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12:01 AM
Some things are just instinctual lol
 
the rule in this house is no gratuitous profanity
 
that seems sensible enough
leslie i assume second edition of concrete mathematics is OK?
 
12:18 AM
yes
 
cool cool :)
 
imagine finishing an exercise about induction and actually learning something beyond "oh, if i order the things i'm assuming in this arbitrary way, this is something about what a natural number is from some other thing about what a natural number is"
 
1:05 AM
@EE18 that makes one direction of the exercise you were posting immediate!
 
Injective implies surjective? That one I got ya
It's the converse which is proving elusive at the moment. Racking my brain for how to do it without induction while watching basketball (so little working getting done as you can imagine...)
 
ah, I missed that in case you already mentioned it
for the other direction, I guess I would pick a right-inverse
 
1:17 AM
Hmm, interesting. For each $a \in A$ there is a $b$ such that $f(b) = a$. Since $A$ finite I can, without choice, define $g: A\ to A$ by $g(a) = b$ for such a pair (I think I would need to formalize this with induction on the size of the set no?). Then $f \circ g = \id$.
 
fun fact functions by default imply if x=y, then f(x)=f(y) hence why in injectivity we only require the other direction and not an if and only if statement
@leslietownes Is induction ever not that lol?
Thats my only experience so far with induction is why i ask
 
1:54 AM
Does the euler-maclaurin formula work for complex numbers too
nvm found it, it's the darboux formula
 
2:26 AM
@Thorgott Still not sure where to go from here even as I think about it. I feel like I should be trying to argue that $g$ as constructed is injective (not even sure if this is true), whence $f$ must be also (I'd have to think about this too)
is that sort of what you had in mind?
 
the first part, yes. the second part requires an additional input, namely that $g$ is then in fact bijective (why? apply the previous case of the claim)
 
2:43 AM
@Thorgott Intensely grateful that this is the first time I've read a (why?) and gotten an answer :)
OK, I will think on the first part. Also gonna think about how to rigorously construct $g$ and will report back. Thank you Thorgott!
 
2:59 AM
@Obliv in some galaxy brain sense, i guess not. i do think a book like the one i recommended above is a useful reference point, for purposes of evaluating thoughts like "ok, if i sit through all this set theory in the belief that it is fundamental to my understanding of the cosmos, how much does actually it help me do these exercises that involve induction, which CS majors who don't care what a set is can do"
there's definitely a world of stuff that those tools are good for beyond "for these choices of axioms, lets give another 20 line proof of the distributive law"
 
I think I am going to buy it Leslie :)
But get to it only after Enderton, dkm...
 
reaching for my revolver
 
Like something out of a western
 
shotgun noise
 
3:28 AM
@Thorgott This is what I came up with FWIW
As discussed with Jakobian...my writing is a little ugly
 
 
3 hours later…
6:02 AM
EE18 you didn't ask, but one unsolicited piece of advice is to remove unnecessary contradictions. somewhere in there you have a proof that if P(x) = P(y) then x = y, styled as a proof that if P(x) = P(y) and x isn't y then blah blah blah contradiction. injecitivty is defined in such a way that maybe you don't need to do that
another thing (and this is very particular to me and maybe others would disagree), there is basically no reason to write "exists" ever in a proof (as in "some m exists such that"). if you can't make do with English "is," rewrite until you can
a lot of people use "exists" because they are mechanically and phonetically unraveling symbolic stuff, without paying really any attention at all to how it sounds in human langauge
"f(A) subset A (proper)" could and should be "f(A) subsetneq A", \subset as a symbol is more or less inherently ambiguous, so to be nice to the person who just walked into the room, \subseteq and \subsetneq for "is a subset of" and "is a proper subset of" literally every time
 
6:30 AM
@Thorgott The point is that in the proof of $S^{2n}$ case, it uses the fact that any fixed point free self map on $S^{n}$ is homotopic to the antipodal map. Unless I have an explicit description of the even dimensional homology sphere, I don't think I can argue it in the same way.
 
