Vectors Year 12 Term 1 1
Vectors - A Vector has Two properties Magnitude and Direction - A vector is usually denoted in bold, like vector a, or a, or many others. In 2D - a = xı + yȷ - a = x, y - where, x = a cos θ and y = a sin θ In 3D - a = xı + yȷ + zk - a = x, y, z Recall: - ı, ȷ & k are unit vectors (length 1) in the directions of their respective axis - a =!! - You should know how to Add, Subtract and Scalar multiply vectors - Note that Scalar Multiplication refers to the enlargement or contraction of a vector (not to be confused with Scalar Product we do later) - You need to be able to do these operations both Mathematically and Geometrically. - Magnitude of a vector is denoted OP = a = r = x! + y! + z! - in 2D, the angle between vectors is given by θ, where tan θ =!! - in 3D, use the dot product to find the angle between 2 vectors. Given a = 2, 1, 3 and b = 1, 1, 4 Calculate the magnitude of the resultant Vector created by a 2b. Then derive a unit vector in the direction of a 2b. Then find a unit vector in the opposite direction. 2
Ans: 34, 0,!,!!!"!" and 0,!!!",!!" The Dot Product (Scalar Product) A dot product is a Scalar Value that is the result of an operation of two vectors with the same number of components. It is the Sum of the Products of each component. a b = x! x! + y! y! + z! z! There is also a Geometric relationship defined by the Dot Product. a b = a b cosθ Therefore, we can say: a b = a b cosθ = x! x! + y! y! + z! z! We use this rule to Find the angle between two vectors in both 2D and 3D. The Dot Product is a good way to see if vectors are perpendicular, because cos 90 = 0, the dot product will be equal to Zero. a b if a b = 0 What is the angle between vector 1, 1 and 2, 2? Draw these vectors on a Cartesian plane to confirm your answer is reasonable. Ans: 90 Text Book, Exercise 3A: This is revision, so doing them ALL will take you no time at all. 3
The Vector Product in 2D Just as two points can define a straight line, two Vectors is one way to describe a Plane. Task: Pick up 2 pens and visualise the only Plane that contains these two vectors. The Vector Product / Cross Product of any two vectors produces a vector at a normal to the plane created by Vectors a and b. There is a weird process for this in the Text on page 104. I have never used this way before, so I will not go through it, but will skip to a quicker, easier and more detailed process using Matrices. (as per page 105) If a = a!, a!, a! and b = b!, b!, b! Then a b = i j k a! a! a! b! b! b! Recall the Determinant of a 3 x 3 Matrix? a b = a! a! b! b! i a! a! b! b! j + a! a! b! b! k and this Resultant Vector will be at a Normal to both initial vectors, and as such will also be at a Normal to the Plane they define. ** there is No need to know the 2 nd vector product rule a b = a b sin θ n, because there is limited application of it. You will see I have Not asked you to do Q1 ** Find a vector that is perpendicular to both 2, 0, 0 and 0, 1, 0. Ans: 0, 0, 2 Consider your answer. Does it seem reasonable? Visualise your initial two vectors. Where do they lie? What is the relationship between all three vectors? 4
An important element of the Vector Product is that the Modulus (Magnitude/Length) of a Vector Product gives the area of the Parallelogram created by the two vectors. So, Area Parallogram ab = a b And, by extrapolation, the area of a triangle created by the two vectors can be calculate as; Area Triangle ab = a b! Recall OP = a = r = x! + y! + z! Draw a Cartesian Plane. By plotting vectors 0, 2 and 3, 0 you have a Parallelogram. Use a Grade 9 technique to find the area of this special parallelogram. Ans: 6 Now calculate the Vector Product (Cross Product) of the two initial vectors, and find the modulus of the resultant vector? Hint: change the initial vectors to have values for i, j & k first. Hopefully you are not surprised at getting the same answer J Ans: 6 ** Care ** For every set of vectors, there are TWO Normal vectors (each vertically opposite to each other). Ensure you understand the righty tighty lefty loosey concept of which direction the Normal vector goes in. Refer to page 103 of the Textbook. 5
On my white board, if i is vertical and j is horizontal, determine 2, 0, 0 0, 5, 0 and describe its direction. Ans: 0, 0, 10, out of the wall. Determine a unit vector in the opposite direction of your result in the previous Task. Ans: 0, 0, 1 I think all of these vectors are perpendicular? What operation can we do to confirm that all of these are in fact perpendicular to each other? Ans: Dot product, and correct, all the Dot products are 0 (accept for the dot product of the two vectors in the opposite direction of each other). We do NOT need to learn anything to do with Force or Torque, as this school does not do the Dynamics chapter (bummer!) So this is a bit of an Easy lesson! Text Book, Exercise 3B 6
The Triple Products There are two triple products, the Scalar Triple Product, and the Triple Vector Product. We are only concerned with the Scalar Triple Product at this year level. The Scalar Triple Product: as the name suggests (scalar), gives us a single number answer, which is the volume of a parallelepiped formed by a, b & c. Scalar Triple Product a (b c) As you can see, we need to perform the vector product first, so that our dot product will give us a scalar answer. Given vectors, a = 1, 1, 2, b = 2, 1, 1 and c = 1, 2, 1, perform the following Triple Products. a (b c) (a b) c b (a c) b (c a) Ans: 12, 12, -12, 12 From your result, what is your hypothesis about the Scalar Triple Product order of operations? And what do you notice about the modulus of your solutions? The Scalar Triple Product, (just like when we multiply 3 sides together on a rectangle), will give us a Volume! To be precise, the Modulus of the Scalar Triple Product will give us the Volume of the Parallelepiped, formed by vectors a, b and c. 7
