Well, it depends on which A-level you are doing. The cross product is what you are after. If you haven't done cross product then post back and there is an algebraic technique which someone could describe to you. Edit: BTW your calculator may well do cross products. In which case it is just finger work
Well, it depends on which A-level you are doing. The cross product is what you are after. If you haven't done cross product then post back and there is an algebraic technique which someone could describe to you. Edit: BTW your calculator may well do cross products. In which case it is just finger work
I am doing Edexcel Core Pure 1, Chapter 7. The question is from a homework. There is nothing on the "cross product" here in the textbook
I am doing Edexcel Core Pure 1, Chapter 7. The question is from a homework. There is nothing on the "cross product" here in the textbook
I assume you're not using the Pearson Edexcel CP1 book dated 2017? In that book vectors are covered in Chapter 9 and there is a specific example of what you want which just requires the dot product and simultaneous equations.
So here is a home brewed example. Assume that we have 3i+2j+k and 2i-3j+4k Our target vector is ai+bj+ck The scalar (dot) product of our original 2 vectors with ai+bj+ck is zero so we can set up
3a+2b+c=0 and 2a-3b+4c =0
Oo dear, two equations and three unknowns however we don't care what length our vector is so we are free to choose one of the variables and the others will scale accordingly so lets do a=1 this gives us 2b+c = -3 and -3b+4c=-2
I get -10/11 for b and -13/11 for c
so our vector is i-10/11 j-13/11 k
So a better one is 11i-10j-13k
obvs, either one would work
So try that on your one. I have no idea if it's the same as the text book method. Long time since I looked at a text book.
So here is a home brewed example. Assume that we have 3i+2j+k and 2i-3j+4k Our target vector is ai+bj+ck The scalar (dot) product of our original 2 vectors with ai+bj+ck is zero so we can set up 3a+2b+c=0 and 2a-3b+4c =0 Oo dear, two equations and three unknowns however we don't care what length our vector is so we are free to choose one of the variables and the others will scale accordingly so lets do a=1 this gives us 2b+c = -3 and -3b+4c=-2 I get -10/11 for b and -13/11 for c so our vector is i-10/11 j-13/11 k So a better one is 11i-10j-13k obvs, either one would work So try that on your one. I have no idea if it's the same as the text book method. Long time since I looked at a text book.
The CP1 book I have uses essentially the same method.
(I like to have a copy of the latest texts for reference as I'm no longer a student, but honestly the current incarnations of the textbooks look more like children's colouring-in books and advertisements for the publisher's website than proper textbooks! Probably why they're so ridiculously expensive now.)
I assume you're not using the Pearson Edexcel CP1 book dated 2017? In that book vectors are covered in Chapter 9 and there is a specific example of what you want which just requires the dot product and simultaneous equations.
Btw, this is not the only way of doing this, you can enter vectors directly to the screen for example.
Menu>1 (run-matrix) f3(mat/vct) f6 (to get the display to Vct mode) Vct A>exe Enter dimensions of vector in order row (probably 3) and column (probably 1)
Enter your first vector I did 4,8,12 Exit Vct B >exe Enter your second vector I did 4,-1,-7
To get to the vector - matrix operations OPTN > f2 (MAT/VCT) f6 f6 ( to scroll along)
Use the CrossP function (f3) to make the screen show CrossP(Vct A, Vct B)
Ah, well a cg50 allows you to do the following key sequences. Btw, this is not the only way of doing this, you can enter vectors directly to the screen for example. Menu>1 (run-matrix) f3(mat/vct) f6 (to get the display to Vct mode) Vct A>exe Enter dimensions of vector in order row (probably 3) and column (probably 1) Enter your first vector I did 4,8,12 Exit Vct B >exe Enter your second vector I did 4,-1,-7 To get to the vector - matrix operations OPTN > f2 (MAT/VCT) f6 f6 ( to scroll along) Use the CrossP function (f3) to make the screen show CrossP(Vct A, Vct B) Which delivers 44, -76, 36 Which you can check by doing DotP(Vct A, Vct C) After storing 44, -76, 36 in C
I might take this opportunity to have a bit of a rant about the role of calculators in Maths education.
Since, your mobile phone could deliver a massively more powerful and also simpler experience for doing maths, why has the arcane Casio Calculator become the de facto monopoly standard? (This is a kind of rhetorical question; I am aware of the exam rules)
The functions in it reduce the complexity of much mathematics to being really easy (especially stats) whilst at the same time doing so in an operating system that is a throw back to about 1985.
Many, Many students fail to learn how to use their calculator and so waste enormous amounts of time in exams trying to figure the damn thing out.
This really needs an influential group of maths teachers to grasp the narrative of what is taught and why. My initial preference would be to either do away with anything other than a basic scientific calculator or start teaching completely new stuff using proper modern Apps. You could do worse than look at a load of the SMP material from the 1970s and work something out from there.
BTW cost really need not be an issue, even the poorest of my students has a pretty fancy phone retailing at >£500 despite often not having their £5.00 calculator.
May I ask what textbook you're actually using. This is a fairly standard approach if you haven't met the cross product yet, so *somewhere* in your book for CP1 there should be a chapter on Vectors which at least covers the dot product, what it means, and how to use it for problems like this