RVHS 2023 JC2 H2MA Lec Test (Std Soln)

RVHS 2023 JC2 H2 Maths Lecture Test 2 1 Name : Index Number : Class : Date : Duration : 50 mins Max. No. of Marks : 30 List of Formulae [Answer all the questions on writing papers. Up to 1 mark will be deducted for poor presentation.] 1. The lines l1 and l2 have equations: 1 2 2:1 2 51 : 5 0 , 01 yl x z l  += = − −       = − +           r (i) Verify that the point A(1, 0, 2) lies on l1. [1] (ii) Find the position vector of the foot of the perpendicular from A to l2. Hence find the coordinates of the point of reflection of A about l2. [5] (iii) Find the angle between l1 and l2. [2] (iv) Determine whether l1 and l2 are intersecting, parallel or skew lines. [3] 2. A baby’s toy teaches babies music by having them place lettered balls into boxes. The toy comprises five boxes arranged in a row. Seven balls each with a different letter from {A, B, C, D, E, F, G} written on it, are placed into the boxes. Each box must have exactly one ball in it. There will be two balls not used. When the “play” button is pushed, the toy will play the notes corresponding to the letters shown on the balls in order. A baby randomly puts five of the balls into the row of boxes. Find the probability that (i) the notes C, E and G are played consecutively, in that order, [3] (ii) the notes C and G are both played but separated by other notes, [2] (iii) the notes D, F and A are played given that the five notes are played in alphabetical order. [4] River Valley High School Integrated Programme 2023 JC2 H2 Mathematics (9758) Lecture Test 2 (Term 1)

RVHS 2023 JC2 H2 Maths Lecture Test 2 2 3. The points A, B, C, and D are such that ABCD is a parallelogram. It is given that AB= a and AD= b . The point P cuts BC such that BP:BC is  for some 01  . The point Q lies on AP produced such that AP:AQ is also  . (i) Show that 1AQ =+ ab . Hence show that D, C and Q are collinear. [3] (ii) It is given that the area of the parallelogram ABCD is equal to the area of the triangle ADQ. Find the value of  . [4] (iii) Given further that =ab , show that AQ is not perpendicular to DQ . [3] ~ The End ~

RVHS 2023 JC2 H2 Maths Lecture Test 2 3 RVHS 2023 JC2 H2 Lecture Test 2 (Student Solution) 1 Vectors in 3-dimensions [11 marks] (i) Substituting (1, 0, 2) into the Cartesian equation: ( ) ( ) ( )021 2 1 2 += = − (verified) (ii) Let B be (5, -5, 0) which lies on l2 and F be the foot of the perpendicular from A to l2. Then, 11 00 11 4 1 1 1 5 0 02 2 1 1 3 0 3 BF BA  −−       =              − −  −           =                 − = 53 50 03 2 5 3 OF OB BF=+ −       = − +           =− Alternative Let F be the foot of the perpendicular. Since it is a point on l2, 5 1 4 5 0 5 22 AF OF OA   =− −−           = − − = −           −      Since it is perpendicular to l2, 41 5 0 0 21 4 2 0 3     −−       −=      −    − + − = =

RVHS 2023 JC2 H2 Maths Lecture Test 2 4 11 05 21 2 5 3 OF OA AF=+        = + −           =− Let A be the point of reflection of A in l2 F is the midpoint of AA ( ) 1 2OF OA OA =+ 2OA OF OA=− 2OA OF OA=− 2 1 3 2 5 0 10 3 2 4 OA            = − − = −                (3, 10,4)A − (iii) Converting l1 into vector form: 1 01 : 2 2 , 11 l         = − +           r Let θ be the angle between the lines. Then 11 20 11cos 0 11 20 11  −                 == −                  Therefore, the angle between the lines is 90°. (iv) Since the direction vectors of the lines are not parallel, the lines are either intersecting or skew lines. Solving,

RVHS 2023 JC2 H2 Maths Lecture Test 2 5 0 1 5 1 2 2 5 0 1 1 0 1 (1) : 5 (2) : 2 0 3 (3) : 1     −               = − + = − +                      += + =− − =− r Solving simultaneous, (1) and (3) gives a solution 2, 3== whereas, (2) gives a solution 1.5 =− . Therefore, the system is inconsistent, and the lines are thus not intersecting. Therefore, they are skew lines.

RVHS 2023 JC2 H2 Maths Lecture Test 2 6 2 Probability and Permutations & Combinations [9 marks] (i) Number of ways to fill 5 boxes with 7 balls 7 5!5 = ( ) ( )( ) ( )( ) 4 3!2P CEG played in order 7 5!5 66 1 21 120 70  =   == Alternatively ( ) 1 1 1P CEG played in order (3) 7 6 5 1 70    =       = (ii) ( ) ( )( )( )( ) ( )( ) 543! 2!32P CG played but separated 7 5!5 10 6 6 2 2 21 120 7           =   == Alternatively ( )P CG played but separated Number of ways with C and G in selection Number of ways with CG together 7 5!5 55(5!) (4!)(2!)33 7 5!5 2 7  −=      −      =   = Alternatively ( ) 21P CG played but separated (3)(2!)76 2 7   =     =

RVHS 2023 JC2 H2 Maths Lecture Test 2 7 (iii) ( ) 7 (1)5P all five alphabetical 7 (5!)5  =   ( )P ADF played and all alphabetical 4 2 7 (5!)5  =   ( ) ( ) ( ) P ADF played | alphabetical P ADF played and all alphabetical P all alphabetical 4 2 2 7 7 5 =  == 

RVHS 2023 JC2 H2 Maths Lecture Test 2 8 3 Abstract Vectors (10 marks) (i) ( ) (1 ) (1 ) AP AC AB   = + − = + + − =+ a b a ab ( ) ( ) 1 11 1 (Shown) AQ AP AP     +−= = = + =+ ab ab DC = a 1 1 1 DQ DA AQ DC    =+ =− + + = = b a b a Thus DQ parallel to DC with D as a common point. D, C and Q are collinear. (ii) Area Area 1 2 11 2 11 2 1 2 1 2 ABCD ADQ AD AQ     =  =  =  + =  +  = = ab b a b b a b b ba

RVHS 2023 JC2 H2 Maths Lecture Test 2 9 (iii) 2 2 2 2 2 11 11 11 cos 1 cos 0 AQ DQ        =+       =+ =+ =+   a b a a a b a a b a Since 1 1 and 1 cos 1   −   Alternatively Using the value of 1 2= obtained in part (ii), ( ) ( ) ( ) 2 2 2 22 42 4 2 cos 2 2 cos 0 AQ DQ   =+ =+ =+ =+  a b a a a b a a b a Since 1 cos 1 −  