What is the parametric form of the unique line which crosses these other three lines?

Question

In the real affine space $\Bbb A^4$ let $A=(1,2,1,0), A'=(1,2,2,-1), B=(1,0,0,0), B'=(2,0,0,0), C=(2,1,1,0), C'=(-2,1,-1,0).$

Now let $a$ be the line that passes through $A$ and $A'$, $b$ through $B$ and $B'$, $c$ through $C$ and $C'$. What is the parametric form of the unique line which crosses $a, b, c$?

This is an exercise of which I was supposed to find the solution on my professor's webpage, as an example for others, but there was no solution actually. I've been trying with what I know from the book but I didn't quite get it, and I can't reach my professor right now... could you help me?

Answer 1

Write the coordinates of the given points $A,A',B,B',C,C'$ in the columns of a matrix and prepend a row with all ones: $$M = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 2 & 2 & -2 \\ 2 & 2 & 0 & 0 & 1 & 1 \\ 1 & 2 & 0 & 0 & 1 & -1 \\ 0 & -1 & 0 & 0 & 0 & 0 \end{bmatrix}$$ This matrix has one more column than rows, therefore it must have a nullspace of dimension at least one. Elementary linear algebra shows that the nullspace is spanned by the vector $$(-2,0,-2,0,3,1)^\top$$ In other words, $$C' = 2A + 2B - 3C$$ Note that the coefficients on the right-hand side sum to $1$. This implies that $C'$ lies in the 2-dimensional plane $f$ containing $A,B,C$. Therefore the line $c$ lies in $f$ as well.

The solution is therefore the line through $A,B$: It lies in $f$ as well and crosses $c$ at $\frac{1}{4}(3C + C') = \frac{1}{2}(A + B)$. Working out a parametrization is now an easy exercise left to the reader.

Generally, if the nullspace of $M$ contains a vector $(\alpha,\alpha',\beta,\beta',\gamma,\gamma')^\top$ with $\alpha+\alpha',\beta+\beta',\gamma+\gamma'$ nonzero, then $$\begin{align} A''\stackrel{\text{def}}{=} \frac{\alpha A + \alpha' A'}{\alpha + \alpha'} &\in a & B''\stackrel{\text{def}}{=} \frac{\beta B + \beta' B'}{\beta + \beta'} &\in b & C''\stackrel{\text{def}}{=} \frac{\gamma C + \gamma' C'}{\gamma + \gamma'} &\in c \end{align}$$ and $A'',B'',C''$ are collinear, e.g. $$\frac{(\alpha + \alpha')A'' + (\beta + \beta')B''} {\alpha + \alpha' + \beta + \beta'} = C''$$ Note that the denominator equals $-(\gamma + \gamma')$ (because of the row of ones in $M$) and is therefore nonzero.

Conversely, if there exist collinear, pairwise distinct $A''\in a, B''\in b, C''\in c$, say $rA'' + sB'' + tC'' = 0$ with nonzero scalars $r,s,t$ such that $r+s+t = 0$, we can write $$\begin{align} A'' &= uA + (1-u)A' & B'' &= vB + (1-v)B' & C'' &= wC + (1-w)C' \end{align}$$ with suitable line parameters $u,v,w$, and set $$\begin{align} \alpha &= ru & \beta &= sv & \gamma &= tw \\ \alpha' &= r(1-u) & \beta' &= s(1-v) & \gamma' &= t(1-w) \end{align}$$ Then $$\begin{align} \alpha + \alpha' &= r \neq 0 & \beta + \beta' &= s \neq 0 & \gamma + \gamma' &= t \neq 0 \end{align}$$ and $$\begin{align} \alpha + \alpha' + \beta + \beta' + \gamma + \gamma' &= r + s + t = 0 \\ \alpha A + \alpha' A' + \beta B + \beta' B' + \gamma C + \gamma' C' &= rA'' + sB'' + tC'' = 0 \end{align}$$ which implies that $(\alpha,\alpha',\beta,\beta',\gamma,\gamma')^\top$ is in the nullspace of $M$.