 
2 hours later…
8:35 AM
Hi!
 
9:18 AM
Given a fixed orthogonal monometric reference in space, consider the following lines $r$ and $s$:
\begin{align*}
r: & \begin{cases} x = 2 \\ z = 0 \end{cases} \\
s: & \begin{cases} x = y \\ y=z \end{cases}
\end{align*}
and the point $A(3,0,1)$. Represent:
$2.$ The line for $A$ incident and orthogonal to $s$.
i tried this
From $x=y$ and $y=z$ $\quad \mathbf{d}_s = (1,1,1)$.
The line through $A$ should have a direction vector perpendicular to $(1,1,1)$.
Let $\mathbf{d} = (a,b,c)$ then $(1, 1, 1) \cdot (a, b, c) = 0$, then $a+b+c=0 \Rightarrow ( a=-b-c)$ therefore $\mathbf{d} = (1,-1,-1)$
The line for $A$ is:
$L:\begin{cases}x=3+t \\ y=-t \\ z=1-t\end{cases}$
I have a doubt whether writing $d=(1,-1,-1)$ is correct
 
you did not use the condition of intersection
moreover, I think you meant $d=(2,-1,-1)$
 
isnt a=1?
ah yes
do I write 2 for the equivalence?
for $L$
$L:\begin{cases}x=3+2t \\ y=-t \\ z=1-t\end{cases}$
for $s$
$s: \begin{cases} x = y \\ y=z \end{cases}$
$3+2t=-t \Rightarrow t=-1$
$(x,y,z) = (1,1,2)$
this is the intersection point ?
 
9:37 AM
I think you want $\psi$ to have domain $f(A)$ in the first part. That said, I don't love how the proof is written. If $f$ is injective, but not surjective, it witnesses that $A$ is equinumerous to a proper subset of itself. However, $A$ is finite, so this immediately violates the pigeonhole principle.
I suspect you may have unfortunately stated the pigeonhole principle only for naturals rather than finite sets. However, the pigeonhole principle holding for a set implies it also holds for any set in bijection with it. This is effectively what you are spelling out, but such an invariance is u
As for the second part, you executed exactly what I had in mind. However, as leslie also commented, the proof of injectivity is a bit confusing. The explicit construction of $g$ actually does not matter for the argument. If you haven't seen this before, I recommend observing that if $f\circ g$ is injective, then so is $g$ and this applies here.
@onepotatotwopotato the point is that any fixed-point free homeomorphism has degree $-1$ (which follows from being homotopic to the antipodal map in the case of $S^{2n}$, but that is immaterial to how you derive the contradiction)
 
$(3,0,1) \cdot (1,1,1) = 3+0+1=4$
 
@Pizza how can it be the intersection point since it does not belong to $s$?
 
@SineoftheTime \begin{cases} 1 = 1 \\ 1=2 \end{cases}
yes its false...
also from my solution there is no perpendicularity
The dot product of the vector is not zero
 
you're always making the same mistake
you can't find directly the direction vector, you have to use both conditions
till now, you only know $a+b+c=0$
 
@SineoftheTime And what does this that you told me refer to?
 
9:48 AM
since a=-b-c, I thought you meant 2 instead of 1
 
@mick solved TPC, please ping me if you'd like to learn the proof from the book
"proof from the book"
 
@SineoftheTime but is the error in writing a=-b-c?
 
no
@HighAsAKiteOnMath twin prime conjecture?
 
Yes, I've come up with the canonical proof from the book I think
 
@SineoftheTime so I also have to consider $(0,1,0) \cdot (a,b,c) = 0$ so $b=0$?
 
9:52 AM
what is $(0,1,0)$?
 
