TMJC Term 4 Revision Chapter 5 & 6 Vectors and 3D Vector Geometry Solutions

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 JC1 H2 Mathematics (9758) Term 4 Revision Topical Quick Check Chapter 5 Vectors Chapter 6 3D Vector Geometry Revision Guide Chapter 5 Page 2-3 Ratio Theorem OA OBOP            A B P | | |   x O Unit Vector ˆv A unit vector is a vector whose magnitude is 1 ˆ 1v and ˆ vv v Scalar (Dot) Product Definition: cosa b a b (0 180 o o ) Uses of Scalar Product Angle between Two Vectors cos  a b a b Proving two non-zero perpendicular vectors a  b 0  a b Length of projection of b onto a: Properties of scalar product: 1.  a b b a 2.       a b c a b a c 3. 2  a a a 4.       a b a b a b

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 Let’s Try Now! 1 CJC Promo 9758/2022/Q9 Relative to the origin O, the position vectors of points A, B and C are a, b and c respectively, where b is a unit vector. It is given that b and b − a are perpendicular and C lies on AB such that : 3:1AC CB  . (i) Show that seca , where  is the angle between a and b. [3] By expressing c in terms of a and b, (ii) find the value of c b and state the geometrical interpretation of c b , [4] (iii) find the value of   b c a b . [2] Vector (Cross) Product Definition: o oˆsin (0 180 )   a b a b n So sin a b a b Normal Vector a b is vector perpendicular to both a and b Zero Vector  a b 0  a is parallel to b OR a 0 OR b 0 Properties of vector product: 1. (  a c + b) = a c + a b 2. c c = 0 (zero vector) 3.   a c = c a 4.       a c = a c a c

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 (i) Method : Let  be the angle between a and b . Since b and b a  are perpendicular,   2 b b a 0 b b b a 0 b b a a b                        Since b is a unit vector, a b 1   a b cos 1   a cos 1 1a seccos        Method : cos 1 since is a unit vector 1 sec (shown)cos OB OA b a ba a            (ii) By Ratio Theorem, 1 3c a b4 4    2 1 3c b a b b4 4 1 3a b b b4 4 1 3= b b b b4 4 b 1                               since a b b b      c b  is the length of projection of c on b . O B A θ O B A θ C 3 1

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 Alternative: If Method 2 was use in (i),  a b a b cos sec 1 cos 1          since from (i) c b  is the length of projection of c on b . By Ratio Theorem, 1 3c a b4 4        1 3c b a b b4 4 1 3a b b b4 4 1 3= 1 14 4 1               