23 Vectors 2 APQ Soln (RVHS)

7 APQ: Vectors II (The Scalar and Vector Products of Vectors) 1. MJC/2017/Prelim/I/Q4(a) The points A and B relative to the origin O have position vectors 3 3 i j k and 3 2 i j respectively. (i) Find the angle between OA  and OB . [2] (ii) Using the result in part (i), find the shortest distance from B to line OA. [2] (i) Let  be the angle between OA  and OB . 1 3 3 1 2 3 0cos 3 3 1 2 3 0 11cos 134.4 (1 d.p) = 2.35 radian (3 s.f)19 13                                           (ii) Let h be the shortest distance from B to line OA. sin134.42 13 sin134.42 2.5752 2.58 units (3 s.f) h h      b

8 2. PJC/2018/MYE/P2/Q1 Referred to the origin O, the points A and B are such that OAa   and OBb  . The point M is on AB produced such that : 3 : 2AB BM and the point N is on OB such that : 1: 2ON NB . It is given that 3a , b is a unit vector and 60AOB   . (i) Find OM in terms of a and b. [2] (ii) Show that the area of triangle OMN can be written as k a b, where k is a constant to be found. Hence evaluate the exact area of triangle OMN. [4] (iii) Find the length of projection OM onto ON. [3] (i) Using Ratio Theorem, 2 3 5 OA OMOB         5 2 3 3 5 2 1 15 2 (5 2 )3 3 OB OA OM OM OB OA OM OB OA         b a                   (ii) Area of triangle OMN  1 2 1 1 1(5 2 )2 3 3 1 1 ,9 9 1 sin9 1(3)(1)sin 609 3 6 OM ON k              b a b a b a b n     (iii) Length of projection of OM on ON =  2 1 1(5 2 )3 3 15 23 15 2 cos 603 1 15 2(3)(1)3 2 2 3 OM ON ON                b a b b b a b b a b       Note: 1 3ON b  and 1 3ON 

9 3. CJC Mid Year 9758/2021/3 (modified) The vectors a and b are such that 2 3 2 3q p p q      a b i j k and 1 8a b where p and q are non-zero constants. It is given that b is a unit vector. (a) If 2p and 3q , state the relationship between a and b. [1] (b) (i) Show that      1 34 2 14 3 7 2 7 32 2 q p p q                 a b a b i j k . [2] (ii) If 1p and 1 34 2 1122 2             a b a b , find the exact values of q. [2] (iii) Given instead that 14 2a b and 32 2a b are perpendicular, find the value of a. [3] (a) The vectors a and b are parallel (b)(i)            1 34 22 2 38 6 4 0 6 0 7 7 2 3 2 3 14 3 7 2 7 3 q p p q q p p q                                            a b a b a a a b b a b b a b a b a b i j k i j k   (ii) When 1p ,                2 2 2 2 2 2 2 2 2 7 2 3 2 3 112 7 2 3 1 3 112 4 3 1 3 16 5 3 1 16 5 3 255 3 51 3 51 q q q q q q q q q q                                 i j k (iii) 1 34 2 02 2            a b a b

10   2 2 2 2 38 6 0 4 38 5 0 4 3 18 1 54 8 1 8 1 8 a a a b b a b b a a b b a a                                    4. TJC Mid Year 9758/ 20201/ P1/ Q5 (modififed) Referred to the origin O, the position vectors of points A and B are a and b respectively. Given a is a unit vector, b = 2 and AOB = 120o. (i) Find 23a b . [3] (ii) Find 3 3a b a b   and state a geometrical interpretation of 3 3a b a b   .[3] A point E lies on the line AB such that OE is perpendicular to AB. (iii) Find the position vector of E, in terms of a and b. [3] (i)            2 2 2 2 3 3 3 3 3 9 6 9 1 6 cos120 9 2 1 6 36 31                   a b a b a b a a a b b a b b a a b b a b       Need to see cos120 1 a b a b  Alternative: Use Cosine rule     2 2 2 2 2 3 3 2 3 cos 60 11 3 2 6 1 2 312              a b a b a b

11 (ii)        3 3 3 9 3 9 110 10 20 2 10 sin120 310 1 2 10 32                                     a b a b a a a b b a b b a b a b a b a b a b a b 3 3  a b a b represents 20 times the area of the triangle OAB OR It is the 10 times the area of the parallelogram formed by adjacent sides OA and OB (iii) Let OE   a b a  ,   since E lies on line AB OR by ratio thm since E lies on line AB A, B and E are collinear: 1OE    b a  Given OE  is perpendicular to AB               0 1 0 4 1 1 1 1 1 0 4 2 2 0 7 2 2 7 OE                       b a b a b a              5 2 7 7OE a b 