$d_r$
 
Note this is my tenth announcement of that, the rest failed and I saw where they failed. This one is holding up to my scrutiny at least
 
@Pizza why are you considering $r$?
you're only asked about s and the point A
 
@SineoftheTime see my ph for the link, I can't post here
post hist
 
I saw the question
 
9:53 AM
I thought you were referring to the mistake from last time, it's not clear to me where I'm going wrong
 
I can teach anyone who knows CRT this proof
The derivation is a mixture of CRT / modular arithmetic / inequality solving
It's quite fascinating there's no canonical paper about this counting formula
They come close to writing it exactly, but use other methods of derivation
So their formulas feel different
This one comes down to a question about a monoid containing a union of other monoids
 
let's see what the experts say
 
Yep
It's gonna be a rough ride. Hold onto your butts - Jurassic park
Hold onto your bots pls
:D
 
@Pizza you can't say $d=(2,-1,-1)$, why not $d=(1,0,-1)$ or $d=(1,-1,0)$?
you have to use the condition of intersection to determine up to a multiplicative constant the vector $d$
@HighAsAKiteOnMath :)
 
LOL
Sounds like I'm on drugs, but really math is to blame
I fuckin proved mon
I can't believe it
 
10:01 AM
@SineoftheTime I was following the same point of last time
 
πŸŒΏπŸπŸ˜ŽπŸ€“
 
Since $x + y = 0$ and $z = 0$, the direction vector of $r$ is $\vec{d}_r = (1, -1, 0)$. So
the line through $A$ should have a direction vector perpendicular to $(1, -1, 0)$.
Let $\vec{d} = (a, b, c)$ be the direction vector of the desired straight line.
Then:$$(a, b, c) \cdot (1, -1, 0) = 0 \implies a - b = 0 \implies a = b$$so $\vec{d} = (1, 1, c)$.
I don't write the coefficients of the variables?
I think I'm thinking wrong
 
@Thorgott You said rational homology sphere is just a space with rational homology of a sphere. Why does that imply any fixed point free Homeo has degree -1?
 
Lefschetz
 
DiRichlet Shetz
 
10:13 AM
@Thorgott Cool.
Hmm ok so as you said now the point is if a $\Bbb Z/2$-action actually exists.
 
@SineoftheTime i think i understand what you mean
wait
$(3,0,1) \cdot (a,b,c) = 0$
$3a+c=0$
$\begin{cases}a+b+c=0 \\ 3a+c=0\end{cases}$
$(a,b,c) = \left(-\frac{c}{3},-\frac{2c}{3},c\right)$
right?
 
$(3,0,1)$ is a point
 
its A
 
ik
why should the dot product be zero?
 
10:29 AM
don't have i to do the system?
I didn't know how to move then
i mean
If the dot product of the vector direction of the line and of the vector
direction of s is zero, perpendicularity is confirmed
but in this case is not $=0$
 
10:48 AM
meaning once I write $3a+c$ , then?
 
11:18 AM
I'm not understanding what you're doing
why $(3,0,1)\cdot(a,b,c)$ must be zero?
 
the dot product between two vectors is zero if and only if the two vectors are orthogonal.
mmm
@SineoftheTime do I have to find any point $P$ belonging to the line $s$ to form a vector connecting $P$ to $A$?
 
11:35 AM
you can find the plane passing through A and s and then the line you're serching is on this plane
 
11:50 AM
@SineoftheTime but shouldn't I have done that?
I tried to do as above, but it's wrong, but now you told me to find the plane etc.
do you intend to use the theorem that expresses the condition of perpendicularity between the line and the
plane?
 
you can use that
 
$$(a,b,c) = \gamma(\lambda,\mu,\nu)$$
@SineoftheTime but what is this?
 
the line you're searching is on the plane
 
mm wait
doesn't the point $A$ belong to the line?
is $A \in s$?
i can check wait
its $A \not \in s$
so
I write the Cartesian of $s$
$\begin{cases}
x - y = 0 \\
y - z = 0
\end{cases}$
@SineoftheTime now i can do this
 
go ahead
 
12:05 PM
$h(x-y)+k(y-z)=0$
$A(3,0,1)$ so $h(3)+k(-1)=0 \Rightarrow h=\frac{k}{3}$
em I got stuck here
$\frac{k}{3}+k(y-z)=0$
$k=0$
Maybe I'm wrong, I'll let you know
 
