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1 Maths Leaving Cert Higher Level Vectors Question 2 Paper 2 By Cillian Fahy and Darron Higgins Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 1
2 1. Geometry based questions, using the properties of vectors Algebra based questions using i and j vectors in the Cartesian plane Vectors in the Cartesian Plane Equality of Vectors Modulus and Unit Vectors The related perpendicular vector Dot Product (Scalar product) Properties of the Dot Product Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 2
3 Paper II Q.2 Vectors The Vectors question is very popular and rightly so. It is a relatively short question allowing you to spend more time on other areas and is one of the easiest questions on Paper II. It is broken down into two main areas: 1. Geometry based questions, using the properties of vectors 2. More algebra based questions using i and j vectors in the Cartesian plane. Students find the second section a lot easier as it is very similar to algebra. However, the first section can cause problems and it is very important to take your time and make sure that everything that you do is double checked. 1. Geometry based questions, using the properties of vectors 1.1 Properties of Vectors in the Coordinate plane. (i) A vector is a translation in a certain direction, a certain distance. The vector from the point a to the point b is written as ab. ab (ii) Equality of Vectors Two vectors ab and cd are said to be equal if they are parallel, have the same length and have the same direction. Therefore, a vector can be moved and still be equal as long as the above facts are still true. (iii) Negative Vectors Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 3
4 A negative vector is simply a reversal in a vectors direction. ab vector ab goes from a to b, where as its negative goes from b to a. a b a b ba, ie the Therefore, ab ba 0. (iv) Addition of Vectors ab cd ab cd This can be done in two ways, (a) The triangle law. To add two vectors using the Triangle law, place the second vector starting where the first vector finishes. Thus the 1 st vector should flow naturally into the 2 nd. The sum of the two vectors is the single vector that goes from the overall beginning to the overall end point. This is known as the Head to Tail method. ab cd x Therefore, ab cd x Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 4
5 (b) The Parallelogram law. To add two vectors using the parallelogram law, position the two vectors so that they start at the same point. The two vectors form 2 sides of a parallelogram, which we can complete. The sum of the two vectors is the diagonal from the beginning point to the opposite corner. ab x cd Therefore, ab cd x These laws are very important and are often used in reverse. Eg. Write bd as the sum of two vectors. d c a b Therefore, bd ba ad.(triangle law) or bd ba bc..(parallelogram law) (v) Single letter vectors. If o is the origin, then it is easier to write the vector oa simply as a. Therefore, if given the vector p we know that it starts at the origin and finishes at the point p. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 5
6 (vi) Single letter vector result. A ab ao ob ao ab oa ob o. b ob oa ba Therefore, any vector can be written in terms of single letter vectors, pq q p, xy y x. The second minus the first. 1.2 Parallel Vectors in the Plane (i) Scalar Multiples (a) Length: The length of k xy is k times that of xy, eg 3ab is three times longer than ab while opposite direction. 2ab is twice as long but goes in the (b) Direction: If k>0 then k xy has the same direction as xy. But if k<0 then k xy has the opposite direction to xy Therefore, if b is parallel to a then b ka This is helpful when looking at vectors on a single, ie collinear vectors. a b c Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 6
7 Each of the vectors ab, ac and bc can be written as a multiple of any of the others. E.g. If a, b and c are collinear and ab : bc 2:1, then we can write any of the following, (i) ab 2bc (ii) ab 2 3 ac (iii) bc 1 3 ac (iv) ac 3 2 ab Note: One common case to remember is that if c is the midpoint of [ab], then ac ab a c b 1 2 c a b c 1 2 ( a b) E.g. opqr is a parallelogram, where o is the origin, y is the midpoint of [qr] and x is the point of intersection of [oq] and [py], dividing [py] in the ratio 2:1 Express in terms of p and r, (i) y (ii) px p o Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 7
8 2 s x 1 p y r Solution: (i) (ii) y oy or ry r px py p ( y p) y ( r p) p p r p p r p Note: Find the start and end of the vector required. Then work out another path that will take you from the start to the end using the vectors required in the question, i.e. p and r in this case. Note: Be careful when given ratios. Remember a line divided in the ratio 1:2 is the same as 1 3 and 2 3 of the length of the line. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 8
