But theres no way to reduce what I have to matrix form? Once you sub each one of the x y and z in to the second equation its just a mess

You don't want matrix form. It is a mess, but you solve for $t$ ...

I didn't make up the stoopid problem.

That partition closely matches how $f$ behaves I think

You need a finite partition that depends on $\epsilon$, @Perturbative.

For example, if I give you $\epsilon = 1/5$, how do you get a partition $P$ so that $U(f,P)<1/5$?

Okay solving for $\alpha = [s- (pq+qp+rc)/(pl+qm+rn)]l +a $

what's $\alpha$?

your t

It was defined at the beginning of the problem instead of t

Oh. So then you need $x,y,z$ in terms of $\alpha$. But your formula doesn't look right at all. How do you get products of $pq$, for example.

px+qy+rz=s

Yeah, and $x=Lt+a$.

So where do you get $pq$?

I subbed in for y and z too

I understand.

Then a bit of rearranging

Oh the pq should be a pa

$p(lt+a) + q(mt+b) + r(nt+c) = d$.

I hit the wrong key

So $t=[d-(pa+qb+rc)]/[pl+qm+rn]$.

It's a stooopid problem, as I said.

yes

Then you put that back in to get $x,y,z$.

But then how do I get it in the Ax=d form?

That's the equation of the plane. You're finding the point ...

No the matrix equaiton

Thats what the problem asks for

imgur.com/a/alnnvcp

It asked where the line meets the plane.

In the form of a set of eqations in a matrix format

I hate this question. It's truly stoopid.

Welcome to cambridge

Here's a slick problem. Let $F$ be a non-Galois extension of $\mathbb{Q}$ of degree 4. Show that the Galois closure of $F$ must be $S_4$, $A_4$, or $D_4$, and it's $D_4$ iff $F$ contains a quadratic extension of $\mathbb{Q}$

So take the equations of the line without parameter in it.

Two of those equations give you the line. The third equation is going to be the equation of the plane. You want to solve that system of three linear equations simultaneously to get the $x,y,z$ that satisfies all of 'em. So forget about the parametric way (which is the obvious way any fool would do this).

Hello!!! I have the following question

What do you mean the third equations gives the plane?

Two linear equations will give that line (use two of the equalities they gave you). The equation of the plane is your third equation. Solve all three simultaneously (in principle) to get the intersection.

Would I need to include $\alpha$ in that then?

Nope.

What do I do about the z in the 3rd equation?

What do you mean?

So I have $(x-a)/L=(y-b)/m$ and $px+qy+rz=s$

Right?

No, you need another equation to get the line.

Oh so use all 3?

Use $(y-b)/m=(z-c)/n$. Of course, rewrite everything with variables on the left.

Whats wrong with using the x?

To get a unique point, you will need three equations in three variables.

You need to use both equations for the line.

another both ...

Im confused

The line is given by two intersecting planes. That takes two equations.

You need THREE equations total.

@TedShifrin $f(0) = 0$ in that question right?

No, $0 = 0/1$, @Perturbative.

But then why havent you used the equation including x?

Jake, you're not reading what I write.

I said another ... then both

Then I said you need three equations.

Im not understanding what you write

Well, I give up. I've got other things I need to do.

ah okay cool

@TedShifrin Okay so I basically have to make the first point $x$ after $0$ in the partition to be really close to $0$ as the sup of $f(x)$ on $[0, x]$ will be $1$, and similary I have to make the last point $x'$ before $1$ in the partition to be really close to the $1$

@Daminark you're using how the subgroups of $S_4$ look like, right?

Yup

Actually no, if $0 < \epsilon < 1$ for a partition $P = \{x_i\}_{i =0}^n$ of $[0, 1]$ since $$M_i = \sup_{x \in [x_{i-1}, x_i]}f(x) \leq 1$$ for all $i = 0, 1, ..., n$ I just need to make the difference $(x_{i-1}, x_i)$ to be less than $ \frac{\epsilon}{n}$

@Perturbative: Note that that is impossible, because you need to cover the whole interval $[0,1]$, which has length way more than $\epsilon$. :P

There are some real ideas to have here. Hint: Where is $f(x)\ge \epsilon/2$?

hi professor @TedShifrin it was nice to see you browsing the JEE maths chat room. Did you find anything interesting?