Look back at your working for the previous Task. When you calculated the Dot product, what did you multiply a! by? If you can t see it, I ll help you. 1 1 1 1 2 and inside those brackets, it looks a lot like (ad-bc) J Ans: a! b! c! b! c! So, what we have seen is that the Scalar Triple Product is simply the Determinant of the 3x3 matrix formed by the 3 vectors. So we can say the scalar Triple Product for vectors a, b and c is: a! a! a! b! b! b! c! c! c! Using the Determinant method, find the volume of a parallelepiped formed by the three vectors, a = 1, 1, 2, b = 2, 1, 1 and c = 1, 2, 1. Ans: 12 Text Book, Exercise 3C 8
Equations to Planes Feel free to read page 116 of your text to give you a starting point. I think that explanation is a little confusing, so I will give you an alternative: Describing a plane is best done by referring to all such vectors that are normal to a given vector, and that pass through some fixed point. : Give this some thought. Don t merely accept it, question how and why we would want to define a plane in this way? Here, we will say r = a position vector to ANY point in the plane n = a position vector NORMAL to the plane a = a position vector of a SET point in the plane We can see that r a lies parallel to the plane. (make sure you CAN see this) Knowing that the dot product of two vectors is 0 if they are at a normal to each other, we can now say, r a n = 0 or r n a n = 0 and we arrive at a Vector Equation for a Plane: r n = a n OK. r, a and n are Position Vectors. Specifically, r is the vector that describes ANY point in the plane, so it is a General vector, so we can say: r = x, y, z and lets set n =, β, γ Now just looking at the left hand side, r n becomes, x, y, z α, β, γ Or, αx + βy + γz And on the right hand side, a n is the Dot product of two vectors, so it is simply a constant, where a and n are known, and the dot product is a scalar, hence we arrive at what is called the: Equation to a Plan in Cartesian Form αx + βy + γz = d 9
Lets start easy Consider the vertical plane that is a distance of 2 in the x direction. Clearly a vector at a normal to this could be n = 2, 0, 0 Clearly a position vector that lies in the plane could be a = 2, 2, 2 What is the equation to the plane? Ans: r 2, 0, 0 = 4 Consider the exact same plane. Show that point 2, 1, 3 lies in the plane. Ans: 2, 1, 3 2, 0, 0 = 4 Satisfies equation, lies in the plane Find 3 other points in the plane and PROVE they lie in the plane. 10
Consider the same plane. What happens if n doesn t connect to the plane? Clearly the position vector 4, 0, 0 is also at a normal to the plane. Given this, what is a revised equation to the plane? Ans: r 4, 0, 0 = 8 Re-prove your above points still satisfy this equation. What s your Hypothesis? Clearly, n does not need to touch / end at the plane, it simply has to be at a Normal to the plane and I am hoping you have hypothesized that; just as; 4x 2y + 6 = 0 is the same as y = 2x + 3 ; we can have plane equations that may look different, but are actually the same Plane! (scalar s of) Create a more difficult plane, and re-perform the previous 5 Tasks with appropriate points. 11
Return to our nice easy vertical plane. We have tried n = 2, 0, 0 and n = 4, 0, 0, and this showed that so long as n is a Normal Vector, the Plane equation will work. Now lets consider what happens when we set n to be a Unit Vector. Clearly n = 1, 0, 0 Put the equation to the plane in Cartesian form αx + βy + γz = d Ans: x + 0y + 0z = 2 What does d represent with regards to the plane? Ans: the distance of the plane to the Origin ** Important, when we have values of the unit vector of n, or n, then, a n is the Perpendicular Distance from the Origin, to the plane, hence this is the shortest distance from the Origin to the Plane. Think back to Yr 9 maths and linear functions... Two points can describe a straight line but we can also describe a straight line by its slope and one point on the line. Similarly we can describe a plane in more than one way as well. 12
: Consider carefully and explain the difference between: Two vectors can define a plane, and three position vectors can define a plane. : What does our above process rely on to create an equation to a plan, a position vector, or a vector? : How do we translate 3 position vectors, to 2 vectors? Do your best to draw 3 position vectors in 3D and convert these to two vectors that lie in the plane. Drawing a set of axes in 3D is hard even the text book struggles (as per the diagram on page 116) but do your best J Hint: if the three points are a, b and c, then two vectors in the plane would be c-a and c-b. 13
Given 3 points described by the position vectors, a = 3ı + 4ȷ + 3k, b = 2ı + 3ȷ + 2k and c = 2ı ȷ + k, or I could write this as a = 3,4,3, b = 2,3,2 and c = 2, 1,1. Derive the equation to the plane that contains all three points in both Vector Form as well as Cartesian Form. Hint, first find two vectors that lie in the plane find a Normal to both of them. Find the Vector equation first, then the Cartesian follows quite directly. Ans: r 3, 1,4 = 1, or 3x y + 4z = 1 Text Book Exercise 3D, 14