2. 2022 Vectors (Lines) Lecture Notes (For Upload)

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 47 H2 MATHEMATICS (9758) TOPIC VECTORS (LINES) 2022/JC1 9 Introduction We’ve all marveled at the incredibly life-like computer generated images in the movies. What most of us don’t realise is that the dinosaurs of Jurassic Park and the wonders of Lord of the Rings – particularly the star turn of Gollum – wouldn’t have been possible without vector geometry and math. But how are these amazing images made? Computer graphics and computer vision are huge subjects. The first step in creating a computer generated movie is to create the characters in the story and the world they live in. Each of these objects is modelled as a surface, made up of connected flat polygons that are usually triangles, with the vertices of each triangle stored in computer memory. Now that the surface of our object is a wire mesh of triangles, we are ready to colour each of its components. Here it’s important to realistically capture the lighting of the scene we’re modelling, and this is done using a process called ray tracing. Starting from our viewpoint, we trace rays backwards towards the object and let them reflect off it. If the ray from our eye reflects off the facet (one of our wire mesh triangles) and intersects a light source, we shade that facet in a bright colour so that it appears lit up the light source. If the reflected ray does not meet the light source, we shade the facet in a darker colour. Trace a ray from your viewpoint to a facet. Does it reflect off and intersect a light source ? To trace a ray back to a particular facet, we need to describe the surface mathematically, and solve geometric equations involving the straight lines described by the ray and the plane described by that facet. This is done using vectors. In the next two chapters, we would be introduced to the mathematical representation of lines and planes. Source: https://plus.maths.org/content/os/issue42/features/lasenby/index Content Outline: Three-Dimensional Vector Geometry • 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) First objects are modelled as wire skeletons made up from simple polygons such as triangles.

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 48 10 Equation of a Line 10.1 Vector Equation of a Line Exploration Activity Let l be the line passing through two points, P and Q , as shown in the diagram below: Suppose the position vector of P is p and the vector PQ = m   . (a) Express the position vector of Q in terms of p and m . pmOQ = +    (b) It is given that the point E lies on the line PQ produced such that 3PE PQ= , the point F is the midpoint of the line segment PQ and the point G lies on QP produced such that : 2:1PG PQ = . (i) Mark E , F and G on the diagram above (ii) Express the position vectors of E , F and G in terms of p and m respectively. p 3mOE = +    1 2pmOF = +    p 2mOG = −    (c) Based on parts (a) and (b), what can you conclude about the position vector of any point R lying on the line l ? p m for some scalar value OR λλ= +∈    O P Q l m E F G p

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 49 General Case: Consider a line in space passing through a fixed point A , with position vector a , and parallel to a given non-zero vector m , usually named direction vector. Let R be a general point on the line with position vector r , then AR= +ra   . Since we have //AR m   , AR λ= m   for some λ ∈  , thus for the general point R on the line, ,λλ= ∈+ mra  . Online Resource: https://www.geogebra.org/m/gGEvauyc Vector equation of a line passing through point A with position vector a in the direction m is given by ,λλ= ∈+ mra  where r is the position vector of a general point on the line, m is a direction vector of the line.  Remarks: • A line is infinitely long. • r represents the position vector of all possible points on the line. • Every value of λ corresponds to a unique point on the line. Conversely, every point on the line has a unique value of λ . • There is more than one way to represent the same line ,λλ= ∈+ mra  . For example, if another point B with position vector b also lies on the line and n is another vector that is parallel to the line (i.e. parallel to m ), another possible vector equation of the line is ,µµ= ∈+ nrb  . R O A

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 50 10.2 Cartesian Equation of a Line Given a line with vector equation ,λλ= ∈+ mra  . Let x y z  =  r , 1 2 3 a a a  =  a and 1 2 3 m m m  =  m , then 11 22 33 xa m ya m za m λ          = +             11 22 33 xa ya z m ma m λ λ λ = + = + = + [a set of parametric equations for the line] In this manner, the vector equation of a line can be seen as a set of parametric functions for x , y and z in terms of the parameter λ (Parametric equations will be covered in later a module). Given a set of parametric equations related to the vector equation of a line 11 2 2 3 3 ,,xa ya z mammλλλ= += += + , λ ∈  , we can make λ the subject. 1 12 3 3 2 zaxa ya mmmλ −−− = == Cartesian equation of the line passing through ( )123,,aaa with direction 1 2 3 m m m    is 321 123 zaxa ya mmm −−− = = .

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 51 Example 20: Find an equation of the line passing through the points ( )3, 4, 7− and ( )5, 4, 6− in vector form and cartesian form. Solution: 532 m4 4 0 67 1      =− −− =     −    Vector equation: 32 r 4 0, 71 λλ    = −+ ∈   −   Let r x y z  =   , then 32 4 0, 71 x y z λλ          = −+ ∈         −      . So we have 32 4 7 x y z λ λ +     = −     −   3 32 2 4 77 x x y zz λλ λλ −== +⇒  = −  = − ⇒= − Cartesian equation: 3 7; 42 x zy− = −= −  Remark: If 1 0m = or 2 0m = or 3 0m = , the c artesian equation of the line would be written differently. For example, if 3 0m = while 1 0m ≠ and 2 0m ≠ , then 3 1 21 2 ;xa ya zamm −− = = .

CJC MATHEMATICS DEPARTMENT 2022 JC1 H2 MATHEMATICS (9758) TOPIC: VECTORS Page | 52 Example 21: A line has equation 5 1 243 x yz− = −= + . Find a vector equation of the line. Solution: Let 5 1 24 ,3 x yz λλ− = −= += ∈  . Then 5 3 x λ− = ⇒ 53x λ= + 1 y λ−= ⇒ 1y λ= − 24z λ+= ⇒ 12 2z λ= −+ Hence 12 53 r 1 1, 2 λλ    = +− ∈  −   56 1r 1 ' 2 , ' where ' 221 λ λ λλ    = +− ∈ =  −    Self-Practice 7: (a) Convert the vector equation of a line 23 3 4, 15 λλ      = +    −   ∈r  to a cartesian equation. (b) Convert the vector equation of a line ( )5,λλ++ ∈= +ikrij  to a cartesian equation. (c) Convert the cartesian equation of a line 1 423 xy z− = = − to a vector equation. (d) Convert the cartesian equation of a line 1 21 ;435 xy z−− = = to a vector equation. (e) Write down a vector equation and cartesian equation for the line through the point A with position vector 32= −+ai j k and is parallel to the vector 5