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Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com GCSE: Vectors Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com Last modified:

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Presentation on theme: "Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com GCSE: Vectors Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com Last modified:"β€” Presentation transcript:

1 Dr J Frost (jfrost@tiffin.kingston.sch.uk) www.drfrostmaths.com
GCSE: Vectors Dr J Frost Last modified: 20th March 2017

2 Coordinates vs Vectors
𝑦 6 ? Coordinates represent a position. We write π‘₯,𝑦 to indicate the π‘₯ position and 𝑦 position. 5 4 ? ? 2,3 Vectors represent a movement. We write π‘₯ 𝑦 to indicate the change in π‘₯ and change in 𝑦. 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 you’ve done transformations; we can use them to describe the movement of a shape in a translation. 1 π‘₯ -1 -2

3 Vectors to represent movements
𝒃 𝒂 𝒂= 𝒃= 𝒄= 2 βˆ’ 𝒅= βˆ’4 2 𝒆= βˆ’4 βˆ’ 𝒇= βˆ’1 1 π’ˆ= 0 βˆ’3 ? ? 𝒅 ? ? 𝒇 ? ? π’ˆ 𝒆 ? 𝒄

4 Writing Vectors You’re used to variables representing numbers in maths. They can also represent vectors! π‘Œ 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. There’s 3 ways in which can represent the vector from point 𝑋 to 𝑍: 𝒂 (in bold) π‘Ž (with an β€˜underbar’) 𝑋𝑍 𝒃 𝑍 ? 𝒂 ? ? 𝑋 ?

5 Adding Vectors ? ? π‘Œ ? 𝑍 ? 𝑋 𝒃 𝒂 𝑋𝑍 =𝒂= 2 5 π‘π‘Œ =𝒃= 5 3 π‘‹π‘Œ = 7 8
𝑋𝑍 =𝒂= π‘π‘Œ =𝒃= π‘‹π‘Œ = 7 8 ? ? π‘Œ ? 𝒃 What do you notice about the numbers in when compared to and ? We’ve simply added the π‘₯ values and 𝑦 values to describe the combined movement. i.e. 𝒂+𝒃= = 𝑋𝑍 + π‘π‘Œ = π‘‹π‘Œ 𝑍 ? 𝒂 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 𝑋 to π‘Œ via 𝑍. 𝑋𝑍 = = 7 8 Route 2: Use the direct line from 𝑋 to π‘Œ: 7 8 𝑋

6 Scaling Vectors We can β€˜scale’ a vector by multiplying it by a normal number, aptly known as a scalar. If 𝒂= , then 2𝒂= = 8 6 What is the same about 𝒂 and 2𝒂 and what is different? ? 𝒂 2𝒂 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.

7 More on Adding/Subtracting Vectors
𝑋 If 𝑂𝐴 =𝒂, 𝐴𝐡 =𝒃 and 𝑋𝐡 =2𝒄, then find the following in terms of π‘Ž, 𝑏 and 𝑐: 𝑂𝐡 =𝒂+𝒃 π‘‚π‘Œ =𝒂+2𝒃 𝐴𝑋 =π’ƒβˆ’2𝒄 𝑋𝑂 =2π’„βˆ’π’ƒβˆ’π’‚ π‘Œπ‘‹ =βˆ’π’ƒβˆ’2𝒄 2𝒄 𝐴 𝐡 𝒃 𝒃 π‘Œ ? 𝒂 ? 𝒂 ? 𝑂 ? 𝒃 ? Note: Since βˆ’ π‘₯ 𝑦 = βˆ’π‘₯ βˆ’π‘¦ , subtracting a vector goes in the opposite direction. Bro Side Note: Since 𝑏+π‘Ž would end up at the same finish point, we can see 𝒃+𝒂=𝒂+𝒃 (i.e. vector addition, like normal addition, is β€˜commutative’)

