brightsky's Maths Thread

[brightsky]

What do mathematicians mean when they say something is 'countable'? Also, what exactly does 'countably infinite' mean?

[TrueTears]

the word countable has a few definitions depending on what field of maths you're talking about but the most common one is easily understood through an example:

the set {1, 2, 3} is countable, so is the set {-1, 2, 3}, {a, b, c} etc

the set {Z} of all integers is also countable, to be more precise, it's countably infinite but {R} is not countable

in general a set, A, is countable, if |A| in

[brightsky]

the word countable has a few definitions depending on what field of maths you're talking about but the most common one is easily understood through an example:

the set {1, 2, 3} is countable, so is the set {-1, 2, 3}, {a, b, c} etc

the set {Z} of all integers is also countable, to be more precise, it's countably infinite but {R} is not countable

in general a set, A, is countable, if |A| in

hmm what do you mean by |A|. is there a more rigid definition?

[Daenerys Targaryen]

I'm gonna hesitate a guess, but when I've learnt these || things, its always meant absolute value. So im guessing |A| means all of the positives of the no. muse be an element of the natural numbers?
Or |A|=
Correct me if im wrong?

[Mao]

Not quite. Even though the notation is the same, in this case it mean 'cardinality'.

Here's a day9 video explaining (not rigorously at all) the concept of bijection and how that relates to cardinality.

[Daenerys Targaryen]

They really need new notation for things...

[brightsky]

is there a quick way to find the equation of the plane on which a line (whose equation is given) and a point (whose coordinates are given) lie? i have a feeling my current method is unnecessarily longwinded.

[Planck's constant]

is there a quick way to find the equation of the plane on which a line (whose equation is given) and a point (whose coordinates are given) lie? i have a feeling my current method is unnecessarily longwinded.

This is testing my memory, but if the line is given in parametric form, it will be of the form (xo, yo, zo) + tv
The given point is (d,e,f)
Form the vector b/w  points (xo, yo, zo) and (d, e, f). Let that vector be w.

The parametric equation of your plane is (xo, yo, zo) + tv + sw

Or so I hope

[Mao]

is there a quick way to find the equation of the plane on which a line (whose equation is given) and a point (whose coordinates are given) lie? i have a feeling my current method is unnecessarily longwinded.

Planck's method is correct, but it's hard to extract a cartesian equation from this.

The easiest method (in my opinion) to find a Cartesian equation for a plane is as follows:

Let the line be described by , and the point be , then we can calculate the normal to the plane by .

The cartesian equation of the plane is therefore , where .

Note that we don't have to worry about using , since would appear on both sides and cancel out. Also, the choice of is arbitrary. This is because any point on the plane and any other point on the plane would satisfy the relationship , thus we can use or any other point on the plane.

The same idea applies if you have three points: calculate a normal from any two coplanar vectors ==> job done.

[Planck's constant]

Planck's method is correct, but it's hard to extract a cartesian equation from this.

Not all that hard.
If you have your two vectors, you cross-product them, which gives you the a, b, c in ax + by + cz = d

All that remains is to substitute either of the two known points to find d.

[FlorianK]

is there a quick way to find the equation of the plane on which a line (whose equation is given) and a point (whose coordinates are given) lie? i have a feeling my current method is unnecessarily longwinded.

OK so I'll just write out my guess, not sure if this is true, cause I never came across that problem. It is an interesting one though

So lets say you have the line and the Point P.

You have a random point on g which I'll call Q, so you could just plug in t=1 or t=0 into the equation of g.

Now on of the normal vectors of the plane will be the cross product of the vector and the vector . Because they both lie on the plane and the cross product of both of them will be orthogonal to the plane. So:

Well now you have the normal vector and you have a Point on the plane so yeah:



or



Hope that's correct

[brightsky]

yep! that was the method i ended up using. thanks anyways

[FlorianK]

anytime again

btw, is this UMEP or spesh?

EDIT:
This would be another method. Even shorter

So lets say you have the line and the Point P.

You have a point on g which I'll call Q which is the point for t=0 and you have R which is on g which is the point for t=1.

Now you have 3 Points (P, Q, R), so you can make a plane

[pi]

Split Deleted User's queries to a more appropriately titled thread of his own Re: Deleted User's Maths Thread

[FlorianK]

For distances can someone check my methods

2 parallel planes:
normalvector into line
-> Intersection on Plane 1 and Plane 2
-> Length of the vector between them

Parallel line and plane:
with normal vector of the plane from a random point on the line to the plane
find the intersection
Length of the vector between the intersection and the point on the line

2 skew lines (here h and g):

find a vector that is orthogonal to both directional vectors through cross product let that vector be
the 2 nearest points on h and g are A and B respectively


solve for r,s and t. Find A and B, find find

How do I find the distance between a point and a line?
How do I find the distance between 2 parallel lines?