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Am I good enough for Oxbridge? Proving it here.

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Original post by Kvothe the arcane
[user=298154]Slowbro93
is. No need to apologise. And thank you.

This week has been pretty bad across the ball. I'm stressed with work. Hope to work rather effectively in the next 2 days in order to meet Wed's targets.


Slowbro93> TeeEm in Maths? :shock:

Let the ultimate maths battle begin! :boxing: :jerry:

Spoiler

(edited 8 years ago)
Original post by Indeterminate
With the correct amount of dedication, you'll have no trouble in meeting your offer :thumbsup:

Self-belief is key! :yep:


I've a slight question.

Question
The transformation T:R3→R3T:\mathbb{R}^3 \rightarrow \mathbb{R}^3 is represented by the matrix T where T = (45−3−121101)\begin{pmatrix} 4 & 5 & -3 \\-1 & 2 & 1 \\1 & 0 & 1 \end{pmatrix}.

The plane Π1\Pi _1 is transformed by T to the plane
Unparseable latex formula:

\PI _2

. The plane Π1\Pi _1 has vector equation r=(011)+s(1−12)+t(304)r=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}+s\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}+t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}, where s and t are real parameters.
Find the equation of Π2\Pi _2 in the form r.n=p\textbf{r.n}=p.

My Answer
r=(011)+s(1−12)+t(304)=(s+3t1−s1+2s+4t)\begin{aligned} r & =\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}+s \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}+t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} \\ & = \begin{pmatrix} s+3t \\ 1-s \\ 1+2s+4t \end{pmatrix} \end{aligned}
r′=Tr=(45−3−121101)(s+3t1−s1+2s+4t)=(2−7s3−s+t1+3s+7t)=(231)+s(−7−13)+t(017)\begin{aligned} r'=Tr & =\begin{pmatrix} 4 & 5 & -3 \\-1 & 2 & 1 \\1 & 0 & 1 \end{pmatrix} \begin{pmatrix} s+3t \\ 1-s \\ 1+2s+4t \end{pmatrix} \\ & = \begin{pmatrix} 2 -7s \\ 3-s+t \\ 1+3s+7t \end{pmatrix} \\ & = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + s \begin{pmatrix} -7 \\ -1 \\ 3\end{pmatrix}+ t \begin{pmatrix}0 \\ 1 \\ 7\end{pmatrix} \end{aligned}

To find vector perpendicular to plane,
Unparseable latex formula:

\begin{aligned}\begin{vmatrix} i & j & k \\-7 & -1 & 3 \\0 & 1 &7\end{vmatrix}& =i \begin{vmatrix}-1 & 3 \\ 1 & 7 \end{vmatrix}-j \begin{vmatrix}-7 & 3 \\ 0 & 7 \end{vmatrix}+k \begin{vmatrix}-7 &-1 \\ 0 & 1 \end{vmatrix} \\ &=i(-7-3)-j(-49)+k(-7)=-10i+49j-7k



(r′−(231))⋅(−1049−7)=0\left( r'- \begin{pmatrix} 2 \\ 3 \\1\end{pmatrix} \right) \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix}=0
r′⋅(−1049−7)=(231)⋅(−1049−7)=−20+147−7=120r' \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\1\end{pmatrix} \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix}=-20+147-7=120

Π2:r′⋅(−1049−7)=120\boxed{\Pi _2:r' \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix} =120}

The final answer is fine but I'm wondering about the layout. In particular the "to find vector..." bit. I know I have to get that to be able to express the plane in the form they want but I was wondering if there's a clearer way to express it.

Thanks
(edited 8 years ago)
Reply 42
Avatar for TeeEm
TeeEm
19
Original post by XxKingSniprxX
Slowbro93> TeeEm in Maths? :shock:

Let the ultimate maths battle begin! :boxing: :jerry:

Spoiler



I am honoured!!!
What is this?
It look like stirring ....

c13.jpg
Original post by TeeEm
I am honoured!!!
What is this?
It look like stirring ....

c13.jpg


Hey, any chance you could help with the above? I know that to find a vector perpendicular to the plane to express it in the form r.n=p, I had to find the cross product of the vectors which lie on the plane. I did this as vectors which lie on the plane form part of the plane equation. I just don't know how to write this.

Thanks
Original post by XxKingSniprxX
Slowbro93> TeeEm in Maths? :shock:

Let the ultimate maths battle begin! :boxing: :jerry:

Spoiler



:colone:

Original post by Kvothe the arcane
I've a slight question.

Question
The transformation T:R3→R3T:\mathbb{R}^3 \rightarrow \mathbb{R}^3 is represented by the matrix T where T = (45−3−121101)\begin{pmatrix} 4 & 5 & -3 \\-1 & 2 & 1 \\1 & 0 & 1 \end{pmatrix}.

The plane Π1\Pi _1 is transformed by T to the plane
Unparseable latex formula:

\PI _2

. The plane Π1\Pi _1 has vector equation r=(011)+s(1−12)+t(304)r=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}+s\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}+t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}, where s and t are real parameters.
Find the equation of Π2\Pi _2 in the form r.n=p\textbf{r.n}=p.