8 Exercise 1 (on provided sheet) 1 State the value of each vector. 2 4 𝒂 𝒃 𝐡𝐴 =βˆ’π‘Ž 𝐴𝐢 =π‘Ž+𝑏 𝐷𝐡 =2π‘Žβˆ’π‘ 𝐴𝐷 =βˆ’π‘Ž+𝑏 ? 𝑍𝑋 =π‘Ž+𝑏 π‘Œπ‘Š =π‘Žβˆ’2𝑏 π‘‹π‘Œ = βˆ’π‘Ž+𝑏 𝑋𝑍 =βˆ’π‘Žβˆ’π‘ ? 𝒄 ? ? ? ? ? ? 𝒅 3 5 𝒆 𝒇 𝑀𝐾 =βˆ’π‘Žβˆ’π‘ 𝑁𝐿 =3π‘Žβˆ’π‘ 𝑁𝐾 =2π‘Žβˆ’π‘ 𝐾𝑁 =βˆ’2π‘Ž+𝑏 ? ? ? 𝐴𝐡 =βˆ’π‘Ž+𝑏 𝐹𝑂 =βˆ’π‘Ž+𝑏 𝐴𝑂 =βˆ’2π‘Ž+𝑏 𝐹𝐷 =βˆ’3π‘Ž+2𝑏 ? ? 𝒂= 𝒃= 3 βˆ’ 𝒄= βˆ’3 0 𝒅= 𝒆= βˆ’ 𝒇= βˆ’3 βˆ’4 ? ? ? ? ? ? ? ? ?

9 Starter Expand and simplify the following expressions: ? 2 𝒂+𝒃 +𝒃 =2𝒂+3𝒃 π’‚βˆ’π’ƒ +𝒃 = 1 2 𝒂+ 1 2 𝒃 𝒂 𝒂+𝒃 = 3 2 𝒂+ 1 2 𝒃 π’‚βˆ’2 π’‚βˆ’2𝒃 =βˆ’π’‚+4𝒃 𝒂+2𝒃 βˆ’π’ƒ= 1 3 π’‚βˆ’ 1 3 𝒃 1 ? 2 ? 3 ? 4 ? 5

10 The β€˜Two Parter’ 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 wasn’t in the original diagram. I like to add the ratio as a visual aid. 𝑆𝑄 =βˆ’π’ƒ+𝒂 ? a For (b), there’s two possible paths to get from 𝑁 to 𝑅: via 𝑆 or via 𝑄. But which is best? In (a) we found 𝑺 to 𝑸 rather than 𝑸 to 𝑺, so it makes sense to go in this direction so that we can use our result in (a). 3 : 2 ? ? 𝑁𝑅 = 2 5 𝑆𝑄 +𝒃 = 2 5 βˆ’π’ƒ+𝒂 +𝒃 = 2 5 𝒂+ 3 5 𝒃 𝑄𝑅 is also 𝑏 because it is exactly the same movement as 𝑃𝑆 . b Bro Workings Tip: While you’re welcome to start your working with the second line, I recommend the first line so that your chosen route is clearer.

11 Test Your Understanding
Edexcel June 2012 𝐴𝐡 = 𝐴𝑂 + 𝑂𝐡 =βˆ’π’‚+𝒃 ? 𝑂𝑃 =𝒂 𝐴𝐡 =𝒂 βˆ’π’‚+𝒃 = 1 4 𝒂+ 3 4 𝒃 ? You MUST expand and simplify.

12 Further Test Your Understanding
𝐡 𝐡 B 𝑂𝑋:𝑋𝐡=1:3 π΄π‘Œ:π‘Œπ΅=2:3 𝒃 𝑋 𝒃 π‘Œ 𝑂 𝒂 𝐴 𝑂 𝒂 𝐴 𝐴𝑋 =βˆ’π’‚ 𝑂𝐡 =βˆ’π’‚ 𝒂+𝒃 =βˆ’π’‚+ 1 4 𝒂+ 1 4 𝒃 =βˆ’ 3 4 𝒂+ 1 4 𝒃 First Step? π‘‚π‘Œ =𝒂 𝐴𝐡 =𝒂 βˆ’π’‚+𝒃 =π’‚βˆ’ 2 5 𝒂+ 2 5 𝒃 = 3 5 𝒂+ 2 5 𝒃 First Step? ? ?

13 Exercise 2 ? ? ? ? ? ? ? ? ? (on provided sheet) 3 1
𝑋 is a point such that 𝐴𝑋:𝑋𝐡=1:4 3 𝐴𝐡 =βˆ’π’‚+𝒃 𝐴𝑋 =βˆ’ 1 5 𝒂+ 1 5 𝒃 𝑂𝑋 = 4 5 𝒂+ 1 5 𝒃 𝐡𝑋 = 4 5 π’‚βˆ’ 4 5 𝒃 ? ? ? [June H Q23] a) Find 𝐴𝐡 in terms of 𝒂 and 𝒃 𝐴𝐡 =βˆ’π’‚+𝒃 π‘œπ‘Ÿ π’ƒβˆ’π’‚ b) 𝑃 is on 𝐴𝐡 such that 𝐴𝑃:𝑃𝐡=3: Show that 𝑂𝑃 = 𝒂+3𝒃 ? ? 2 π‘Œ is a point such that π‘Œπ΅=2π΄π‘Œ ? π΄π‘Œ =βˆ’ 1 3 𝒂+ 1 3 𝒃 π‘‚π‘Œ = 2 3 𝒂+ 1 3 𝒃 π‘Œπ‘‚ =βˆ’ 2 3 π’‚βˆ’ 1 3 𝒃 ? ? ?

