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Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com
GCSE: Vectors Dr J Frost Last modified: 20th March 2017
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Coordinates vs Vectors
π¦ 6 ? Coordinates represent a position. We write π₯,π¦ to indicate the π₯ position and π¦ position. 5 4 ? ? 2,3 Vectors represent a movement. We write π₯ π¦ to indicate the change in π₯ and change in π¦. 3 ? 4 1 2 Note we put the numbers vertically. Do NOT write ; there is no line and vectors are very different to fractions! You may have seen vectors briefly before if youβve done transformations; we can use them to describe the movement of a shape in a translation. 1 π₯ -1 -2
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Vectors to represent movements
π π π= π= π= 2 β π
= β4 2 π= β4 β π= β1 1 π= 0 β3 ? ? π
? ? π ? ? π π ? π
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Writing Vectors Youβre used to variables representing numbers in maths. They can also represent vectors! π What can you say about how we use variables for vertices (points) vs variables for vectors? We use capital letters for vertices and lower case letters for vectors. Thereβs 3 ways in which can represent the vector from point π to π: π (in bold) π (with an βunderbarβ) ππ π π ? π ? ? π ?
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Adding Vectors ? ? π ? π ? π π π ππ =π= 2 5 ππ =π= 5 3 ππ = 7 8
ππ =π= ππ =π= ππ = 7 8 ? ? π ? π What do you notice about the numbers in when compared to and ? Weβve simply added the π₯ values and π¦ values to describe the combined movement. i.e. π+π= = ππ + ππ = ππ π ? π Bro Important Note: The point is that we can use any route to get from the start to finish, and the vector will always be the same. Route 1: We go from π to π via π. ππ = = 7 8 Route 2: Use the direct line from π to π: 7 8 π
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Scaling Vectors We can βscaleβ a vector by multiplying it by a normal number, aptly known as a scalar. If π= , then 2π= = 8 6 What is the same about π and 2π and what is different? ? π 2π Same: Same direction / Parallel Different: The length of the vector, known as the magnitude, is longer. ? ? Bro Note: Note that vector letters are bold but scalars are not.
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More on Adding/Subtracting Vectors
π If ππ΄ =π, π΄π΅ =π and ππ΅ =2π, then find the following in terms of π, π and π: ππ΅ =π+π ππ =π+2π π΄π =πβ2π ππ =2πβπβπ ππ =βπβ2π 2π π΄ π΅ π π π ? π ? π ? π ? π ? Note: Since β π₯ π¦ = βπ₯ βπ¦ , subtracting a vector goes in the opposite direction. Bro Side Note: Since π+π would end up at the same finish point, we can see π+π=π+π (i.e. vector addition, like normal addition, is βcommutativeβ)
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Exercise 1 (on provided sheet) 1 State the value of each vector. 2 4 π π π΅π΄ =βπ π΄πΆ =π+π π·π΅ =2πβπ π΄π· =βπ+π ? ππ =π+π ππ =πβ2π ππ = βπ+π ππ =βπβπ ? π ? ? ? ? ? ? π
3 5 π π ππΎ =βπβπ ππΏ =3πβπ ππΎ =2πβπ πΎπ =β2π+π ? ? ? π΄π΅ =βπ+π πΉπ =βπ+π π΄π =β2π+π πΉπ· =β3π+2π ? ? π= π= 3 β π= β3 0 π
= π= β π= β3 β4 ? ? ? ? ? ? ? ? ?
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Starter Expand and simplify the following expressions: ? 2 π+π +π =2π+3π πβπ +π = 1 2 π+ 1 2 π π π+π = 3 2 π+ 1 2 π πβ2 πβ2π =βπ+4π π+2π βπ= 1 3 πβ 1 3 π 1 ? 2 ? 3 ? 4 ? 5
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The βTwo Parterβ exam question
Many exams questions follow a two-part format: Find a relatively easy vector using skills from Lesson 1. Find a harder vector that uses a fraction of your vector from part (a). Edexcel June H Q27 Bro Tip: This ratio wasnβt in the original diagram. I like to add the ratio as a visual aid. ππ =βπ+π ? a For (b), thereβs two possible paths to get from π to π
: via π or via π. But which is best? In (a) we found πΊ to πΈ rather than πΈ to πΊ, so it makes sense to go in this direction so that we can use our result in (a). 3 : 2 ? ? ππ
= 2 5 ππ +π = 2 5 βπ+π +π = 2 5 π+ 3 5 π ππ
is also π because it is exactly the same movement as ππ . b Bro Workings Tip: While youβre welcome to start your working with the second line, I recommend the first line so that your chosen route is clearer.