9 2. Algebra based questions using i and j vectors in the Cartesian plane 2.1 Vectors in the Cartesian Plane Standard Basis Vectors (i) i is the vector from 0,0 to 1,0 (ii) j is the vector from 0,0 to 0,1 That is, j Note: Think of it in the same way as an x and y axis. Therefore, any vector, v, can be written in terms of i and j i xy, v xi y j v yj xi We can easily convert from point form to i, j form. E.g. The vector from 0,0 to 5,7 can be written as 5i 7j. This topic is very useful as vectors expressed in terms of i and j can be added, subtracted and multiplied by scalars using the ordinary laws of algebra. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 9
10 E.g. If x 3i 2j and y i 3 j express the following in terms of i and j, (i) x 2y (ii) 2x y Solution: (iii) (i) xy x 2y 3i 2 j 2 i 3 j 3i 2 j 2i 6 j 5i4j (ii) 2x y i j i j i 4 j i 3 j 5i7j (iii) xy yx i 3 j 3i 2 j 2i 5j 2.2 Equality of Vectors As in Algebra, Complex Numbers, and Trigonometry equations containing vectors must be equal form each individual term. Therefore, if ai b j ci d j then a c and b d. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 10
11 2.3 Modulus and Unit Vectors This is the same as Complex Numbers where the modulus of a vector is its length from the origin to its end point xy,. Therefore, if v xi y j then 2 2 v x y The unit vector in the direction of a vector v is a vector which (i) (ii) has the same direction as v, and is one unit long, i.e. has modulus of one. The unit vector, u, in the direction of the vector v is given by u v v 2.4 The related perpendicular vector This is the method of finding perpendicular vectors that are the same length. If r is the vector ai b j, then the related perpendicular vector is r bi a j E.g. If r i 2 j find its related perpendicular vector. Solution: If r i 2 j Then r 2i j E.g. Express the vector 8i j as a combination of two vectors, one of which is parallel to 2i 3j and the other is perpendicular to 2i 3jNote:. Parallel vectors have the same direction but can have different lengths, hence we Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 11 multiply 2i 3j by k. Similar method for perpendicular vectors
12 Solution: A vector parallel to 2i 3j is k 2i 3j A vector perpendicular to 2i 3j is l 3i 2j k 2i 3 j l 3i 2 j 8i j 2ki 3k j 3li 2l j 8i j 2k 3l i 3k 2l j 8i j using equality of vectors 2k 3l 8 and 3k 2l 1 Solving these equations simultaneously we get k 1 and l 2 The two vectors are k 2i 3 j 1 2i 3 j 2i 3 j l 3i 2 j 2 3i 2 j 6i 4 j We can check this: 2i 3 j 6i 4 j 8i j 2.5 Dot Product (Scalar product) The product of two vectors can be described in many ways but the one the Leaving Cert. Syllabus deals with is called the Dot product. (i) The Dot product of the vectors x and y is given by x. y x. y cos where is the (smaller) angle between x and y. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 12
13 (ii) The Dot product (Co-ordinate definition) If x x1i x2 j and y y1i y2 j then x. y ( x i x j) y i y j x. y x y x y Properties of the Dot Product (i) The Dot product satisfies most of the usual laws of algebra e.g. (a) (b) x. y y. x ax bycx d y axcx d y by cx d y acxx ad bc xy bd y y Therefore it is commutative (a) and it is distributive (b). But if a. b a. c we can not divide by a to give b c and indices have no meaning in vectors. x. x 2 x as (ii) x. x 2 x Note: This is not the same as before And can be easily verified, x. x x. x cos0 x. x x 2 (iii) If x and y are non-zero vectors, then x y if xy. 0 This is also easily verified, Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 13
14 If x y then 90 x. y x. y cos90 Note: This is probably the most important property of the dot product. x. y.0 0 This can also be used to show that two vectors are perpendicular by calculating their dot product and showing that it is equal to zero. It can also be used to find a missing vector if told that they are perpendicular then we simply let dot product equal to zero and solve. (iv) To find the angle between two vectors If is the angle between x and y, then cos xy. x y The following example deals with most of the areas that can be asked involving vectors in the Cartesian plane. {2007, Paper II Q2 (c)} p 3i 4j and q 5i 12 j r 65t p q 16 p q, where t 0. (i) Express r in terms of i and j. (ii) Find p. r and q. r (iii) Hence, show that r is on the bisector of poq, where o is the origin. Ans. (i) p 3i 4j and q 5i 12 j Note: Don t get caught up in how complicated the question looks. Just do what you are told. There was some confusion here as r still contained t values. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 14
15 r 65t p q 16 p q Therefore, 65t 3i 4 j 5i 12 j r t 39i 52 j 25i 60 j r t r 64i 112 j 16 r 4ti 7t j (ii) p. r 3i 4 j. 4ti 7t j 12t 28t 40t q. r 5i 12 j. 4ti 7 j Note: See 2.5 above for the dot product rule used here. 20t 84t 104t (iii) Let por cos p. r pr 40t t 5t 65 t t 2 2 Note: If r is on the bisector of poq then por must equal qor. You must have a plan before you start this question. As always a quick diagram will always help. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 15
16 8 65 Now let qor q. r cos qr 104t 13t Therefore, cos cos Hence, por qor indicating that r is on the bisector. Mocks.ie Maths LC HL Vectors Question 2 mocks.ie Page 16
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