My Answer
r=(011)+s(1−12)+t(304)=(s+3t1−s1+2s+4t)\begin{aligned} r & =\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}+s \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}+t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} \\ & = \begin{pmatrix} s+3t \\ 1-s \\ 1+2s+4t \end{pmatrix} \end{aligned}
r′=Tr=(45−3−121101)(s+3t1−s1+2s+4t)=(2−7s3−s+t1+3s+7t)=(231)+s(−7−13)+t(017)\begin{aligned} r'=Tr & =\begin{pmatrix} 4 & 5 & -3 \\-1 & 2 & 1 \\1 & 0 & 1 \end{pmatrix} \begin{pmatrix} s+3t \\ 1-s \\ 1+2s+4t \end{pmatrix} \\ & = \begin{pmatrix} 2 -7s \\ 3-s+t \\ 1+3s+7t \end{pmatrix} \\ & = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + s \begin{pmatrix} -7 \\ -1 \\ 3\end{pmatrix}+ t \begin{pmatrix}0 \\ 1 \\ 7\end{pmatrix} \end{aligned}

To find vector perpendicular to plane,
Unparseable latex formula:

\begin{aligned}\begin{vmatrix} i & j & k \\-7 & -1 & 3 \\0 & 1 &7\end{vmatrix}& =i \begin{vmatrix}-1 & 3 \\ 1 & 7 \end{vmatrix}-j \begin{vmatrix}-7 & 3 \\ 0 & 7 \end{vmatrix}+k \begin{vmatrix}-7 &-1 \\ 0 & 1 \end{vmatrix} \\ &=i(-7-3)-j(-49)+k(-7)=-10i+49j-7k



(r′−(231))⋅(−1049−7)=0\left( r'- \begin{pmatrix} 2 \\ 3 \\1\end{pmatrix} \right) \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix}=0
r′⋅(−1049−7)=(231)⋅(−1049−7)=−20+147−7=120r' \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\1\end{pmatrix} \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix}=-20+147-7=120

Π2:r′⋅(−1049−7)=120\boxed{\Pi _2:r' \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix} =120}

The final answer is fine but I'm wondering about the layout. In particular the "to find vector..." bit. I know I have to get that to be able to express the plane in the form they want but I was wondering if there's a clearer way to express it.

Thanks


I would have said that was fine tbh, took me a while to remember how to do that as I haven't touched that topic for over 3 years :tongue:

Original post by TeeEm
I am honoured!!!
What is this?
It look like stirring ....

c13.jpg


Hiya :gah:
Reply 45
Avatar for TeeEm
TeeEm
19
Original post by Kvothe the arcane
Hey, any chance you could help with the above? I know that to find a vector perpendicular to the plane to express it in the form r.n=p, I had to find the cross product of the vectors which lie on the plane. I did this as vectors which lie on the plane form part of the plane equation. I just don't know how to write this.

Thanks


Please post a thread in maths with the question and tag me there
Original post by TeeEm
Please post a thread in maths with the question and tag me there


Sure :smile:
Reply 47
Avatar for TeeEm
TeeEm
19
Original post by Slowbro93
:colone:

...


hello

c7.jpg
Good Luck @Kvothe the arcane Fantastic start to your blog!


You seem so motivated and dedicated- I'm sure you'll get the grades you need :h:
Tutor day changed to Sunday so that's when this week's targets are due!

Original post by Tank Girl
Good Luck @Kvothe the arcane Fantastic start to your blog!

You seem so motivated and dedicated- I'm sure you'll get the grades you need :h:


Thank you. I'm afraid your assessment of me is false :frown:. I'm pretty demotivated and dedication is something I need to work on. I mean to say I am rarely motivated to do anything. It's just me rather than anything being wrong. I survive on stress of deadlines and other things to get stuff done.

I am, however, confident in my ability to sustain a habit so once I get into a habit of studying regularly, I should be fine. I do actually like working and it's a wonderful feeling getting into the zone. The trouble is starting to be productive due to anxiety or distractions.
(edited 8 years ago)
Very impressed that you got an offer :biggrin: hopefully you make it!

:excited: to see how this all turns out, might be able to learn a few things from you :tongue:
Original post by SeanFM
Very impressed that you got an offer :biggrin: hopefully you make it!

:excited: to see how this all turns out, might be able to learn a few things from you :tongue:

Thank you :smile:.

What do you mean?

Sent from my SM-G925F using Tapatalk
Good luck mate, and congratulations on the offer, you seem like you deserved it and you sound determined!
Original post by Ninjakam119
Good luck mate, and congratulations on the offer, you seem like you deserved it and you sound determined!

Thanks

Sent from my SM-G925F using Tapatalk
Studied for 2 days in a row. Tiny amounts but progress.

Anxiety is reducing and I'm no longer feeling as stupid and incompetent. Having an all nighter to chill, clean and then catch a bit of study as my Friday afternoons area always unproductive.
Original post by Kvothe the arcane
I've a slight question.

Question
The transformation T:R3→R3T:\mathbb{R}^3 \rightarrow \mathbb{R}^3 is represented by the matrix T where T = (45−3−121101)\begin{pmatrix} 4 & 5 & -3 \\-1 & 2 & 1 \\1 & 0 & 1 \end{pmatrix}.