12:54 PM
$r:\begin{cases}x+y=0 \\ 2x-z-3=0\end{cases} \quad A=(0,0,1)$
Determine the plane that passes through the point $A$ and the line $r$, using the plane sheaf equation:
$$h(x+y)+k(2x-z-3)=0$$ I impose the passage through the point $A$:
$$h(0)+k(-1-3)=0 \Rightarrow k=0$$ Choose $h=1$ and then we get the equation: $x+y=0$
It's not clear to me why $h=1$ is chosen
 
1:26 PM
In my book, it says the $\mathrm{Cov}(X,Y)=E(XY)-E(X)E(Y)$ is defined for $X,Y\in L^2$. I don't really see why we require $X,Y\in L^2$. Is it so that $\mathrm{Cov}(X,X)=\mathrm{Var}(X)$, the variance which is only defined for $X\in L^2$?
I guess we require it not only for $Y=X$, but also in general, if we define $$\langle X,X\rangle=\|X\|^2:=E(|X|^2)=\int_\Omega |X|^2\,dP.$$
 
@psie don't overthink it. Its one possibility when Cov(X, Y) is defined. There could be other possibilities but probably not important to consider
But sure - you want to think of Cov as bilinear form
And this is on L^2
 
ok πŸ‘
 
1:48 PM
Can anyone understand why $h=1$ is chosen above? I still do not understand...
 
Is there a classification of the 3-manifolds that can be obtained by identifying opposite sides of polyhedrons? For example, using for cube you obtain a flat torus
 
choose $h=1$ to get $x+y=0$,ok
 
2:05 PM
I mean, every 3-manifold is essentially a polyhedron
but it sounds like you're moreso looking for something like arxiv.org/pdf/0806.1912
 
2:18 PM
It's quite surprising that every closed connected orientable 3-manifold $M$ admits a 3-fold branched covering $M\to S^3$ branched along a knot $K\subset S^3$.
and it implies any such 3-manifold is parallelizable.
 
Hello. I have a question, to be sure. In a general (non-finite) measure space, convergence almost everywhere and convergence in measure are independent, in the sense that one does not imply the other. My question is: suppose that \{f_n\} sequence of measurable functions is regular and has limit $f$. Then, if it converges in measure, does the limit in measure have to be $f$?
 
2:57 PM
@onepotatotwopotato interesting, how does the argument go? I don't know how branched covers interact with tangent bundles
 
3:11 PM
@leslietownes "Further, we have that $g$ is injective. This is because $a \neq b$ implies $\min \{m \in n \mid f(\varphi(m)) = a\} \neq \min \{m \in n \mid f(\varphi(m)) = b\}$, because if they were equal then for this $l \in n$ we'd have $a = f(\varphi(l)) = f(\varphi(l)) = b$, a contradiction." Is it this part you mean here?
Everything else is well taken. I probably should omit the (proper) because Enderton uses $\subset$ for proper subset, I mostly put it there because I was going to send this in the chat. But your comment is well taken, I've never really seen $\subsetneq but probably it should be used more
 
@Mr.Feynman what is definition of a regular sequence of measurable functions?
 
By regular I mean existence of pointwise limit
 
this is not a standard definition as far as I know
 
I avoided "convergence" because I'm considering $f_n:X\to\overline{\mathbb{R}}$ measurable
 
@Thorgott Thank you, will rewrite this part. As for the comment above, I think I will need to reread to fully understand :)
 
3:15 PM
@Mr.Feynman what does this has to do with anything
 
@Jakobian because in the message before I said pointwise limit and not pointwise convergence. Not quite relevant, though
 
@Mr.Feynman huh?
to me, you're making little to no sense
 
Let me rephrase my original question. This detail was irrelevant and I should have brought it up
In other words, the question is: suppose that $\exists\lim_{n\to\infty} f_n(x)=:f(x)\quad\mathrm{a.e.}$. Then, can it happen that $\{f_n\}$ converges in measure to $g\neq f$?
 