14 Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? (on provided sheet)
[Nov H Q27] 𝑀 is the midpoint of 𝑂𝑃. 𝑂𝐴𝐢𝐡 is a parallelogram. 𝑅 is a point such that 𝐴𝑅:𝑅𝐡=2:3. 𝑆 is a point such that 𝐡𝑆:𝑆𝐢=1:3. 4 5 ? 𝑂𝑅 = 3 5 𝒂+ 2 5 𝒃 𝐡𝑆 = 1 4 𝒂 𝑂𝑆 = 1 4 𝒂+𝒃 𝑅𝑆 =βˆ’ 7 20 𝒂+ 3 5 𝒃 ? ? ? 6 Express 𝑂𝑀 in terms of 𝒂 and 𝒃. 𝑂𝑀 = 1 2 𝒂+𝒃 π‘œπ‘Ÿ 1 2 𝒂+ 1 2 𝒃 Express 𝑇𝑀 in terms of 𝒂 and 𝒃 giving your answer in its simplest form. 𝑇𝑀 =βˆ’π’‚ 𝑂𝑀 =βˆ’π’‚ 𝒂+𝒃 = 1 2 π’ƒβˆ’ 1 2 𝒂 𝑀 is the midpoint of 𝐢𝐷, 𝐡𝑃:𝑃𝑀=2:1 ? 𝐷𝐢 =βˆ’π’›+π’š 𝐷𝑀 =βˆ’ 1 2 𝒛+ 1 2 π’š 𝐴𝑀 = 𝐴𝐷 + 𝐷𝑀 = 1 2 𝒛+ 1 2 π’š 𝐡𝑀 = 𝐡𝐴 + 𝐴𝑀 =βˆ’π’™+ 1 2 𝒛+ 1 2 π’š 𝐡𝑃 = 2 3 𝐡𝑀 =βˆ’ 2 3 𝒙+ 1 3 𝒛+ 1 3 π’š 𝐴𝑃 = 𝐴𝐡 + 𝐡𝑃 = 1 3 𝒙+ 1 3 𝒛+ 1 3 π’š ? ? ? ? ? ? ?

15 Exercise 2 ? ? ? ? ? ? ? ? ? ? ? ? ? ? (on provided sheet) 8 7
𝐴𝐡 =βˆ’π’‚+𝒃 𝐡𝐢 =𝒂 𝑀𝐡 =βˆ’ 1 2 𝒂+ 1 2 𝒃 𝑀𝑋 =βˆ’ 1 6 𝒂+ 1 2 𝒃 𝑋𝐴 = 2 3 π’‚βˆ’π’ƒ 𝐢𝑀 =βˆ’ 1 2 π’‚βˆ’ 1 2 𝒃 𝑋𝑂 =βˆ’ 1 3 π’‚βˆ’π’ƒ ? 𝑂𝐡 =𝒂+2𝒃 𝐡𝐢 =π’‚βˆ’π’ƒ 𝐴𝑀 = 1 2 𝒂+𝒃 𝑂𝑀 = 3 2 𝒂+𝒃 ? ? ? ? ? ? ? ? ? 9 ? 𝑂𝐢 =βˆ’π’‚+𝒃 𝑋𝐢 =βˆ’ 3 4 𝒂+ 1 2 𝒃 𝐴𝑋 =βˆ’ 5 4 𝒂+ 1 2 𝒃 ? ? ?

16 Recap ?

17 Types of vector β€˜proof’ questions
β€œShow that … is parallel to …” β€œProve these two vectors are equal.” β€œProve that … is a straight line.”

18 Showing vectors are equal
Edexcel March 2013 ? ? 2 5 π‘‚π‘Œ = 𝑏+5π‘Žβˆ’π‘ =2π‘Ž+2𝑏 𝑂𝑋 =3π‘Ž βˆ’3π‘Ž+6𝑏 =2π‘Ž+2𝑏 ∴ 𝑂𝑋 = 2 5 π‘‚π‘Œ There is nothing mysterious about this kind of β€˜prove’ question. Just find any vectors involved, in this case, 𝑂𝑋 and π‘‚π‘Œ . Hopefully they should be equal! With proof questions you should restate the thing you are trying to prove, as a β€˜conclusion’.