The plane Π1\Pi _1 is transformed by T to the plane
Unparseable latex formula:

\PI _2

. The plane Π1\Pi _1 has vector equation r=(011)+s(1−12)+t(304)r=\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}+s\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}+t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix}, where s and t are real parameters.
Find the equation of Π2\Pi _2 in the form r.n=p\textbf{r.n}=p.

My Answer
r=(011)+s(1−12)+t(304)=(s+3t1−s1+2s+4t)\begin{aligned} r & =\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}+s \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix}+t\begin{pmatrix} 3 \\ 0 \\ 4 \end{pmatrix} \\ & = \begin{pmatrix} s+3t \\ 1-s \\ 1+2s+4t \end{pmatrix} \end{aligned}
r′=Tr=(45−3−121101)(s+3t1−s1+2s+4t)=(2−7s3−s+t1+3s+7t)=(231)+s(−7−13)+t(017)\begin{aligned} r'=Tr & =\begin{pmatrix} 4 & 5 & -3 \\-1 & 2 & 1 \\1 & 0 & 1 \end{pmatrix} \begin{pmatrix} s+3t \\ 1-s \\ 1+2s+4t \end{pmatrix} \\ & = \begin{pmatrix} 2 -7s \\ 3-s+t \\ 1+3s+7t \end{pmatrix} \\ & = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + s \begin{pmatrix} -7 \\ -1 \\ 3\end{pmatrix}+ t \begin{pmatrix}0 \\ 1 \\ 7\end{pmatrix} \end{aligned}

To find vector perpendicular to plane,
Unparseable latex formula:

\begin{aligned}\begin{vmatrix} i & j & k \\-7 & -1 & 3 \\0 & 1 &7\end{vmatrix}& =i \begin{vmatrix}-1 & 3 \\ 1 & 7 \end{vmatrix}-j \begin{vmatrix}-7 & 3 \\ 0 & 7 \end{vmatrix}+k \begin{vmatrix}-7 &-1 \\ 0 & 1 \end{vmatrix} \\ &=i(-7-3)-j(-49)+k(-7)=-10i+49j-7k



(r′−(231))⋅(−1049−7)=0\left( r'- \begin{pmatrix} 2 \\ 3 \\1\end{pmatrix} \right) \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix}=0
r′⋅(−1049−7)=(231)⋅(−1049−7)=−20+147−7=120r' \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix} = \begin{pmatrix} 2 \\ 3 \\1\end{pmatrix} \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix}=-20+147-7=120

Π2:r′⋅(−1049−7)=120\boxed{\Pi _2:r' \cdot \begin{pmatrix} -10 \\ 49 \\ -7 \end{pmatrix} =120}


= the reason I never bothered with FP3!!!!!!!!!! (Also, dedicated texing!!!)
Original post by mobbsy91
= the reason I never bothered with FP3!!!!!!!!!! (Also, dedicated texing!!!)


:P

It's not really any difficult than C4. Hardly anything new and you discovered matrices in Fp1.

The texing is annoying but I'm trying to get more used to it.
Original post by Kvothe the arcane
:P

It's not really any difficult than C4. Hardly anything new and you discovered matrices in Fp1.

The texing is annoying but I'm trying to get more used to it.


Can you imagine doing it without tex :redface:
I have a confession to make.

Despite starting this thread with a lot of gusto, revision has been minimal. I did, however, start last week but progress has been slow and disorganised. I've been making my way slowly through the M2 60 question review as well as the Fp3 Vectors chapter but most of the study periods have been spent on random maths questions.

Biochem hasn't been done. My goals are however long term so I'm not stressing too much about it. I know what I want to achieve by Sunday and will make sure I do it at the end of this week. I'm going to the library after work today so will have a solid 3 hrs to dedicate to revision. I hope to give some time to the abandoned sciences and perhaps some maths if I grow bored
(edited 8 years ago)
Had a good productive 2 hr study session yesterday.

Spent an hr working my way through Fp3 exercise 7E (matrices are time-consuming) and then I made a start on Bio. Read and made some notes on .1-.4 of Chapter one. I'll finish the chapter this evening and probs do ATP. Then those past paper questions :eek:. Do some maths as well. I'll aim to finish Fp3 Vectors Chapter this evening.

Regarding M2, I feel like the 60 question exercise is too time consuming so I'll skip to some WEP questions and COM. Projectiles I'm solid on. I might come back to it when I've finished the module but I think it's important to get everything learnt. I'm still not too sure how I'll integrate the B+C still but I plan to use them as reference guides and for their examples.

I had errands to run in town yesterday - hence the 2 hrs.

I have to do the gym this afternoon but aiming for 3.5-4hrs of study but I'm fine with 3.

If I'm efficient with my time and complete my Bio4 and Fp3 goals on time then I plan to make a start with Chemistry. Don't want to do two modules of Further Maths in the same day :wink:. Thankfully I'm solid on solid on most of this week's targets so will mainly be doing questions!
(edited 8 years ago)

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