well the relevant theorem is this one
Theorem. Let $(f_n)$ be a sequence converging in measure to $f$. Then there is a subsequence $(f_{n_k})$ convergent a.e. to $f$
This is irrelevant of if your measure space is finite or not
 
Oh shoot, you're right. The subsequence converging almost uniformly assures that the limit is the same
Well, thanks
 
3:21 PM
@EE18 yeah, that
 
I wasn't quite sure how to show $\min \{m \in n \mid f(\varphi(m)) = a\} \neq\min \{m \in n \mid f(\varphi(m)) = b\}$ without a contradiction
 
EE18: in showing that the sets are equal implies that a = b, does that not show that g is injective?
i.e. if the contradiction arises from "assume that g isn't injective, prove that g is injective, contradiction" you can just take out the stuff before the first comma and after the second comma
maybe that's not what's going on here, but it's something people do a lot
 
@Thorgott The actual statement of the branched covering theorem is more stronger which says that a branched covering is branched along a single knot in the covering manifold $M$ that bounds an embedded disc. One can find a 3-ball $D$ in $M$ that contains the branched set in its interior. Since $S^3$ is parallelizable, the 3-frame on $S^3$ can be lifted to a frame field on $M\setminus\mathrm{Int}(D)$.
The bundle $TM|_{D^3}$ is trivial so with the chosen trivialization, the lifted frame field determines an element of $SO(3)$ at each point of $\partial D$. Since $\pi_2(SO(3)) =0$, that frame field extends over $D$.
It's actually written in the book Balarka introduced to me before.
 
@leslietownes I could be wrong but I think that's not what's going on here. When I say "if they were equal" in between the commas there I am referring to the minima of the sets (which I then call some $l$), not to $a,b$
Not sure if that absolves me of this sin though
 
the book actually contains 3 different proofs and I think you like the first proof more.
 
3:28 PM
EE18: okay, i guess my comment should have been "in showing that the minima of those sets are equal implies that a = b, does that not show that g is injective" and maybe the answer is no. divorced from that problem (which has probably been done to death), do be on the lookout for proofs by contradiction that are a contradiction wrapper around a direct proof
 
I didn't understand the proof but I think you can
 
@leslietownes well taken on all fronts, especially the done to death ;)
 
EE18: recall that this is the thing that ted has some good rants on, if you search the chat for "pleonastic" or "unnecessary contradiction"
 
That's because the proof by contradiction is a hammer in search of a nail :)
 
it's a good example of a larger issue, which is, any time you've finally worked out a proof, there are often things one can simplify or dispense with entirely in an argument, without changing its overall form, and yet a lot of the time, people never take that extra step (whether out of time pressure in a school setting, or just because it's hard to budge one's mind out of the exact route one took to get somewhere)
and sifting through people doing that for 35 years will change a man
 
4:05 PM
I definitely should look at my proofs more
It’s just I’m so eager to get on to the next section that it’s a battle to do the exercises in the first place, let alone reread them!
 
you should have strong words with that jerk who is making you work through enderton in the first place
 
πŸ’€
 
my daughter completely ignored a question that my wife asked her, and when she answered it the second time, i said "it's helpful if you listen, so people don't have to ask you the same thing over and over before you answer them," and my daughter said "what do you mean over and over, she only asked two times"
 
same emoji Sine just did
Your daughter is just asking for precision in your language!
 
while also making it clear that she did hear my wife the first time
kind of an oopsie, strategy wise
anyway, i found that funnier than my wife did
 
4:11 PM
I'm sensing this was your Lt Kaffee moment with your daughter as Colonel Jessup
Did you hear your mother the first time?
YOURE GOD DAMN RIGHT I DID
 
yeah, exactly
 
4:26 PM
so as not to send a very long message, I made a help, if anyone can and wants to come and help me pls
0
Q: Determining the Plane for a Given Point and Line Using the Equation of a Plane

PizzaI was working on an exercise following an example that required exactly what I had to do as well, but I encountered some difficulties. Below, I have indicated 2 titles, the first one is the example I was following, the second one is my attempt at doing a similar exercise. In bold, I have written ...

