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 and Vector Product (Students’ version) / Pg 3 Example 3 The equation of a straight line is given by 11 12 13         =+          r ,    . Find the value of λ that will give the position vectors below: (a) 1 1 1  =  r , therefore 1 1 1 1 1 2 1 1 3             =+                  =  (b) 3 5 7  =  r , therefore 3 1 1 5 1 2 7 1 3             =+                  =  (c) 0 1 2  =−− r , therefore 0 1 1 1 1 2 2 1 3             − = +          −        = − Note: If there is a unique solution for , the point P lies on the line, otherwise the point is not on the line. Example 4 Determine whether the points P(2, 3, 4) and Q(–1, –3, –6) lie on the line )32()( kjikjir +++++=  ,    . Solution Vector equation of line: 11 12 13         =+          r    Consider point P(2, 3, 4) 2 1 1 3 1 2 4 1 3             = = +                r Equate components and solve for  in each equation: 1 2 = 1 - (1) 1 2 3 = 1 - (2) 1 3 4 = 1 - (3)    + =  + =  + =  Conclusion: Since an unique solution of  is found, the point P(2, 3, 4) _______on the line.

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines and Vector Product (Students’ version) / Pg 4 Consider Point Q(–1, –3, –6) 1 1 1 3 1 2 6 1 3  − − −            = = +                r Equate components and solve for  in each equation: 1 1 + = −  =2 − 1 2 3 + = −  =2 − 1 3 6 + = −  7= 3 − Conclusion: Since the solution of  ___________, the point Q(–1,–3,–6)__________ on the line. Note: All 3 equations must be used to check the uniqueness / non- uniqueness of . If there is a unique value of  , the point lies on the line. Otherwise, the point is not on the line. 2. Cartesian Equation of a Line The Cartesian equation of a line can be derived from its vector equation. Generally, from the vector equation, we have           +           =           3 2 1 3 2 1 b b b a a a z y x  so 33 22 11 baz bay bax    += += += Making λ the subject, we have 1 1 xa b −= , 2 2 ya b −= , 3 3 za b −= The Cartesian equation of the line is: 312 1 2 3 zax a y a b b b −−− ==

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines and Vector Product (Students’ version) / Pg 5 Example 5 Find the C artesian equation of the line passing through the point A with position vector OA  = i + 2j – k and parallel to the vector 2 i – 3j + k. Hence check whether the point B with position vector OB  = i + 2j lies on the line. Solution Vector equation of line is: = x y z   = r 12 23 11         +−      −    ,    x y z  =  12 23 1    +  −−+ Equate components: x = 1 + 2  , y = 2 – 3  , z = –1 +  Making  the subject: 1 2 x −= , 2 3 y −= , 1z =+ Cartesian equation is 13 2 2 1 +=−=− zyx Check if point B(1, 2, 0) is on the line: Now, x = 1 and 1 02 x − = , z = 0 and 11z += . Since 1 12 x z− + , the point B(1, 2, 0) ____________ on the line. L R ( )1, 2, 1A − 2 3 1  − O

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines and Vector Product (Students’ version) / Pg 6 Example 6 Express the vector equation 0 12 = 7 5 14 0         −+          r ,    in Cartesian form. Solution = x y z   = r 0 12 12 7 5 7 5 14 0 14              − + = − +                x = 12  y = –7 + 5  z = 14. 12 x = , 7 5 y += , 14z = ,  The Cartesian equation is 7 , 1412 5 +==xy z . Geometrically, what does the Cartesian equation in Example 6 mean? The line lies on a plane parallel to both the x and y axes, such that the z-coordinates of every point on the line is 14. plane O y x z 14

Jurong Pioneer Junior College H2 Mathematics (9758) JC1-2026 Chapter 1(b) Vectors − Lines and Vector Product (Students’ version) / Pg 7 Example 7 Express the Cartesian equation 33 4 2 3 x y z+− == in vector equation form. Solution Let  = 33 4 2 3 x y z+− == 3 4 , 3 2 , 3x y z   = − + = + = x y z   =