Maths Ext 2 Predictions (11 Viewers)

[appleali]

nah it was in my school trials, u basically let some parameter equal each of the components, so for x => 2a+2 then u get ur vector line, and show direction vectors are parallel which implies same plane

elaborate

 

[K2Trappy]

elaborate

on which part? finding the line or co-planarity (is that even a word)

 

[K2Trappy]

elaborate

also out of the papers u have done, which one have u found challenging? i need to go to sleep crying and suffering

 

[Edward34245]

Actually, the word "plane" is used everywhere in HSC Maths e.g. "complex plane", "x-y plane" or "inclined plane". It's even in one of the vector topic dot points:
View attachment 45037
Seems pretty obvious that students are expected to know what the word "plane" actually means.

Even if they don't explicitly use the word "plane" there is no reason why they can't just synonymously label it as a region or a set of points in the number space that satisfies a set of conditions.

What I mean in my earlier post is that there is nothing stopping them technically asking something like this:
View attachment 45039

If we honestly think they can't use the word "plane", then what's stopping them replacing the word "plane P" with "region R" or "a set of points P"?

The point is that students can answer this question with the tools they have within the syllabus and it doesn't require knowledge outside the syllabus (i.e. notice it's not asking you to recall the equation of a plane). All you need to do is compute the dot product of two perpendicular vectors (with a bit of work to figure out what those vectors are) to derive the result. This is simply an "application" of something in the syllabus to explore something unfamilar.

If you want to dismiss this as being "outside of syllabus", then you do so at your own risk. The HSC exams have time and time again proven otherwise throughout history because they can sneakily lean on this "application" side of the syllabus (which is why Ext2 has this reputation for being so challenging in the first place).

Any chance you have worked solutions for this? How fo you find the two perpendicular vectors with the given information?

 

[C2H6O]

elaborate

I think it goes like this:

set the line equal to λ
(x-2)/2=(Y+1)/-1=(z-4)/3=λ

x=2λ+2, y=-λ-1, z=3λ+4

Now just put that in ijk notation and you've got your line equation

line = (2 -1 4) + λ(2 -1 3)

then prove coplanarity using what you said earlier, prove the vectors intersect or are parallel

no clue how to do ii though

 

[K2Trappy]

I think it goes like this:

set the line equal to λ
(x-2)/2=(Y+1)/-1=(z-4)/3=λ

x=2λ+2, y=-λ-1, z=3λ+4

Now just put that in ijk notation and you've got your line equation

line = (2 -1 4) + λ(2 -1 3)

then prove coplanarity using what you said earlier, prove the vectors intersect or are parallel

no clue how to do ii though

im prob very wrong but for ii could u use one of the points on either line, and have the difference of direction vectors be the resultant direction? or have any point on both lines and just have that difference as the direction vector?

 

[appleali]

also out of the papers u have done, which one have u found challenging? i need to go to sleep crying and suffering

bruh i am not that good at maths imma get 52 tmrw
probably ruse/nsb are the hardest ones i've done? abbotsleigh was kinda hard as well

 

[K2Trappy]

bruh i am not that good at maths imma get 52 tmrw
probably ruse/nsb are the hardest ones i've done? abbotsleigh was kinda hard as well

brother ur getting high 80s I'm lucky if I touch 70s

 

[K2Trappy]

im on my knees for an e4 currently

 

[Dontcallmedude]

Ruse, north Sydney boys, north Sydney girls. Probably skip north syd girls. It's so different It's prolly not worth it

 

[Trebla]

Any chance you have worked solutions for this? How fo you find the two perpendicular vectors with the given information?

This is from the BoS trials 2024, see

BoS Trials Maths and Business Studies 2024

Here's my attempt. I've probably over-complicated it or am just plain wrong: Base case (n=0) – Trivial Inductive hypothesis – Assume true for some non-negative integer k: \text{i.e. } \exists p, q \in \mathbb{Z} : w_{2k} = 11 p \rightarrow u_{2k} = 11q \qquad (1) Inductive step – RTP true...

boredofstudies.org

 

[ringading]

how many calculators are we bringing

 

[I Love maths So much]

4

 

[ringading]

4

HELP i thought 3 was enough. damn u rich to buy that many

 

[Hejjwheh23]

Nice to see everyone is active on the day before the exam

View attachment 45049

I got this question in one of my school's past papers... No clue how to do any of it, so if it comes up in the exam tmr I'm cooked

I attempted this, not 100% confident tho.