19 What do you notice? ? 𝒂+2𝒃 𝒂 2𝒂 2𝒂+4𝒃
! Vectors are parallel if one is a multiple of another. 𝒂 βˆ’ 3 2 𝒂

20 Parallel or not parallel?
Vector 1 Vector 2 Parallel? (in general) π‘Ž+𝑏 π‘Ž+2𝑏 3π‘Ž+3𝑏 βˆ’3π‘Žβˆ’6𝑏 π‘Žβˆ’π‘ βˆ’π‘Ž+𝑏 No οƒΌ Yes   No Yes οƒΌ  No Yes οƒΌ No  οƒΌ Yes For ones which are parallel, show it diagrammatically.

21 How to show two vectors are parallel
𝑨 π‘ͺ 𝒂 𝑿 𝑴 𝑢 𝑩 𝒃 𝑋 is a point on 𝐴𝐡 such that 𝐴𝑋:𝑋𝐡=3:1. 𝑀 is the midpoint of 𝐡𝐢. Show that 𝑋𝑀 is parallel to 𝑂𝐢 . ? For any proof question always find the vectors involved first, in this case 𝑋𝑀 and 𝑂𝐢 . 𝑂𝐢 =π‘Ž+𝑏 𝑋𝑀 = 1 4 βˆ’π‘Ž+𝑏 π‘Ž= 1 4 π‘Ž+ 1 4 𝑏 = 1 4 π‘Ž+𝑏 𝑋𝑀 is a multiple of 𝑂𝐢 ∴ parallel. The key is to factor out a scalar such that we see the same vector. The magic words here are β€œis a multiple of”.

22 Test Your Understanding
Edexcel June 2011 Q26 𝐴 Find 𝐴𝐡 in terms of 𝒂 and 𝒃. βˆ’2𝒂+3𝒃 𝑃 is the point on 𝐴𝐡 such that 𝐴𝑃:𝑃𝐡=2:3. Show that 𝑂𝑃 is parallel to the vector 𝒂+𝒃. 𝑃 ? 2𝒂 𝐡 𝑂 3𝒃 ? Bro Exam Note: Notice that the mark scheme didn’t specifically require β€œis a multiple of” here (but write it anyway!), but DID explicitly factorise out the 6 5

23 Proving three points form a straight line
Points A, B and C form a straight line if: 𝑨𝑩 and 𝑩π‘ͺ are parallel (and B is a common point). Alternatively, we could show 𝑨𝑩 and 𝑨π‘ͺ are parallel. This tends to be easier. ? C B A C B A

24 Straight Line Example ? ? a βˆ’3𝒃+𝒂 b 𝐴𝑁 =2𝒃, 𝑁𝑃 =𝒃
𝑁𝑀 =𝑏 π‘Žβˆ’3𝑏 =𝑏+ 1 2 π‘Žβˆ’ 3 2 𝑏 = 1 2 π‘Žβˆ’ 1 2 𝑏 = 𝟏 𝟐 (π’‚βˆ’π’ƒ) 𝑁𝐢 =βˆ’2𝑏+2π‘Ž =𝟐 π’‚βˆ’π’ƒ 𝑡π‘ͺ is a multiple of 𝑡𝑴 ∴ 𝑡π‘ͺ is parallel to 𝑡𝑴 . 𝑡 is a common point. ∴ 𝑡𝑴π‘ͺ is a straight line. b 𝐴𝑁 =2𝒃, 𝑁𝑃 =𝒃 𝐡 is the midpoint of 𝐴𝐢. 𝑀 is the midpoint of 𝑃𝐡. a) Find 𝑃𝐡 in terms of 𝒂 and 𝒃 b) Show that 𝑁𝑀𝐢 is a straight line.

25 Test Your Understanding
November H Q24 𝑂𝐴 =π‘Ž and 𝑂𝐡 =𝑏 𝐷 is the point such that 𝐴𝐢 = 𝐢𝐷 The point 𝑁 divides 𝐴𝐡 in the ratio 2:1. (a) Write an expression for 𝑂𝑁 in terms of 𝒂 and 𝒃. 𝑢𝑡 =𝒃+ 𝟏 πŸ‘ βˆ’π’ƒ+𝒂 = 𝟏 πŸ‘ 𝒂+ 𝟐 πŸ‘ 𝒃 (b) Prove that 𝑂𝑁𝐷 is a straight line. 𝑢𝑫 =𝒂+πŸπ’ƒ 𝑢𝑡 = 𝟏 πŸ‘ (𝒂+πŸπ’ƒ) 𝑢𝑡 is a multiple of 𝑢𝑫 and 𝑢 is a common point. ∴ 𝑢𝑡𝑫 is a straight line. ? ?