@SineoftheTime I tried to move forward but then I got stuck
 
4:38 PM
@Pizza I've added a couple of comments
 
@leslietownes Is all this set theory truly useless for other more applicable mathematics? I had this vague sense that in topology in particular it might be helpful
But cardinal arithmetic and ordinal operations etc. is useless outside set theory?
it's interesting in and of itself at the moment so im gonna stick it out, but figured id ask
 
@SineoftheTime regarding $k=3, h=1$, then $1(x-y)+3(y-z) = x+2y-3z$
 
yes
the plane is $\pi: x+2y-3z=0$
 
mm ok
 
ee18 i think a general familiarity with the concepts is helpful as mathematical background culture, and in very limited areas maybe it is an inoculant against thinking that the world is nicer than it is, but basically nobody ever uses that stuff. if you are constructing some counterexample in topology with the set of all countable ordinals, you are well outside the mainstream of what anybody ever actually thinks about
 
4:46 PM
@SineoftheTime does what you wrote to me next refer to case 2, or is it something else?
 
and yes there are set theorists in real life but it's like 50 people on planet earth, not one in every math department
 
I'm talking about the same exercise
 
I figure after reading Enderton I will have satisfied my set theory inclinations for at least a decade
But probably not for life :)
(Hopefully)
 
elements of set theory?
 
ee18 a useful counterpoint to all of that is, like, some people have spent their entire lives studying 3-manifolds, probably more people than have spent their lives on set theory, and nobody pretends that detailed knowledge of 3-manifolds is some kind of prerequisite to anything else
 
4:49 PM
set theory isn't really an active area of research anymore but is still used as a foundation of math
 
@SineoftheTime okok im seeing
 
in early 1900s you'd have a lot more people working on it
 
ill try to understand , i will let you know
 
@Obliv that one exactly
 
@Pizza now you have two conditions about $d_t$ so you can determine the coefficients of the line
 
4:57 PM
@SineoftheTime (a,b,c) = (1,2,-3)?
so $\lambda = 1, \mu = 2, \nu = -3$
 
I get the same equation as the $\pi$ plane
 
@EE18 Check out the long line, it uses the first uncountable ordinal omega_1
 
@SineoftheTime mm wait
 
5:04 PM
Very cool
I'm only going to get the faintest sniff of topology from my analysis text
one day looking forward to studying it. I bought Munkres a while back when i thought i might soon study it but realized my analysis background was too weak to appreciate
 
@SineoftheTime like from what i write in the help then$$\underbrace{1}_a \cdot (x-3) \underbrace{2}_b \cdot (y) \underbrace{-3}_c \cdot (z-1) = 0$$
so $x+2y-3z=0$
 
@EE18 You don't need analysis to study topology
 
Certainly, but what I sensed (and have read from others) is that one ought to know some analysis in order to appreciate topology. perhaps that's mistaken though
 
this is the equation of the plane
 
@SineoftheTime yes i mean
 
5:08 PM
you could have just substitute $h$ and $k$
 
Why doesn't this happen in the example I followed?
 
are you talking about the plane perpendicular to s?
 
@SineoftheTime i mean look at the conclusion of the example
 
it's a different ex
 
why did it find the line with the same steps and I didn't?
 
