TMJC H2 Chp 5 Vectors Learning Package 2024

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Chapter 5 Vectors TMJC 2024 Page 1 of 28 H2 Mathematics (9758) Chapter 5 Vectors Core Concept Notes Success Criteria: Surface Learning Deep Learning Transfer Learning  Identify position vectors, displacement vectors and direction vectors.  Convert a vector in cartesian form x y z i j k into column vector form x y z         .  Express a displacement vector in terms of the position vectors of its end points. (e.g. AB OB OA        )  Carry out addition and subtraction of vectors, multiplication of a vector by a scalar, and interpret these operations in geometrical terms.  Calculate the magnitude of a vector.  Find the unit vector of a given vector.  Determine if 2 given vectors are parallel.  Calculate dot (scalar) product of two vectors in component form.  Find angle between 2 vectors using the definition of dot product cosa b a b .  Calculate cross (vector) product of two vectors in component form.  Find a vector perpendicular to two non-parallel vectors using cross product.  Use of properties of dot and cross product of equal vectors. (i.e. 2 a a a ,  a a 0 )  Find a vector of a specified magnitude that is parallel to a given vector.  Apply ratio theorem to find vectors.  Determine whether three points with given coordinates are collinear.  Determine if 2 given vectors are perpendicular using dot product (i.e. 0a b ).  Find length of projection of a vector a onto a vector b  Find projection vector of a vector a onto a vector b.  Find the perpendicular/shortest distance from a point to a line.  Use cross product to find area of triangle and parallelogram.  Use properties of dot and cross product to solve problems.  Use ratio theorem or collinearity in geometrical applications.  Visualise using a diagram and use vector concepts to solve problems.  Give geometrical interpretation of a b , a b and a b .  Solve vector related questions involving unknowns.

Chapter 5 Vectors TMJC 2024 Page 2 of 28 §1 Introduction to Vectors At the ‘O’ Level, you have learnt about vectors in 2-dimensional space. We will be now exposed to vectors in 3 -dimensional space, which is more realistic since we live in a 3 -dimensional world. All operations in 2-D also apply in 3-D. Basic Vector Operations (3 dimensional) (i) 1 2 1 2 1 2 1 2 1 2 1 2 x x x x y y y y z z z z                               (ii) , x x y y z z                         (iii) If 1 2 1 2 1 2 x x y y z z                  , then 1 2x x , 1 2y y and 1 2z z . 1.1 3-Dimensional Vectors in Cartesian Form In the 3-dimensional Cartesian space with the x, y, z axes, i, j and k are mutually perpendicular unit vectors along the x, y and z axes respectively and i.e. 1 0 0        i , 1i ; 0 1 0        j , 1j ; 0 0 1        k , 1k . Let  , ,P x y z be any point in space, and let r be its position vector. O i j k P (x, y, z) z r y D (0, 0, z) B A (x, 0, 0) C (0, y, 0) x

Chapter 5 Vectors TMJC 2024 Page 3 of 28 In Cartesian form, the vector OP   is given by . . . OP OA AB BP OA OC OD x y z           r i j k               In column vector form, 1 0 0 . 0 . 0 1 0 1 . 0 OP x y z x y z x y z                                          r i j k   . Exercise: Given the diagram below such that OA, OC and OG are 5, 4 and 3 units respectively, write down, in column vector form, the position vectors of the points A, B, D and E. Solution: 5 0 0 OA          , 5 4 0 OB          , 0 4 3 OD          , 5 4 3 OE          O i j k E z y D B A C x F G 5 units 4 units 3 units

Chapter 5 Vectors TMJC 2024 Page 4 of 28 The magnitude of the vector OP   is given by 222 zyx r . The unit vector in the direction of r is   2 2 2 2 2 2 1 1ˆ x x y z y x y z x y z z                 rr i j kr Example 1 Find the unit vector in the direction of .OD   Solution: 2 22 , 2 4 5 0 4 3 50 3OD OD                 Hence, the unit vector in the direction of 0 1 4 .5 3 OD           Exercise: Write down a unit vector parallel to OD   . 1 5 0 4 3         or 3 1 5 0 4         . Note: (a) If P is a point in the x-y plane, then 0 x OP x y y         = i + j =   (i.e. 0z ) (b) If Q is a point in the x-z plane, then 0 x OQ x z z         = i + k =   (i.e. 0y  ) (c) If R is a point in the y-z plane, then 0 OR y z y z         = j + k =   (i.e. 0x )

Chapter 5 Vectors TMJC 2024 Page 5 of 28 Example 2 The position vectors of points A, B and C are 2i  j + k , 3i + 5j  4k , and 4i + j respectively. Find (i) the length of AB , (ii) the unit vector in the direction of AC , (iii) the position vector of D so that ABCD is a parallelogram, (iv) the position vector of E such that 2BE AC     . Solution: (i) 3 2 1 5 1 6 4 1 5 AB                                  AB      2 2 21 6 5 62    ( ) (ii) 4 2 2 1 1 2 0 1 1 AC OC OA                                        Unit vector in the direction of AC   2 2 2 2 1 2 2 2 ( 1) 1 2 1 2 3 1                     (iii) If ABCD is a parallelogram, then AB DC      or AD BC      1 4 6 1 5 0 OC OD OD                            4 1 3 1 6 5 0 5 5 OD                                 (iv) 2BE AC     3 2 5 2 2 4 1 OE OB OE                             4 3 7 4 5 9 2 4 6 OE                                  D C B A Recall that: AB OB OA        Convert to column vectors first > >

Chapter 5 Vectors TMJC 2024 Page 6 of 28 §2 Basic Vector Algebra 2.1 Triangle Law of Vector Addition Consider three vectors OA   , OB   and AB   formed by the points O, A, and B. Observe that these vectors make up the sides of a triangle as shown and that OB ABAO        This is known as the triangle law of addition. OB OABA        For any two column vectors, p q     a and x y     b , p x q y       a b . Example: 2 5 3 1 3 4                   2.2 Parallelogram Law of Vector Addition Consider the parallelogram OBCA below. If OA a   , OB b   , then OC  a b   , where OC is a diagonal of the parallelogram OBCA. Proof: OC OA AC   a b       Note: AC OB  b     as AC   and OB   are equal vectors. They have same magnitude and are in the same direction. Exercise: Write down AB   (the other diagonal) in terms of a and b. AB AO O OA B OB       ab 