RI Vectors C2A Lect Notes

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 _____________ Chapter 2A: Vectors I Page 1 of 20 Chapter 2A: Vectors I Vector Algebra, Ratio Theorem, Scalar and Vector Products SYLLABUS INCLUDE • Addition and subtraction of vectors, multiplication of a vector by a scalar, and their geome trical interpretations • Position vectors, displacement vectors and direction vectors • Magnitude of a vector • Unit vectors • Distance between two points • Collinearity • Use of the ratio theorem in geometrical applications • Concepts of scalar product and vector product of vectors and their properties • Angle between two vectors • Geometrical meaning of ˆ⋅an and ˆ ,×an where ˆn is a unit vector CONTENT 1 Vectors in Three Dimensions (3D Vectors) 1.1 Modulus of a Vector 2 Some Fundamental Results 2.1 Parallel Vectors and its Applications 2.2 Linear Combinations of Non- Parallel Vectors 2.3 The Ratio Theorem 3 The Scalar Product (or Dot Product) 3.1 Definition of Scalar Product 3.2 Basic Properties of Scalar Product 3.3 Perpendicular Vectors 3.4 A Formula in Cartesian Form for Scalar Product 3.5 Applications of Scalar Product 4 The Vector Product (or Cross Product) 4.1 Definition of Vector Product 4.2 Properties of Vector Product 4.3 Vector Product for Vectors in Cartesian Form 4.4 Applications of Vector Product We will now move on to study vectors in three dimensions. When it comes to operations like addition, subtraction and scalar multiplication , the situation is totally analogous to what you have learn in secondary school. The only difference is the introduction of a third (z) component within the vectors. We shall start with a recap of the types of vectors.

Raffles Institution H2 Mathematics 2025 Year 5 ________________________________________________________________________________________________ _____________ Chapter 2A: Vectors I Page 2 of 20 In general, there are two types of vectors – position vectors and free vectors. A position vector is used t o denote the location of a point. It defines the position of one point relative to another point. A free vector (also known as displacement vector) is a vecto r with no associated position. OP    and OQ   are the position vectors of points P and Q with respect to the origin O . On the other hand, the vectors a and b are free vectors. In fact, a can be represented by PR    , or any directed line segment of length a and having the same direction as a . Finally, a direction vector i s a vector that describes the direction of a line segment (this will be discussed more in the next chapter) 1 V ectors in Three Dimensions (3D Vectors) To describe a vector in three dimensions, coordinates axes are chosen in the ,xy and z−directions as shown on the right. Each vector can then be represented by giving its “components” in each of the chosen directions. In the figure on the right, the vector OQ   is 3 units in the x direction, 5 units in the y direction, and 4 units in the z direction. We write 354OQ=++ijk   , where ,ij and k are vectors of magnitude 1 in the ,xy and z directions respectively. In column vector form, 3 5, 4 OQ  =   where 1 0, 0  = i 0 1 0  = j and 0 0. 1  = k Note: Vectors of magnitude 1 are known as unit vectors which will be discussed in details in section 2.1.2. Q (3, 5, 4) 3 5 4 R

Raffles Institution H2 Mathematics 2025 Year 5 ________________________________________________________________________________________________ _____________ Chapter 2A: Vectors I Page 3 of 20 Example 1 The diagram shows a cuboid where 5OA= units, 3OC = units and 2OP= units. Using O as the origin, unit vectors i, j and k are taken along OA, OC and OP respectively. State the corresponding column vectors for OQ   , SQ    and BP    . Solution 5 0 2 OQ  =   , 5 3 0 SQ  = −    , 5 3 2 BP − = −    1.1 M odulus of a Vector The magnitude of a vector a , denoted by a , is also known as its modulus. If x y z  = a , then 2 22xyz= ++a (by Pythagoras’ Theorem applied two times). 3 5 2

Raffles Institution H2 Mathematics 2025 Year 5 ________________________________________________________________________________________________ _____________ Chapter 2A: Vectors I Page 4 of 20 Example 2 With reference to the origin O , the position vectors of A , B and C are given by 2 3 4 OA  =−    , 5 1 2 OB  = −    and 11 14 OC λ  =   . (i) Find the exact distance between A and B . (ii) Find the position vector of D in terms of λ if ABCD is a parallelogram. Solution (i) 52 3 13 4 2 46 AB OB OA      = − = −− = −    −            Required distance between A and B 2 223 ( 4) 6 61 AB= = +− + =    . (ii) 11 3 8 44 14 6 8 AB DC OC OD OD OC AB λλ = = −        = − = −− = +                        2 Some Fundamental Results 2.1 Parallel Vectors and its Applications 2.1.1 Parallel Vectors Let a and b be two non-zero vectors. We say that a is parallel to b , denoted by ab , if and only if λ=ba for some { }\0λ∈  . In other words, two non-zero vectors are parallel if one is a scalar multiple of another. 2.1.2 Unit Vectors A unit vector a is a vector whose magnitude is unity, i.e. 1=a . Recall 10 0 0 , 1 and 0 00 1              i= j= k= are unit vectors. Other vectors such as 0.6 0 10.8 , 5 130 12      −−       a= b= are also unit vectors. For any non-zero vector a , the unit vector in the direction of a is denoted by ˆa .

Raffles Institution H2 Mathematics 2025 Year 5 ________________________________________________________________________________________________ _____________ Chapter 2A: Vectors I Page 5 of 20 If b is a vector parallel to a , then ˆλ=ba , where λ is the magnitude of b . In particular, ˆ=a aa or 1ˆ =aa a . Remark: In short, to obtain a unit vector from a given vector, it suffices to divide the vector by its magnitude (length). Example 3 If 3= −c ab , where 2= +−a i jk and 6=−+bij , find a unit vector parallel to c . Hence write down the vector of length 10 and in the same direction as c . Solution ~ ~~ 1 14 3 32 6 0 10 3 c ab −   = −= − =   −−  A unit vector parallel to ~ c is given by ^ 22 2~~ ~ 44 11 1 00 540( 3 ) 33 cc c    = = = + +−  −−  Another unit vector parallel to ~ c is 4 1 05 3  − − The vector of length 10 and in the same direction as ~ c is 44 110 0 2 05 33    ×=   −−  2.1.3 Collinear Points Three points ,AB and C are collinear (i.e. lie on a straight line) if and only if AB AC      , with A as the common point. In other words, AB ACλ=     for some { }\0λ∈  . We can also take B or C as the common point instead of A by showing BA BC     or CA CB     respectively.

Raffles Institution H2 Mathematics 2025 Year 5 ________________________________________________________________________________________________ _____________ Chapter 2A: Vectors I Page 6 of 20 Example 4 With reference to the origin O , the position vectors of A , B and C are given by 23OA= −ij    , 434OB=+−ijk    and 10 21 16OC =+−i jk   . Show that A , B and C are collinear. Solution 42 2 3 36 40 4 AB      = −− =    −−      10 4 6 21 3 18 16 4 12 BC        = −= =      − −−      2 36 3 4 AB   =−  