5:15 PM
@onepotatotwopotato yeah, I've seen the characteristic classes proof before (though I never worked out the details), but the geometric one you mentioned is neat, too
 
in the example was asking to determine the line s that passes through point A and is incident perpendicular to the line r;
for me the line through A incident and orthogonal to s;
 
@Thorgott As a first crutch towards sensible writing I think I may try to start putting the details in footnotes in my notes. It's tough because I want to supply the details in order to verify that I actually understood/to exercise my brain with the definitions/theorems I just learned. But point taken, especially as it relates to going back and reading the notes at some point
 
I'm not understanding
 
isnt orthonoal = perpendicular?
@SineoftheTime I mean the request was identical to mine
 
@EE18 as a topologist, set theory is not relevant outside of some rare exceptions (the Shelah paper about spaces with fundamental group $\mathbb{Q}$ comes to mind)
 
5:20 PM
@Pizza yes
 
the occasional set theory I use pops up in algebraic settings instead: vector spaces of large infinite dimension, there's an argument in Boardman's famous paper on conditionally convergent spectral sequences that I only managed to prove using transfinite induction, the theory of inaccessible cardinals as way of resolving size issues in category theory, etc.
 
@Pizza what request? you have to be clear
 
typically, though, the distinction between finite, countably infinite and uncountably infinite is all one needs
 
In the example: determine the line $s$ that passes through point $A$ and is incident perpendicular to the line r;
in my case: determine the line through $A$ incident and orthogonal to $s$
is not the same thing?
How does the exercise differ?
 
it's the same
but you're asking about too many different exercises in your post
that's not clear
just focus on one ex at the time
 
5:26 PM
no no wait
in the example it does like that to determine the line s
that passes through point A and is incident perpendicular to the line r;
so can I do it this way too?
 
actually
@SineoftheTime .
in the example the plane is x+y=0?
 
what example and what plane?
dude...
just work on one exercise
 
@SineoftheTime the one i posted
 
you have different requests
 
5:34 PM
but you said its the same
$(1)$ determine the line $s$ passing through the point $A$ and incident perpendicularly to the line $r$; in the example it solve this
they are not different exercise?
 
@Pizza the structure is the same but I can't solve all the ex
 
@SineoftheTime no but its finished
Conclusion: the equation of line $s$ is
$$s:\begin{cases}x+y=0 \\ x-y+2z-2=0\end{cases}$$
 
if you chose $h=1$ you get $x+y=0$
 
yes
 
so what's the question?
 
5:38 PM
are you referring to the example?
 
ok
 
what's your question about $x+y=0$ ??
 
its the plane?
yes
 
@leslietownes Do u think I could get funding to be a matrix theory specialist with a strong theoretic background in set theory? I study boxes and brackets
U should fund me with leslie coin
 
5:40 PM
$x+y=0$ is the plane passing through point $A$ and the line $r$
 
ok for me
it was
 
by strong theoretic background in set theory i mean theoretically a strong background in set theory
 
you can draw it in 3d with geogebra if you struggle
 
$x+2y-3z=0$
this is what i found instead right?
 
5:42 PM
@Obliv this is exactly what the world needs
 
@Pizza yes
so what?
 
Determine the plane passing through point $A$ that is perpendicular to the line $r$, using the equation of the star of planes:
in the example it does this and then finds the desired line
Isn't it enough for me to do this too?
 
yes
just follow the same method
 
yes
 
as I suggested a few hours ago
 
5:45 PM
for the example: $$\lambda = 1, \quad\mu = -1, \quad\nu = 2$$ hence we have
$$\underbrace{1}_a \cdot (x) \underbrace{-1}_b \cdot (y) \underbrace{2}_c \cdot (z-1) = 0 \Rightarrow x-y+2z-2=0.$$Conclusion: the equation of line $s$ is
$$s:\begin{cases}x+y=0 \\ x-y+2z-2=0\end{cases}$$
 
again, what's your question?
 
no but there is something different
@SineoftheTime I was showing you the steps
but now for my case
i dont have $\rho$ coefficient
 
I don't have all the time in the world, just ask your question
in your case I've wrote the equation of the plane an hour ago
$x+2y-3z=0$
 
ok
I find this in the previous step
so now theoretically, by doing this step which also gives the example, should I find the desired straight line?
 
yes
read my second comment in your post
I have to go
bye
 
5:51 PM
ok, bye im going to see
 
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