Answer 2

The three line equations can be written as follows:

$$a(u) = A + u(A'-A) = (1,2,1,0) + u(0,0,-1,1)$$

$$b(v) = B + v(B'-B) = (1,0,0,0) + v(-1,0,0,0)$$

$$c(w) = C + w(C'-C) = (2,1,1,0) + w(-4,0,-2,0)$$

The parametric form of the requested line:

$$x(t) = (x_0, x_1, x_2, x_3) + t(d_0, d_1, d_2, d_3)$$

The intersection between $x(t_a)$ and $a(u_x)$ leads to four equations:

$$\begin{aligned} x_0 + t_a d_0 &= 1 \\ x_1 + t_a d_1 &= 2 \\ x_2 + t_a d_2 &= 1 - u_x \\ x_3 + t_a d_3 &= u_x \end{aligned}$$

The intersection between $x(t_b)$ and $b(v_x)$ gives us:

$$\begin{aligned} x_0 + t_b d_0 &= 1 - v_x \\ x_1 + t_b d_1 &= 0 \\ x_2 + t_b d_2 &= 0 \\ x_3 + t_b d_3 &= 0 \\ \end{aligned}$$

The intersection $x(t_c)$ and $c(w_x)$ leads to:

$$\begin{aligned} x_0 + t_c d_0 &= 2 - 4w_x \\ x_1 + t_c d_1 &= 1 \\ x_2 + t_c d_2 &= 1 - 2w_x \\ x_3 + t_c d_3 &= 0 \\ \end{aligned}$$

I have used the Excel Solver to solve this set of $12$ non-linear equations. The resulting line equation in parametric form:

$$(1,0,0,0) + t(0,2,1,0)$$

The intersection points: $$a: (1,2,1,0)$$ $$b: (1,0,0,0)$$ $$c: (1,1,0.5, 0)$$

There is a total of $14$ unknowns:

• The base vector of the line ($4$ coordinates $x_0, x_1, x_2, x_3$)
• The direction vector of the line ($4$ components $d_0, d_1, d_2, d_3$)
• The three distance parameters $u_x, v_x, w_x$
• The three intersection point parameters $t_a, t_b, t_c$ of the line

The set of equations to be solved consists of $3 \times 4$ nonlinear equations. Rather than using a general solver, one could probably make use of the fact, that some of the direction components are zero. The direction vector can be scaled.

It remains to be shown that this solution is unique.

Answer 3

An elementary and intuitive solution

Parametric equations for the three given lines can be

$$a(t)=\begin{bmatrix}0\\0\\-1\\1\end{bmatrix}t+\begin{bmatrix}1\\2\\1\\0\end{bmatrix},\ b(u)=\begin{bmatrix}1\\0\\0\\0\end{bmatrix}u+\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\ c(v)=\begin{bmatrix}2\\0\\1\\0\end{bmatrix}v+\begin{bmatrix}2\\1\\1\\0\end{bmatrix}.$$

Consider the following figures now:

(Here $x$, $y$, $z$, and $w$ are the first, the second, the third, and the fourth coordinates, respectively -- the upper one being $x$.)

(1) The first figure depicts the graphs of $\color{red}a,b$, and $c$ in the $xy$ plane. In this view one can see that a line ($\color{blue} d$ ) crossing all these three lines can be a line having $1$ at its $x$ coordinate independently form the other coordinates. In other words the view of the crossing line has to go throug the red point (the image of $\color{red}a$), the vertical black line (the image of $b$), and the horizontal black line (the image of $c$.)

(2) The second picture shows that the $w$ coordinate of the crossing line has to be $0$ otherwise there could not exist three common points.

(3) The third figure shows clearly that the crossing line has to go through the point $(0,0)$. The equation of this line in the $yz $ plane is

$$\begin{bmatrix}1\\m\end{bmatrix}s.$$ Such a line could meet all the lines virtually for any $m$. at least in the $yz$ plane.

Summarize what we have for the crossing line:

$$d(s)=\begin{bmatrix}0\\1\\m\\0\end{bmatrix}s+\begin{bmatrix}1\\0\\0\\0\end{bmatrix}.$$ One can check that $m=\frac12$ is the unique solution for which all the three lines are met. (When I say unique, I mean that the direction vector can be still multiplied by any nonzero real and it remains the same.)