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                       (iii) Method :   1 3b a bb c 4 4 a b a b 1 b a4 , where b b 0a b a b1= , where b a a b4 a b 1 4                                     Method : 1 b cb c 2 1a b a b2 area of area of 1 2 1 2 1 4 OBC OAB BC h AB h                  O B A θ C 3 1 O B A θ C 3 1

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 Revision Guide Chapter 6 Page 6 Let’s Try Now! 2 ACJC Promo 9758/2022/Q5 The line L has vector equation 6 2 : 0 1 1 1 L                     r ,   . (i) Find the position vector of the point N, the foot of perpendicular from the point A with coordinates (3, 6, 1) to the line L. [3] (ii) Find the position vector of the point 'A , the reflection of point A in the line L. [2] (iii) Find the coordinates of the points on the line L that are 3 3 units away from point A. [2] Foot of the Perpendicular from a Point to a Line P A N Step 1: Since N lies on l, for som e O N    a m    ( is unknown, to be found later) Step 2: Find PN   , using PN ON OP        . Step 3: PN l   implies 0PN m    . Use this to find  . Reflection of a point P about a line  l P P X Step 1: Find the position vector of X , the foot of the perpendicular from P to l Step 2: Use the Ratio Theorem to obtain the position vector of the reflection OP  . or '2 OP OPOX PX XP           

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 (i) Since N lies on L, 6 2 , for some 1 ON                6 2 3 3 2 6 6 1 1 2 1 0 1 3 2 2 6 1 0 1 6 4 6 0 2 2 2 1 AN AN ON                                                                                                   (ii) ' 2 2 3 1 2 2 6 2 1 1 3 OA ON OA                                      (iii) Let D denote the point(s) that are 3 3 units away from A on line L. Then 6 2 1 OD             and 3 2 6AD               . (can be taken from (i)).          2 2 2 2 2 3 2 6 3 3 3 2 6 27 6 24 18 0 4 3 0 3 1 0 1 or 3                                             4,1,0 or 0,3, 2D D  A (3, 6, 1) N ൭ −2 1 −1 ൱ L L A (3, 6, 1) D1 D2 3√3 3√3

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 Revision Guide Chapter 6 Pages 7 and 8 Equation of a plane ( π ) in three forms: Vector equation form: 1 2 , ,    r a m m  position vector of a general point on the plane position vector of a particular point on the plane 1 2 and m m are two non- parallel vectors that are parallel to the plane real parameters Scalar product form: π :  r n a n π : , d r n where d is a scalar constant. if the point P lies on π , then the position vector of P will satisfy OP d n  . Note: 1 2 n m m , Cartesian form: Set , , x a y b d z c                     r n a n , : ax by cz d    Angle between two planes Acute angle between and Foot of the Perpendicular from a Point to a Plane n P Q  Step 1: Form the equation of the line PQ Note that the line PQ  passes through the point P, and  is parallel to n (the normal vector of π ) Equation of line PQ ,OP    r n    Step 2: Solve for the point of intersection OQ   of the line PQ and the plane π . line PQ Length of Projection and Shortest distance Length of projection of AP onto the plane AQ AP  ^ n   Shortest ( ) distance from a point P to plane PQ AP ^ n    P

Term 4 Revision Topical Quick Check: Vectors & 3D Vector Geometry TMJC 2024 Revision Guide Chapter 6 Page 10 Method 1: Using GC Equation of Planes: 1 1 1π : p r n and 2 2 2π : p r n . Step 1: Convert 1 2π and π into Cartesian Form Step 2: Use GC. o Press APPS and select PlySmlt2 o Select 2: SIMULT EQN SOLVER Step 3: Interpret the solutions given by the GC 10 3 10 3 2 = 2 By letting The equation of line of intersection is : 10 3 2 0 , . 0 1 x z y z z z l                                                      r  Method 2: Using 1 2n n Given a point A that lies on both planes 1 2π and π, Step 1: Find the direction v