RI Vectors C2B Lecture Notes

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 ________________ Chapter 2B: Vectors II Page 1 of 17 Chapter 2B: Vectors II – Equations of Straight Lines SYLLABUS INCLUDE  Vector and cartesian equations of lines  Finding the foot of the perpendicular and distance from a point to a line  Finding the angle between two lines  Relationships between two lines (coplanar or skew) CONTENT 1 Equations of Straight Lines 1.1 Vector Equation of a Straight Line 1.2 Parametric Form of Equation of a Straight Line 1.3 Cartesian Equation of a Straight Line 2 Calculations Involving a Point and a Line 2.1 Determining whether a Point lies on a Line or a Line passes through a Point 2.2 Finding the Foot of Perpendicular from a Point to a Line and the Corresponding Perpendicular Distance 3 Calculations Involving a Pair of Lines 3.1 Parallel Lines, Intersecting Lines and Skew Lines 3.2 Acute Angle between Two Lines Appendix A: Guide on Solving Simultaneous Equations with/without a GC INTRODUCTION In this chapter, we shall see how the equation of a straight line can be expressed in three forms : vector, parametric and cartesian. We will then use the equation to solve problems involving distances, intersections and angles. 1 Equations of Straight Lines 1.1 Vector Equation of a Straight Line Recall that to find the equation of a straight line in the xy plane, we need a point which lies on the line and the gradient of the line. Similarly, to obtain a vector equation of a line, we need (i) the position vector of a point on the line, and (ii) a vector parallel to the line (known as the direction vector).

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ________________ Chapter 2B: Vectors II Page 2 of 17 A straight line l which passes through a fixed point A with position vector a and is parallel to a vector b has vector equation given by :,l   ra b  , where r represents the position vector of any point on the line l . Each real value of  gives the position vector of a point on the line l . To see how this formula arises, consider any point P on the line that is different from A . Since l is parallel to b , the vector AP  is also parallel to b  b  AP for some \{ 0 }  . Thus OP OA AP  ra b        for some \{ 0 }  . When point P is A , clearly OPra  (i.e. 0  ) Example 1 Relative to the origin O , the points , A B and C have position vectors given by 732aij k , 23 bi j k and 10 s t ci j k respectively, where s and t are constants. (a) Find a vector equation of the line passing through A and .B (b) Find a vector equation of the line passing through O and parallel to .a (c) The line passing through A and C has equation given by (7 3 2 ) ( 3 ),  ri j k i j   . Find s and t . Solution (a) 27 5 13 2 32 1 AB             ba  A vector equation of the line passing through A and B is 75 32 , 21              r  , P B A P O AB 

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ________________ Chapter 2B: Vectors II Page 3 of 17 or 25 12 , 31              r  , or … Remark: Vector equation of a line is not unique! (b) A vector equation of the line passing through O and parallel to a is 07 03 , 02         r  or 7 3, 2     r  (c) 3 3 2 AC s t      ca   Since AC   is parallel to 1 3 0    , we have 31 33 20 sm t       for some 0m 3 3 3 3(3) 3 12 20 2 m sms tt          1.2 Parametric Form of Equation of a Straight Line Let the vector equation of a straight line be :,   ra bl  . By writing 11 22 33 , and          ra b abx ya b z ab , we have 11 22 33          abx ya b z ab 11 22 33 ,         xa b ya b za b  This set of three equations is known as the parametric form of the equation of the line l . B A P O BA 

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ________________ Chapter 2B: Vectors II Page 4 of 17 1.3 Cartesian Equation of a Straight Line From the parametric form of the equation of line l , by making  the subject, we have 11 22 33        x ab ya b za b 1 1 2 2 3 3            x a b ya b za b , provided 123,, 0 bbb . The equation 312 123 zaxa ya bbb   is known as the cartesian equation of the line l . Remarks 1. In 2D-plane, the cartesian equation of a line ym x c can be written as ,.1 yc x x ym cm      The vector equation of the line is thus 01 ,.x yc m         ra b  We can see that a is the position vector of a point on the line and b , the direction vector of the line, is essentially the gradient (for every increase in 1 unit in the x-direction, there is an increase by m units in the y-direction). 2. In 3D, when 1 0b , 11 1 2 22 2 3 33 3 xa b xa yaya b b zaza b b                  The cartesian equation of l becomes 32 1 23 ,  zayaxa bb , which is a line parallel to the yz plane. Similar results for 2 0b  or 3 0.b  3. When 12 0bb , 11 1 22 2 33 33 xa b xa ya b ya za b za b                Since   , z  . The cartesian equation of l becomes 12,, xayaz  , which is a line perpendicular to the xy plane (or parallel to the z axis). Similar results for the cases where 13 0bb or 23 0bb .

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ________________ Chapter 2B: Vectors II Page 5 of 17 Example 2 (a) Find the cartesian equation of the line 1 23 :5 1 , 04              rl  . (b) Find the vector equation of the line 2 31:2 142   xylz . Solution (a) From 23 51 04 x y z             23 5 4      , we get 23 255 31 44             x x yzy z Cartesian equation of 1l is 2 534  x zy (b) Let 31 2142 xy z     for   and solve for , and x yz , we get 34 12 1 (1 )2 x y z            11 22 34 12 , x y z               Vector equation of 2l is 1 2 38 14 , 1 r           2 Calculations Involving a Point and a Line 2.1 Determining wheth er a Point lies on a Line or a Line passes through a Point To determine if a point C lies on a line : ,l   ra b  , we let OC ab   and see if we can solve for a unique value of . If it is possible, it means that C lies on l (or l passes through C ). Otherwise, C does not lie on l (or l does not pass through C ).

Raffles Institution H2 Mathematics 2025 Year 5 _________________________________________________________________________________________ ________________ Chapter 2B: Vectors II Page 6 of 17 Example 3 The points A and B have position vectors 233aij k and 3bi k . The line 1l passes through A and B while the line 2l passes through A and is parallel to the vector 35 ci j k . Determine if the lines 1l or 2l passes through the point 13,, 322