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I am given a line $l$ by the intersection of two planes $x-2y-5t=0$ and $y+z+3t=0$. I need to find the infinite point of the line $l$. I think that I can find vectors orthogonal to each of the planes and then their cross product is a vector parallel to $l$. And in this way to find the infinite point of the line $l$. Am I right?

I need to find the plane formed by the points $(0,0,1,-1), (1,0,0,-1), (2,-1,0,0)$ which are given in homogeneous coordinates. Can I use a method similar to this with the determinant?

Thanks!

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  • $\begingroup$ Explain more of what you mean by using the determinant to find the plane and how you consider this similar to using a cross product to solve the first problem. $\endgroup$
    – amd
    Commented May 20, 2019 at 5:48
  • $\begingroup$ I find two vectors parallel to the plane and a point which is in the plane, make a determinant(equal to zero) with their coordinates. It is not similar to the first part. $\endgroup$
    – spyer
    Commented May 20, 2019 at 6:05
  • $\begingroup$ Yet you asked “Can I use a method similar to this with the determinant?” $\endgroup$
    – amd
    Commented May 20, 2019 at 6:07
  • $\begingroup$ Anyway, in doing that you’re not really working in homogeneous coordinates. You’re converting back and forth between homogeneous coordinates and inhomogeneous Cartesian coordinates. There’s a common way to solve both problems that works directly with the former. $\endgroup$
    – amd
    Commented May 20, 2019 at 6:09
  • $\begingroup$ Can you explain it to me, please? $\endgroup$
    – spyer
    Commented May 20, 2019 at 7:17

1 Answer 1

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You can certainly solve these problems by converting to inhomogeneous Cartesian coordinates as you’ve done for the first one, but this misses an important duality between the two problems that allows them both to be solved in exactly the same way.

Recall that the point at infinity of a line is the line’s intersection with the plane at infinity, which I’ll assume is given by $t=0$. Thus in the first problem, you’re seeking a nontrivial solution to the homogeneous system $$\begin{align}x-2y-5t&=0\\y+z+3t&=0\\t&=0\end{align}.$$ The solutions to this system are the null space of the matrix $$\begin{bmatrix}1&-2&0&-5\\0&1&1&3\\0&0&0&1\end{bmatrix},$$ which is easily found by row-reducing this matrix a bit further to produce $$\begin{bmatrix}1&0&2&1\\0&1&1&3\\0&0&0&1\end{bmatrix}$$ from which the null space is spanned by $(2,1,-1,0)$, which is the point at infinity that corresponds to the intersection line. This gibes with your method: $(0,1,1)\times(1,-2,0)=(2,1,-1)$.

If you represent the planes by homogeneous vectors of coefficients in their equations, you can then view the intersection of the planes as the null space of the matrix that has these vectors as its rows. This is the same as the orthogonal complement of the span of the homogeneous vectors that represent the planes.

Note,too, that we could’ve instead started with just the two rows of the coefficient matrix that correspond to the two given planes: in that case, the null space is two dimensional and is the span of the vector found above and $(-1,-3,0,1)$—i.e., a parameterization of the line in inhomogeneous Cartesian coordinates is $(-1,-3,0)+\lambda(2,1,-1)$. This gives you two basic ways to represent a line in $\mathbb P^3$: as the meet of two planes or the join of two points.

For the second problem, substituting the coordinates of the three given points into the generic equation $ax+by+cz+dt=0$ generates the homogeneous system $$\begin{align}c-d&=0\\a-d&=0\\2a-b&=0\end{align}$$ and again the solution space is the null space of the coefficient matrix. Just as in the first problem, we find the vector that represents the plane that passes through three points by computing the null space of the matrix that has the homogeneous coordinate vectors of these points as rows: the plane that passes through three points is the orthogonal complement of their span.

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    $\begingroup$ Incidentally, the determinant method alluded to in the question follows directly from the solution to the second question. The plane is the join of the three points—the set of all nonzero linear combinations of them—so a point $(x,y,z,t)$ lies on the plane iff $\det\small{\begin{bmatrix}x&y&z&t\\0&0&1&-1\\1&0&0&-1\\2&-1&0&0\end{bmatrix}}=0$. $\endgroup$
    – amd
    Commented May 21, 2019 at 16:58

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