26 Exercise 3 1 2 𝑇 is the point on 𝐴𝐡 such that 𝐴𝑇:𝑇𝐡=5:1. Show that 𝑂𝑇 is parallel to the vector 𝒂+2𝒃. 𝑂𝑇 =2𝒃 βˆ’2𝒃+5𝒂 =2π’ƒβˆ’ 1 3 𝒃+ 5 6 𝒂 = 5 6 𝒂+ 5 3 𝒃 = 5 6 𝒂+2𝒃 𝑀 is the midpoint of 𝑂𝐴. 𝑁 is the midpoint of 𝑂𝐡. Prove that 𝐴𝐡 is parallel to 𝑀𝑁. 𝐴𝐡 =βˆ’2π’Ž+2𝒏 =2 βˆ’π’Ž+𝒏 𝑀𝑁 =βˆ’π’Ž+𝒏 𝐴𝐡 is a multiple of 𝑀𝑁 ∴ parallel. ? ?

27 Exercise 3 3 4 𝐴𝐢𝐸𝐹 is a parallelogram. 𝐡 is the midpoint of 𝐴𝐢. 𝑀 is the midpoint of 𝐡𝐸. Show that 𝐴𝑀𝐷 is a straight line. 𝐷𝑀 =βˆ’π’ƒ 𝒃+𝒂 =βˆ’π’ƒ+ 3 2 𝒃+ 1 2 𝒂 = 1 2 𝒂+ 1 2 𝒃= 1 2 𝒂+𝒃 𝐷𝐴 =2𝒃+2𝒂=2 𝒂+𝒃 𝐷𝐴 is a multiple of 𝐷𝑀 and 𝐷 is a common point, so 𝐴𝑀𝐷 is a straight line. 𝐢𝐷 =𝒂, 𝐷𝐸 =𝒃 and 𝐹𝐢 =π’‚βˆ’π’ƒ i) Express 𝐢𝐸 in terms of 𝒂 and 𝒃 𝒂+𝒃 ii) Prove that 𝐹𝐸 is parallel to 𝐢𝐷 𝐹𝐸 =π‘Žβˆ’π‘+π‘Ž+𝑏=2𝒂 which is a multiple of 𝐢𝐷 iii) 𝑋 is the point on 𝐹𝑀 such that such that 𝐹𝑋:𝑋𝑀=4:1. Prove that 𝐢, 𝑋 and 𝐸 lie on the same straight line. 𝐢𝑋 =βˆ’π’‚+𝒃 π’‚βˆ’ 1 2 𝒃 = 3 5 𝒂+ 3 5 𝒃 = 3 5 𝒂+𝒃 𝐢𝐸 =𝒂+𝒃 ? ? ? ?

28 Exercise 3 5 6 𝑂𝐴 =3𝒂 and 𝐴𝑄 =𝒂 and 𝑂𝐡 =𝒃 and 𝐡𝐢 = 1 2 𝒃. 𝑀is the midpoint of 𝑄𝐡. Prove that 𝐴𝑀𝐢 is a straight line. 𝐴𝑀 =𝒂 βˆ’4𝒂+𝒃 =βˆ’π’‚+ 1 2 𝒃 𝐴𝐢 =βˆ’3𝒂+ 3 2 𝒃=3 βˆ’π’‚+ 1 2 𝒃 𝐴𝐢 is a multiple of 𝐴𝑀 and 𝐴 is a common point, therefore 𝐴𝑀𝐢 is a straight line. 𝑂𝐴𝐡𝐢 is a parallelogram. 𝑃 is the point on 𝐴𝐢 such that 𝐴𝑃= 2 3 𝐴𝐢. i) Find the vector 𝑂𝑃 . Give your answer in terms of 𝒂 and 𝒄. 𝑂𝑃 =6𝒂 βˆ’6𝒂+6𝒄 =2𝒂+4𝒄=2 𝒂+2𝒄 ii) Given that the midpoint of 𝐢𝐡 is 𝑀, prove that 𝑂𝑃𝑀 is a straight line. 𝑂𝑀 =3𝒂+6𝒄=3 𝒂+2𝒄 𝑂𝑀 is a multiple of 𝑂𝑃 and 𝑂 is a common point, therefore 𝑂𝑃𝑀 is a straight line. ? ? ?


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