Can 2D Vector Subspaces Form 3D Surfaces and How to Find Their Intersection?

[rabbed]

If each component of a 2D vector have multiple variables/dimensions, does that make two subspaces?
For example the vector (1 - b/a, 1 - a/b)

Can I convert both components surfaces into two 3D surfaces in the same space and find their intersection?

 

[Mark44]

If each component of a 2D vector have multiple variables/dimensions, does that make two subspaces?
For example the vector (1 - b/a, 1 - a/b)

I think you need to specify more clearly what you mean by a "2D vector." A vector belongs to some vector space. One such space is $\mathbb{R^2}$, a two-dimensional space whose coordinates come from the field of real numbers.

Can I convert both components surfaces into two 3D surfaces in the same space and find their intersection?

 

[rabbed]

Yes, the vector originated from a two dimensional space.
I would like to find some transformation that turns the vector back into two dimensions, having the same curve (not necessarily the same algebraic expression) as this destination vector: (cot(k) + 1, tan(k) + 1)
I was thinking either by intersection of the subspaces(?) surfaces (or whatever they are) of the source vector
or by some projection back to an x-y plane, creating a curve that coincides with the destination vector curve.

 

[Mark44]

Yes, the vector originated from a two dimensional space.

$\mathbb{R^2}$?

I would like to find some transformation that turns the vector back into two dimensions

The vector is still two-dimensional. a and b are just parameters here.

, having the same curve (not necessarily the same algebraic expression) as this destination vector: (cot(k) + 1, tan(k) + 1)

Can't you set 1 - b/a to cot(k) + 1, and set 1 - a/b to tan(k) + 1? I don't know if it's possible to solve these equations. Are you trying to solve for k?

I was thinking either by intersection of the subspaces(?) surfaces (or whatever they are) of the source vector
or by some projection back to an x-y plane, creating a curve that coincides with the destination vector curve.

 

[rabbed]

R^2, yes, two real-valued dimensions.
The thing is that normally the destination vector would be unknown, which is why I want to find a transformation which gives that curve.
The reason I got into subspaces was that Wolfram alpha displayed two surfaces when I gave it the source vector, and it looks like you can perform some intersection to get the destination curve. I guess 'a' would be in the x-dimension, 'b' would be in the y, and (1-b/a) and/or (1-a/b) would be in the z?
Maybe I need to explain what I'm trying to do? In that case, I'll come back.

 

[Mark44]

Maybe I need to explain what I'm trying to do?

Excellent idea...

 

[FactChecker]

If each component of a 2D vector have multiple variables/dimensions, does that make two subspaces?
For example the vector (1 - b/a, 1 - a/b)

Can I convert both components surfaces into two 3D surfaces in the same space and find their intersection?

Not usually. For a particular example, the set of possible vectors of that form is not necessarily a subspace. If you take two vectors of that form (in general), a linear combination of them is not guarantied to be of that form. You would have to check the particular example.
It would be interesting to determine the conditions that would make it a complete vector space. But that was not your specific question.

 

[rabbed]

Ok, if we drop the whole subspace thing, is it possible to find an expression of the figure representing the intersection of
z = 1 - y / x
and
z = 1 - x / y
?

 

[FactChecker]

Sure. You can graph them and see their intersection. In the general case, going from a graphed intersection to a math expression can be difficult. An algebraic approach is better. In this case, you can just set the two expressions for z equal and see what that says about x and y.

 

[rabbed]

I get
y = +/- x
should I insert that into for example z = 1 - y / x to get the intersecting figure?
Or maybe the "source vector"?
(1 - (+/-a)/a, 1 - a/(+/-a))

 

[FactChecker]

That just depends on what you want to get from it. A graph? An algebraic expression (x, ±x, ±1)

CORRECTION: (x, ±x, 1-(±1))

 

[rabbed]

An algebraic expression that can be used to make a graph..
Cool, how did you get that expression? Does that represent the intersection?
I guess to graph it you would need to do it piecewise for each combination of + and -?

