2026 Chp 1B (Student) - JPJC

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines and Vector Product (Students’ version) / Pg 1 O r direction vector b R A L a Chapter 1(b): Lines and Vector Product 1. Vector Equation of a Line Consider the line l passing through the point A with position vector a and parallel to the vector b. Let R1, R2 and R3 be points on l. Then we have 1 1 1OR OA AR  → = + = + ab  , for some 1 . (since 1AR  // b). Similarly, we have, 2OR → = 22OA AR + = + ab  , for some 2 . 3OR → = 33OA AR + = + ab  , for some 3 . Hence, the position vector of any point on a line can be obtained by identifying a point that the line passes through and a vector that is parallel to the line. Therefore, the vector equation of a line is as follows: Vector equation of a line: =+r a b  ,    Note: 1.  is a real value that gives a particular point on the line. 2. r is the position vector of a point on the line corresponding to a  value. 3. a is the position vector of any known point on the line. 4. b is a vector (called direction or displacement vector) parallel to the line. 5. Vector equation of a line is not unique because of the different choices of a and b. l R3 O A R1 R2 b a

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines and Vector Product (Students’ version) / Pg 2 Example 1 Write down a vector equation of the line passing through the point (1, 3, –2) and parallel to the vector           − 3 2 1 . Solution The position vector of a point on the line is 1 3 2   − Direction vector of the line is 1 2 3  −  Vector eqn of the line is 11 3 2 , 23         = + −       −    r  Example 2 Find a vector equation of the line passing through the points A(1, 5, 6) and B(–1, 1, 4). Solution 11 15 46 AB OB OA −       = − = − =            21 4 2 2 21 −       − = −      −    Direction vector of the line is : 1 2 1    The position vector of a point on the line is : 11 5 or 1 64 −                  Vector eqn of the line is 1 1 1 1 = 5 2 or = 1 2 6 1 4 1  −               ++                      rr    Note: (1) The direction vector can also be found using BA  . (2) Only directional vector can be reduced. O R 1 3 2   − r 1 2 3  − L A ( )1, 5, 6 A ( )1, 1, 4B − O AB  L R r

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines