part question. At the end of the examination, fasten all your work securely together. Total marks for this paper is 100 marks. This question paper consists of 7 printed pages and 1 blank page.

2 [Turn Over Section A: Pure Mathematics [40 marks] 1 The line ℓ with Cartesian equation 4 43 xz  , y = 1 contains the point B with position vector j + 3k. A point A, not lying on ℓ, has position vector 2i + ( 15 )j ‒ k. (i) Given that c denotes a unit vector parallel to ℓ, find AB  c and give a geometrical interpretation of this quantity. [3] (ii) Hence find the shortest distance from A to ℓ. [2] The foot of perpendicular from A to ℓ is denoted by F and the foot of perpendicular from F to AB is denoted by G. (iii) Write down the ratio between the area of ∆AGF and area of ∆BGF. [1] (iv) Hence, deduce the ratio AG:GB and find the position vector of G. [2] Solution (i) 4 4 x = 3 z , y = 1 ⇒ 40 43 xz  , y = 1. Vector equation of ℓ is 44 1 0 , 03                     r . A unit vector c parallel to ℓ = 4 1 05 3    (OR 4 1 05 3    ) 204 11 1 5 0 53 1 3 AB                        c 2 4 1 505 43 4        AB c is the length of projection of AB onto ℓ (or onto c). (iii) 2 55 4 AB      A B ℓ

3 [Turn Over By Pythagoras’ theorem, The shortest distance from A to ℓ = 225 4 3 (iv) AGF and ∆BGF are similar triangles. Since AF corresponds to BF, 2 2 Area of AGF 3 9 Area of BGF 4 16   (v) 1 ( )( )Area of 2 1Area of ( )( )2 AG GFAGF AG BGF BG BG GF