 

[FactChecker]

I just made a correction. The intersection points are (x, ±x, 1-(±1)), where the ± signs of y=±x and z=1-(±1) are the same.
Proof: If the X-coordinate is x, setting z=1-x/y=1-y/x leads to y=±x for points in the intersection. So the Y-coordinate values of the intersection are y=±x. And putting y=±x back into z=1-x/y gives Z-coordinate values of the intersection of z=1-(±1).

 

[rabbed]

Okay, thanks.
When looking at the expression I don't think it will produce the destination vector figure, so it's not the transformation I'm looking for.
But that made me try to find a similar expression and I think y = 1-1/(1-x), or (x, 1-1/(1-x)), looks like (cot(k) + 1, tan(k) + 1), although I'm not sure how to check it.
I know I'm on thin ice here, but maybe by assigning different axis than x to a and y to b (and not necesserily have a or b represent the same dimension in both expressions) for (1 - b/a) and (1 - a/b), and then intersect the different combinations, you could end up with something like (x, 1-1/(1-x))..
It has to do with finding an arc length parameterization for r^2-x^2-y^2=0, and the destination vector comes from (cos(k),sin(k)).
I can explain more later to give some context if you're interested.

 

[rabbed]

Some context:

For an implicit function f(x,y) = 0, at the point (a,b):
- The curve normal is (df[x=a,y=b]/dx, df[x=a,y=b]/dy)
- The equation of the curve tangent is (df[x=a,y=b]/dx)*(x-a) + (df[x=a,y=b]/dy)*(y-b) = 0

Solving the equation for delta_x = x-a and setting y = b+delta_y gives delta_x = -(df[x=a,y=b]/dy) / (df[x=a,y=b]/dx) * delta_y, which is the contribution that a small change in y has on the movement in x along the tangent. Adding delta_x to the result gives the contribution that a small change in x _plus_ a small change in y has on the movement in x along the tangent. Solving the equation also for the movement in y gives a "movement contribution vector" at the point (a,b):
(delta_x - (df[x=a,y=b]/dy) / (df[x=a,y=b]/dx) * delta_y, delta_y - (df[x=a,y=b]/dx) / (df[x=a,y=b]/dy) * delta_x)

As delta_x -> dx and delta_y -> dy, we have a relation for the difference in arc length -> ds^2 = dx^2 + dy^2, making the movement contribution vector:
(sqrt(ds^2-dy^2) - (df[x=a,y=b]/dy) / (df[x=a,y=b]/dx) * sqrt(ds^2-dx^2), sqrt(ds^2-dx^2) - (df[x=a,y=b]/dx) / (df[x=a,y=b]/dy) * sqrt(ds^2-dy^2))
As the contribution of dx and dy goes to zero, the movement contribution vector becomes:
(ds - (df[x=a,y=b]/dy) / (df[x=a,y=b]/dx) * ds, ds - (df[x=a,y=b]/dx) / (df[x=a,y=b]/dy) * ds) or
ds * (1 - (df[x=a,y=b]/dy) / (df[x=a,y=b]/dx), 1 - (df[x=a,y=b]/dx) / (df[x=a,y=b]/dy))

If f(x,y) = r^2-x^2-y^2 = 0, the vector becomes ds*(1 - (-2*b)/(-2*a), 1 - (-2*a)/(-2*b))
and to find a way to arc length parameterize any function, we want to find a transformation of the non-arc length parameterized circle giving the arc length parameterized circle:
g(x(t), y(t)) = (r*cos(t), r*sin(t)), which has the movement contribution vector:
dt*(1 - (dg[x,y]/dy[t=k]) / (dg[x,y]/dx[t=k]), 1 - (dg[x,y]/dx[t=k]) / (dg[x,y]/dy[t=k])) = dt*(1 - (r cos(k))/(-r*sin(k)), 1 - (-r*sin(k))/(r*cos(k))) = dt*(cot(k)+1, tan(k)+1)

Does it make sense?

 

[rabbed]

Maybe it's necessary to set b = yf(a) also?

 

[rabbed]

If f(x,y) = r^2-x^2-y^2, what is f'(a, g(a)) if g(x) = +/- sqrt(r^2-x^2)? (I would like both the expression when f(x,y) and g(x) are known and the expression when they are not known)
Is it then possible to express the derivative of the result but this time with respect to a?