# Further Maths

Watch
Announcements
#1
Hi

I have some questions which I am a bit confused about. I have completed question 5a and 5b and 6a (I think). But the other questions I have no clue about.

All help is appreciated. 😊
0
3 years ago
#2
Holy sh*t I wish I could help I do edexcel further pure but wow I have no clue as to how to do 6b or 5c onwards I've always struggled with the geometric expression so matrices.
For question 7 though although it's 3D vectors which I despise I'm pretty sure you'd have to form 2 straight line equations then work out the perpendicular distance but I can only see a way to form one straight line equation meaning that isn't the correct approach. Since they give you two points confident you have to make a vector equation of them so a+¥(b-a) ¥ being my excuse of lander. I recognise a similar question to that except there were two lines

I'm gonna follow this threat cause I'm itirgued as to actually solve these

Sorry I couldn't be of use. Also where do you get these questions they see great
0
#3
(Original post by Ismaeel Ali)
Holy sh*t I wish I could help I do edexcel further pure but wow I have no clue as to how to do 6b or 5c onwards I've always struggled with the geometric expression so matrices.
For question 7 though although it's 3D vectors which I despise I'm pretty sure you'd have to form 2 straight line equations then work out the perpendicular distance but I can only see a way to form one straight line equation meaning that isn't the correct approach. Since they give you two points confident you have to make a vector equation of them so a+¥(b-a) ¥ being my excuse of lander. I recognise a similar question to that except there were two lines

I'm gonna follow this threat cause I'm itirgued as to actually solve these

Sorry I couldn't be of use. Also where do you get these questions they see great
I am doing OCR and these are practice papers made by a company 'crash maths' I believe. 😊
1
3 years ago
#4
Actually for 6b maybe try converting the two equations into Cartesian form.
So Z=2 would be x^2+y^2=4
And Argz=k would be y/x=tan(k) which would be arctan(y/x)=k
Then sub one equation into the other.
I have no idea if this is correct just a suggestion

Or since it satisfies the region you shaded in part a) maybe try something with that

This was awful help I'm sorry
0
3 years ago
#5
(Original post by Alexia_17)
I am doing OCR and these are practice papers made by a company 'crash maths' I believe. 😊
Thank you so much for that, new spec so not that many past papers the struggle D:
1
#6
(Original post by Ismaeel Ali)
Thank you so much for that, new spec so not that many past papers the struggle D:
Yeah I completly understand. Mine is also a new spec but OCR didn't publish some of the text books untill December and only a few ago published there specimin papers. We've had to use AQA and Edexcell and questions from the previous syllambus.
0
3 years ago
#7
(Original post by Alexia_17)
Yeah I completly understand. Mine is also a new spec but OCR didn't publish some of the text books untill December and only a few ago published there specimin papers. We've had to use AQA and Edexcell and questions from the previous syllambus.
Same story for edexcel further Pure was available from the start of the year but we only got our decision and further stats books near the end of December so only started them in 2018, hopefully we are favoured with lower grade boundaries
0
3 years ago
#8
For question 7 you find the vector in the direction of the line AB. (-12i 9j 4k) - (10i - j 18k) = (-22i 8j - 14k). You use this to get the vector for the line which is: (-12i 9j 4k) t(-22i 8j -14k) = (-12-22t)i (9 8t)j (4 - 14t)k. This forms an equation for a general point on the line. This should be perpendicular to the gradient of the first line so their dot product should be 0. You dot product (-12-22t)i (9 8t)j (4 - 14t)k and (-22i 8j - 14k). t should be about 482/780 but I can't really remember. You then substitute it into the equation for the line and calculate the magnitude.Hopefully this helps a bit. Sorry if it doesn't make much sense.
0
3 years ago
#9
Also heres the mark scheme:
https://crashmaths.com/wp-content/up...7-MS-FINAL.pdf
1
3 years ago
#10
(Original post by Alexia_17)
Yeah I completly understand. Mine is also a new spec but OCR didn't publish some of the text books untill December and only a few ago published there specimin papers. We've had to use AQA and Edexcell and questions from the previous syllambus.
On OCR as well, it's a madness really