 

[OD6]

I've done the 2022 and 2021 hsc papers over the last couple of days and got a 60 and a 65, aligning to around 88-89, so just below an e4.

I would love to get up to an e4 but idk how to go about it. Should I focus on perfecting q11-14 or answering more of q15-16? for context I'm usually at or close to full marks for 11 and 12, around 8-12/15 for q13 and 14 and q15-16 varies a bit, mostly around 4-6/15. Also, how do you guys judge when to move on from a question? A couple of times in these papers I have made a small mistake/oversight on a question (usually a proof or something where I can double check my answer) and then gotten stuck trying to work out what I did wrong to no avail.

Is it worth my time to go after more marks in q15/16, or should I not worry about it and maximise what i get in the easier questions

 

[Hejjwheh23]

Ignore the shading, Im not sure If im headed on the right track or not tho for this one aswell

 

[ringading]

I've done the 2022 and 2021 hsc papers over the last couple of days and got a 60 and a 65, aligning to around 88-89, so just below an e4.

I would love to get up to an e4 but idk how to go about it. Should I focus on perfecting q11-14 or answering more of q15-16? for context I'm usually at or close to full marks for 11 and 12, around 8-12/15 for q13 and 14 and q15-16 varies a bit, mostly around 4-6/15. Also, how do you guys judge when to move on from a question? A couple of times in these papers I have made a small mistake/oversight on a question (usually a proof or something where I can double check my answer) and then gotten stuck trying to work out what I did wrong to no avail.

Is it worth my time to go after more marks in q15/16, or should I not worry about it and maximise what i get in the easier questions

same i get around that too lol atp i just maximise the amount of marks i can get, though I'm sure it'll look better if u can get decent in e4 style questions

since it is tomorrow, maybe reviewing the topics you cant do well? cause in the exam you can be like oh i saw this yesterday and attempt it (attempts always get at least 1 if you write the right stuff) my teacher also said even if you cant do it there's possibility of just writing the steps you were going to do or what your thinking of how to solve it and can get a mark cause hsc markers are always trying to give out marks. if your getting high marks for q 11 and 12s but not full marks they're most likely silly mistakes. in the exam just go by the exam of what you can - you have the liberty to start anywhere cause of the different booklets. maybe starting with 11 and 12 for the easy marks you can get and make sure you don't rush cause whats the point of rushing and losing marks or losing time and not doing the questions you probably wont be able to solve yk?

i suggest practicing q13 and 14 to try and maximise those marks too - maybe also look at q15 and 16 (sometimes they have some questions for easy marks which u can maximise and get as many marks as u can)

whenever i do exams i try and attempt it and think of different ways i can solve it, attempt it each time (try not crossing out even if u think its wrong) just leave it there until you def know how to do it and write it). i generally move on when i absolutely lost hope and cannot figure it out and ill just come back to it later (i highlight it and maybe at the end of the exam i might think of smt and just write it down. it has worked b4)

 

[epicmaths]

im prob very wrong but for ii could u use one of the points on either line, and have the difference of direction vectors be the resultant direction? or have any point on both lines and just have that difference as the direction vector?

There are two ways I know about in finding the equation of a plane:
1. The dot product between a perpendicular vector and a vector between (x y z) and the foot of that perpendicular should be 0. Not suitable for this question.
2. If you already have two lines on that plane, then any point can be found by a linear combination of the direction vector of both lines, starting from their intersection point as the "centre" of the plane. At uni they call the two non-parallel vectors the basis (not necessarily at right angles (orthogonal)) of the plane.

In your question, the equation of the plane will be:
line 1 r= (2 03) + λ (1 -1 2)
line 2 r= (2 -1 4) + λ(2 -1 3)
Find find where they meet; say (a b c)
Then the equation of all points on that plane can b e given by (x y z) = (a b c) + lamba1 (1 -1 2) +lamba2 (2 -1 3)

I think really anyone of the other given points can be used in place of (a b c) as your choice of "centre" of plane is arbitrary as long as its on the plane.
Looks gross though!

 

[user1207]

HELP i thought 3 was enough. damn u rich to buy that many

stop im bringing 2 i lost my 3rd one