equation of a plane passing through a line and parallel to another line verification

Question

In homogeneous coordinates I have the points $M(2,3,1,1), N(3,3,2,1), P(1,3,1,1), Q(0,2,1,0)$. I have to find equation of a plane which contains the line PQ and it's parallel to MN.

What I have so far:

$$PQ: \left\{ \begin{array}{c} x=1\mu+0\lambda \\ y=3\mu+2\lambda \\ z=1\mu+1\lambda \\ t=1\mu+0\lambda \end{array} \right. \text{ and } MN: \left\{ \begin{array}{c} x=2\mu+3\lambda \\ y=3\mu+3\lambda \\ z=1\mu+2\lambda \\ t=1\mu+1\lambda \end{array} \right. $$

From the first answer of this question I know that I can form a determinant with rows the directions of the lines.

$$ \begin{array}{|aaa|} x & y & z\\ 0& 2 & 1\\ 3 & 3 & 2\\ \end{array}=0 $$

From where $\alpha: x-3y+3z-a=0$.

Since $\alpha$ passes through $P(1,3,1,1)$ I substitute it's coordinates in the equation of $\alpha$ and I get:

$\alpha: x-3y+3z+5=0$.

Is my reasoning correct? I have the feeling that being in homogeneous coordinates the approach is entirely different.

Thanks in advance!

Answer 1

The plane we are looking for in the projective space corresponds to a 3d linear subspace of $\Bbb R^4$.
It contains $P$ and $Q$, and it contains the ideal point $I$ of the line $MN$, which is now $I=N-M=(1,0,-1,0)$.

We thus have $3$ vectors of $\Bbb R^4$, and are looking for a common orthogonal vector to them (the coefficients of the equation of the plane).
For this, the same determinant formula can be applied: $$\left\vert\matrix{x&y&z&1\\ 1&3&1&1\\ 0&2&1&0\\ 1&0&-1&0}\right\vert\ =\ 0$$

Answer 2

With $p = (x,y,z,t)^{\top}$ given the lines

$$ L_{PQ}\to p = P + \lambda_1(P-Q)\\ L_{MN}\to p = M + \lambda_2(M-N) $$

and the plane

$$ \Pi\to < p-p_0, \vec n > = 0 $$

If $L_{PQ}\in \Pi\Rightarrow < P-p_0+\lambda_1(P-Q),\vec n > = 0$ or $< P-p_0,\vec n > = 0$ and $ < P-Q, \vec n > = 0$

If $L_{MN} || \Pi \Rightarrow < M-N, \vec n > = 0$ then we have

$$ \vec n \perp (P-Q)\\ \vec n \perp (M-N)\\ <P,\vec n > = < p_0,\vec n > $$

then choosing $p_0 = P$ we have the plane equation as

$$ \Pi\to < p-P,\vec\omega> = 0 $$

where $\vec\omega$ can be one of the solutions for

$$ < P-Q,\vec \omega > = 0\\ < M-N, \vec \omega > = 0\\ ||\vec\omega|| = 1 $$

for instance

$$ \vec\omega = \left\{\frac{1}{3} \left(-\sqrt{3-5 \lambda ^2}-\lambda \right),\frac{1}{3} \left(\sqrt{3-5 \lambda ^2}-2 \lambda \right),\frac{1}{3} \left(\sqrt{3-5 \lambda ^2}+\lambda \right),\lambda \right\} $$

for any $\lambda \in \Re$ such that $3-5\lambda^2 \ge 0$