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Test Your Understanding
Edexcel June 2012 π΄π΅ = π΄π + ππ΅ =βπ+π ? ππ =π π΄π΅ =π βπ+π = 1 4 π+ 3 4 π ? You MUST expand and simplify.
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Further Test Your Understanding
π΅ π΅ B ππ:ππ΅=1:3 π΄π:ππ΅=2:3 π π π π π π π΄ π π π΄ π΄π =βπ ππ΅ =βπ π+π =βπ+ 1 4 π+ 1 4 π =β 3 4 π+ 1 4 π First Step? ππ =π π΄π΅ =π βπ+π =πβ 2 5 π+ 2 5 π = 3 5 π+ 2 5 π First Step? ? ?
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Exercise 2 ? ? ? ? ? ? ? ? ? (on provided sheet) 3 1
π is a point such that π΄π:ππ΅=1:4 3 π΄π΅ =βπ+π π΄π =β 1 5 π+ 1 5 π ππ = 4 5 π+ 1 5 π π΅π = 4 5 πβ 4 5 π ? ? ? [June H Q23] a) Find π΄π΅ in terms of π and π π΄π΅ =βπ+π ππ πβπ b) π is on π΄π΅ such that π΄π:ππ΅=3: Show that ππ = π+3π ? ? 2 π is a point such that ππ΅=2π΄π ? π΄π =β 1 3 π+ 1 3 π ππ = 2 3 π+ 1 3 π ππ =β 2 3 πβ 1 3 π ? ? ?
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Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? (on provided sheet)
[Nov H Q27] π is the midpoint of ππ. ππ΄πΆπ΅ is a parallelogram. π
is a point such that π΄π
:π
π΅=2:3. π is a point such that π΅π:ππΆ=1:3. 4 5 ? ππ
= 3 5 π+ 2 5 π π΅π = 1 4 π ππ = 1 4 π+π π
π =β 7 20 π+ 3 5 π ? ? ? 6 Express ππ in terms of π and π. ππ = 1 2 π+π ππ 1 2 π+ 1 2 π Express ππ in terms of π and π giving your answer in its simplest form. ππ =βπ ππ =βπ π+π = 1 2 πβ 1 2 π π is the midpoint of πΆπ·, π΅π:ππ=2:1 ? π·πΆ =βπ+π π·π =β 1 2 π+ 1 2 π π΄π = π΄π· + π·π = 1 2 π+ 1 2 π π΅π = π΅π΄ + π΄π =βπ+ 1 2 π+ 1 2 π π΅π = 2 3 π΅π =β 2 3 π+ 1 3 π+ 1 3 π π΄π = π΄π΅ + π΅π = 1 3 π+ 1 3 π+ 1 3 π ? ? ? ? ? ? ?
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Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? (on provided sheet) 8 7
π΄π΅ =βπ+π π΅πΆ =π ππ΅ =β 1 2 π+ 1 2 π ππ =β 1 6 π+ 1 2 π ππ΄ = 2 3 πβπ πΆπ =β 1 2 πβ 1 2 π ππ =β 1 3 πβπ ? ππ΅ =π+2π π΅πΆ =πβπ π΄π = 1 2 π+π ππ = 3 2 π+π ? ? ? ? ? ? ? ? ? 9 ? ππΆ =βπ+π ππΆ =β 3 4 π+ 1 2 π π΄π =β 5 4 π+ 1 2 π ? ? ?
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Recap ?
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Types of vector βproofβ questions
βShow that β¦ is parallel to β¦β βProve these two vectors are equal.β βProve that β¦ is a straight line.β
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Showing vectors are equal
Edexcel March 2013 ? ? 2 5 ππ = π+5πβπ =2π+2π ππ =3π β3π+6π =2π+2π β΄ ππ = 2 5 ππ There is nothing mysterious about this kind of βproveβ question. Just find any vectors involved, in this case, ππ and ππ . Hopefully they should be equal! With proof questions you should restate the thing you are trying to prove, as a βconclusionβ.
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What do you notice? ? π+2π π 2π 2π+4π
! Vectors are parallel if one is a multiple of another. π β 3 2 π
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Parallel or not parallel?
Vector 1 Vector 2 Parallel? (in general) π+π π+2π 3π+3π β3πβ6π πβπ βπ+π No οΌ Yes ο ο» ο ο» No Yes οΌ ο ο» No Yes οΌ No ο ο» οΌ Yes For ones which are parallel, show it diagrammatically.
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How to show two vectors are parallel
π¨ πͺ π πΏ π΄ πΆ π© π π is a point on π΄π΅ such that π΄π:ππ΅=3:1. π is the midpoint of π΅πΆ. Show that ππ is parallel to ππΆ . ? For any proof question always find the vectors involved first, in this case ππ and ππΆ . ππΆ =π+π ππ = 1 4 βπ+π π= 1 4 π+ 1 4 π = 1 4 π+π ππ is a multiple of ππΆ β΄ parallel. The key is to factor out a scalar such that we see the same vector. The magic words here are βis a multiple ofβ.
