Can Two Intersecting Planes be Found from a Parametric Line Equation in ℝ^{3}?

[Bipolarity]

Suppose you have the parametric form of a line in ℝ^{3}. Given this equation, can we find two planes that intersect at this line? I know that there are infinite such ways we can do this, but what is one way this can be done? Essentially what I am asking is if you can reverse the procedure <system of two equations in 3-space> --> <solution vector (in 3-space)>.

BiP

 

[mfb]

It is trivial to construct planes which use the same point and share one direction vector with the line.
You can convert them to other expression methods afterwards.

 

[mathwonk]

do you understand this? mfb is saying to find the parametric equations for the planes, in terms of p,v,w, by starting from the parametric data p,v for the line, and choosing two different and independent choices of w.

 

[Erland]

There is a general method to "unsolve" a system of linear parametric equations, that is, to go from this system to a system of linear equations whose solution set is the affine manifold which is expressed by the original system. I found this method by myself, but it is likely that somebody else came up with the same method before me.

Suppose we have an affine manifold in R^n given in linear parametric form by

x=b+At,

where x is an n-column vector which gives a point on the manifold, b a constant n-column vector, A an n x m -matrix, and t an m-column-vector of parameters.

Now, write down the n x (m+1) - matrix (b A). On this matrix we perform a sequence of elementary column operations. Such an operation is like an elementary row operation but using columns instead of rows. But we exclude all such operations which involve the first column (originally b), except that we allow adding a multiple of another column to the first column (but not the other way round). One can verify that such an operation does not change the manifold given by the corresponding system of linear parametric equations.

We choose this sequence of column operations in a way that corresponds to Gauss-Jordan elimination, but using columns instead of rows (and vice versa) and starting in the bottom right corner (instead of the upper left one) and going left and upwards.
Let us write the resulting matrix as

(b' A'),

where b' is its first column.

The submatrix A' will now be in what we may call upside down reduced column echelon form, which is as reduced row echelon form with columns instead of rows (and vice versa) and bottom/right instead of upper/left (and vice versa).
Also, if a row in A' is a unit row (one element is 0 and the other 0:s), the element in b' at this row is also 0.

Next, if there is a set of 0-columns collected at the left of A', we remove these, so that we obtain new matrix

(b' A'')

of type n x (k+1), with k\le\max(m,n).
With t' as a k-column vectors of parameters, the system of linear parametric equations

x'=b'+A''t' (*)

gives the same manifold as the original one. m-k (which may be 0) parameters turned out to be superfluous. The dimension of the manifold is k.

This new system (*) of linear parametric equations has precisely the standard form which is obtained if we start from a system of linear equations, solve it by Gauss-Jordan elimiation and write the solution in parametric form in the standard way.
Now, we can, from (*), go "backwards" and (re)construct an (n-k) x (n+1) - matrix in reduced row echelon form, which is the augmented matrix of a system of linear equations, whose solution set is the given manifold.

Ok, I know I should give an example to illistrate this, but it is so cumbersome doing it in TeX-notation here... Maybe some day...

 

[Erland]

However, in the simple case OP originally mentioned, to obtain equations for two planes which intersects each other in a parametrically given line, we don't need to go through this procedure (although the procedure covers all cases).
In this case, solve the parametric equations for t (the parameter) wherever possible and set the expressions for t equal to each other.

For example, if we have

x=2+t
y=3+3t
z=4-2t

this gives (t=) (x-2)/1=(y-3)/3=(z-4)/(-2).

From this we obtain e.g 3(x-2)=y-3 <==> 3x-y = 3, and (-2)(y-3)=3(z-4) <==> 2y+3z=18.
The planes 3x-y=3 and 2y+3z=18 intersects in the given line.

This must be modified if not all the parametric equations can be solved for t, for example

x=2
y=3+3t
z=4-2t

Then, as before (t=) (y-3)/3=(z-4)/(-2) <==> 2y+3z=18, and as the other equation, we take x=2.