23 Vectors 2 BMQ Soln (RVHS)

1 Tutorial 8B: Vectors II (The Scalar and Vector Products of Vectors) Basic Mastery Questions 1. The position vector of A relative to an origin O is 3 5 8 i j k . Given that 8 2 5AB  i j k  , find AOB . 3 11 5 , 7 8 13 OA OB OA AB                            33 35 104 172OA OB        1 1 172cos 172 cos 9 25 64 121 49 169 19.3 (to 1d.p.) AOB OA OB               2. The position vectors of A and B relative to an origin O are 6 4 i j k and 3 2p i j k respectively. Express AO AB    in terms of p and hence find (i) the value of p for which AO is perpendicular to AB , (ii) OAB when 6p . 6 3 4 , 1 2 3 4 3 OA OB p AB OB OA p                                     ( 6)( 3) ( 4)( 4) (1)(3) 37 4AO AB p p              (i) If AO AB   , then 0AO AB     37 4 0 37 4 p p      (ii) When p = 6, 3 13 and 2 3 AO AB AB                1 1 13cos 13 cos 36 16 1 9 4 9 67.6 (to 1 d.p.) OAB AO AB                      

2 3. The position vectors of points A and B are 2 0.5  a i j k and 4 3 b i k . Find (i) the perpendicular distance from A to the line segment passing through O and B, (ii) the exact area of the triangle OAB. (i) Required distance 2 2 1 4 2 0 0.5 3 61 54 3 8 125 5 units5 OA OB                                      (ii) Required area 2 1 2 1 41 2 02 0.5 3 61 1255 units2 28 OA OB                                Tutorial Questions 1. N99/1/Q6 The angle between the vector 3 6 i j k and the vector i is 120. Find the exact value of the constant . 2 2 2 2 2 2 1 3 0 9 36 (1) cos120 6 0 1 45 - - - - - (1)2 ( 2 ) 45 3 45 15 15 (rej.) or 15 ( is ve, from (1))                                            O B A