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Test Your Understanding
Edexcel June 2011 Q26 π΄ Find π΄π΅ in terms of π and π. β2π+3π π is the point on π΄π΅ such that π΄π:ππ΅=2:3. Show that ππ is parallel to the vector π+π. π ? 2π π΅ π 3π ? Bro Exam Note: Notice that the mark scheme didnβt specifically require βis a multiple ofβ here (but write it anyway!), but DID explicitly factorise out the 6 5
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Proving three points form a straight line
Points A, B and C form a straight line if: π¨π© and π©πͺ are parallel (and B is a common point). Alternatively, we could show π¨π© and π¨πͺ are parallel. This tends to be easier. ? C B A C B A
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Straight Line Example ? ? a β3π+π b π΄π =2π, ππ =π
ππ =π πβ3π =π+ 1 2 πβ 3 2 π = 1 2 πβ 1 2 π = π π (πβπ) ππΆ =β2π+2π =π πβπ π΅πͺ is a multiple of π΅π΄ β΄ π΅πͺ is parallel to π΅π΄ . π΅ is a common point. β΄ π΅π΄πͺ is a straight line. b π΄π =2π, ππ =π π΅ is the midpoint of π΄πΆ. π is the midpoint of ππ΅. a) Find ππ΅ in terms of π and π b) Show that πππΆ is a straight line.
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Test Your Understanding
November H Q24 ππ΄ =π and ππ΅ =π π· is the point such that π΄πΆ = πΆπ· The point π divides π΄π΅ in the ratio 2:1. (a) Write an expression for ππ in terms of π and π. πΆπ΅ =π+ π π βπ+π = π π π+ π π π (b) Prove that πππ· is a straight line. πΆπ« =π+ππ πΆπ΅ = π π (π+ππ) πΆπ΅ is a multiple of πΆπ« and πΆ is a common point. β΄ πΆπ΅π« is a straight line. ? ?
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Exercise 3 1 2 π is the point on π΄π΅ such that π΄π:ππ΅=5:1. Show that ππ is parallel to the vector π+2π. ππ =2π β2π+5π =2πβ 1 3 π+ 5 6 π = 5 6 π+ 5 3 π = 5 6 π+2π π is the midpoint of ππ΄. π is the midpoint of ππ΅. Prove that π΄π΅ is parallel to ππ. π΄π΅ =β2π+2π =2 βπ+π ππ =βπ+π π΄π΅ is a multiple of ππ β΄ parallel. ? ?
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Exercise 3 3 4 π΄πΆπΈπΉ is a parallelogram. π΅ is the midpoint of π΄πΆ. π is the midpoint of π΅πΈ. Show that π΄ππ· is a straight line. π·π =βπ π+π =βπ+ 3 2 π+ 1 2 π = 1 2 π+ 1 2 π= 1 2 π+π π·π΄ =2π+2π=2 π+π π·π΄ is a multiple of π·π and π· is a common point, so π΄ππ· is a straight line. πΆπ· =π, π·πΈ =π and πΉπΆ =πβπ i) Express πΆπΈ in terms of π and π π+π ii) Prove that πΉπΈ is parallel to πΆπ· πΉπΈ =πβπ+π+π=2π which is a multiple of πΆπ· iii) π is the point on πΉπ such that such that πΉπ:ππ=4:1. Prove that πΆ, π and πΈ lie on the same straight line. πΆπ =βπ+π πβ 1 2 π = 3 5 π+ 3 5 π = 3 5 π+π πΆπΈ =π+π ? ? ? ?
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Exercise 3 5 6 ππ΄ =3π and π΄π =π and ππ΅ =π and π΅πΆ = 1 2 π. πis the midpoint of ππ΅. Prove that π΄ππΆ is a straight line. π΄π =π β4π+π =βπ+ 1 2 π π΄πΆ =β3π+ 3 2 π=3 βπ+ 1 2 π π΄πΆ is a multiple of π΄π and π΄ is a common point, therefore π΄ππΆ is a straight line. ππ΄π΅πΆ is a parallelogram. π is the point on π΄πΆ such that π΄π= 2 3 π΄πΆ. i) Find the vector ππ . Give your answer in terms of π and π. ππ =6π β6π+6π =2π+4π=2 π+2π ii) Given that the midpoint of πΆπ΅ is π, prove that πππ is a straight line. ππ =3π+6π=3 π+2π ππ is a multiple of ππ and π is a common point, therefore πππ is a straight line. ? ? ?
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