we got our FP books in October

and we didn't get our Mech and Stats books until January rip
1
#11
(Original post by 177239)
For question 7 you find the vector in the direction of the line AB. (-12i 9j 4k) - (10i - j 18k) = (-22i 8j - 14k). You use this to get the vector for the line which is: (-12i 9j 4k) t(-22i 8j -14k) = (-12-22t)i (9 8t)j (4 - 14t)k. This forms an equation for a general point on the line. This should be perpendicular to the gradient of the first line so their dot product should be 0. You dot product (-12-22t)i (9 8t)j (4 - 14t)k and (-22i 8j - 14k). t should be about 482/780 but I can't really remember. You then substitute it into the equation for the line and calculate the magnitude.Hopefully this helps a bit. Sorry if it doesn't make much sense.
I started to do this however got a different direction vection to you. When I calculated i got (22i, -10j, 14k). After that I got a bit lost and i am still slightly confused by your explanation. Sorry
0
3 years ago
#12
(Original post by Alexia_17)
I started to do this however got a different direction vection to you. When I calculated i got (22i, -10j, 14k). After that I got a bit lost and i am still slightly confused by your explanation. Sorry
I stuffed that up slightly. Your vector is right. You add your direction vector, multiplied by t, to one of the points A or B. This will give you something like
(-12i,9j,4k)+t(22i ,-10j,14k). This is the equation for any point along the line AB. To get the distance between a point and a line you subtract the vector of the point from the line. However, the point is the origin so you can skip this step. You can then do the dot product of the direction vector and the vector of the line. As they are perpendicular the dot product will equal 0. This means that you can solve it to find t. Then substitue your value for t back into the equation for the line and you will get thee closest point on the line which you can use to find the distance.

Heres a youtube video which will probably help a lot more than I can:

Sorry, I am kinda useless at explaining things
0
3 years ago
#13
(Original post by thotproduct)
On OCR as well, it's a madness really

we got our FP books in October

and we didn't get our Mech and Stats books until January rip
doing edexcel and I've got a test in about a month. (AS) The book isn't even out...
0
3 years ago
#14
(Original post by RickHendricks)
doing edexcel and I've got a test in about a month. (AS) The book isn't even out...
that's mad still, but im jelly cos y'all have more resources than we do on Physicsandmathstutor

the grade boundaries will be chill dw fam, and you're probably most likely going to boss it m9
1
3 years ago
#15
(Original post by thotproduct)
that's mad still, but im jelly cos y'all have more resources than we do on Physicsandmathstutor

the grade boundaries will be chill dw fam, and you're probably most likely going to boss it m9
I guess this is why my school decided to start teaching FM to me when im in year 13.

How hard is further maths?
0
3 years ago
#16
(Original post by stoyfan)
I guess this is why my school decided to start teaching FM to me when im in year 13.

How hard is further maths?
not that hard tbh, granted, AS FM isn't really designed to be extremely tougher than AS normal, though the challenge is certainly noticeable, can't really speak much for A2 but i hear AS -> A2 FM transition is some next level juju.

personally it depends on what kind of person you are, modules etc, im quite confident on all 3 of my modules, though there can be gods of statistics that crash and die on pure or mech, etc.
0
3 years ago
#17
(Original post by thotproduct)
not that hard tbh, granted, AS FM isn't really designed to be extremely tougher than AS normal, though the challenge is certainly noticeable, can't really speak much for A2 but i hear AS -> A2 FM transition is some next level juju.

personally it depends on what kind of person you are, modules etc, im quite confident on all 3 of my modules, though there can be gods of statistics that crash and die on pure or mech, etc.
Oh wow, I guess I am expeted to make the transition from AS to A2 in less than a year, lol.

I've nearly completed A level maths and it isn't too bad.

Also, just saying, CGP released their revision book for AQA further maths 12 days ago.
0
3 years ago
#18
(Original post by stoyfan)
Oh wow, I guess I am expeted to make the transition from AS to A2 in less than a year, lol.

I've nearly completed A level maths and it isn't too bad.

Also, just saying, CGP released their revision book for AQA further maths 12 days ago.
you'll be fine big man if you did AS to A2 maths in a year you'll be ready for whatever FM has to throw at you, being honest I'd rather have done Maths early too, get it out of the way ygm.

cgp lmaooo, i haven't used one of those for a level yet, but they were peng bois back in gcse
0
X

new posts
Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Are you tempted to change your firm university choice on A-level results day?

Yes, I'll try and go to a uni higher up the league tables (20)
28.17%
Yes, there is a uni that I prefer and I'll fit in better (7)
9.86%
No I am happy with my choice (40)
56.34%
I'm using Clearing when I have my exam results (4)
5.63%