Intersections and unions of points, lines, and planes

Now that we have a representation for lines, we can proceed to more complex relations. It should not surprise the reader that the coordinates of the plane passing through the three points $\ensuremath{{\bf p}} _i=(X_i,Y_i,Z_i,W_i)$, i=1,2,3, is obtained from the determinants of the four $3 \times 3$ submatrices of

$$\left[\matrix{ X_1 & X_2 & X_3 \cr Y_1 & Y_2 & Y_3 \cr Z_1 & Z_2 & Z_3 \cr W_1 & W_2 & W_3}\right].$$

Paying careful attention to the order of the submatrices then yields the plane's coordinates:

$$\ensuremath{{\bf n}} = ( \left\vert\matrix{ Y_1 & Y_2 & Y_3 ... ... X_3 \cr Y_1 & Y_2 & Y_3 \cr Z_1 & Z_2 & Z_3}\right\vert).$$

Note that, as a result of duality, the same formula can be used to find the point defining the intersection of three planes.

A point $\ensuremath{{\bf p}} = (X,Y,Z,W)$ lies on a line $\ensuremath{{\bf u}} = (l_{41}, l_{42}, l_{43}, l_{23}, l_{31}, l_{12})$if and only if the vectors $\ensuremath{{\bf p}} _1$, $\ensuremath{{\bf p}} _2$, and $\ensuremath{{\bf p}}$ are collinear, that is, the determinants of the four $3 \times 3$ submatrices of the following matrix

$$\left[\matrix{ X_1 & X_2 & X \cr Y_1 & Y_2 & Y \cr Z_1 & Z_2 & Z \cr W_1 & W_2 & W}\right]$$

are zero. In terms of the Plücker coordinates, this concurrence can be expressed as follows:

$$A\ensuremath{{\bf p}} = \ensuremath{{\bf0}} ,$$

where

$$A = \left[\matrix{ l_{23} & l_{31} & l_{12} & 0 \cr 0 & -l_... ...& l_{41} & -l_{31} \cr -l_{42} & l_{41} & 0 & l_{12}}\right].$$

An intuitive way to think about A is to realize that a line can be defined as the intersection of two planes. Therefore, a point lies on the line if it lies in the two planes. The equation above says that a point lies on the line if it lies in four planes. Only two of A's rows are important for any given line (indeed, A is of rank two), but all four rows are necessary to ensure that degenerate cases are handled properly.

Two lines $\ensuremath{{\bf u}}$ and $\ensuremath{{\bf u}} '$ intersect if and only if their Plücker coordinates satisfy the equation

(l₄₁l'₂₃+l'₄₁l₂₃) + (l₄₂l'₃₁+l'₄₂l₃₁) + (l₄₃l'₁₂+l'₄₃l₁₂) = 0,

which arises from the fact that the $4 \times 4$ determinant $\left\vert\matrix{\ensuremath{{\bf p}} _1 & \ensuremath{{\bf p}} _2 & \ensuremath{{\bf p}} '_1 & \ensuremath{{\bf p}} '_2}\right\vert$ is zero, where $\ensuremath{{\bf p}} _1$ and $\ensuremath{{\bf p}} _2$ lie on $\ensuremath{{\bf u}}$ and $\ensuremath{{\bf p}} '_1$ and $\ensuremath{{\bf p}} '_2$ lie on $\ensuremath{{\bf u}} '$.

In this section we have shown how to find the line containing two points and the plane containing three points. Symmetrically, we have shown how to find the point at the intersection of three planes but have not shown how to find the line at the intersection of two planes. This latter problem can be solved by computing the determinants of the submatrices, but it appears according to my calculations that a different choice of submatrices will have to be made. We have shown how to determine whether a point lies on a line or in a plane, and we have shown whether two lines intersect, though we have not calculated their intersection point. Other remaining problems include finding the intersection of a line and a plane, calculating the plane defined by two lines, and computing the plane defined by a point and a line.