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Projective geometry - Intersection of two planes


New member
Apr 2, 2013

I have another question which is closely related to my previous question.
Suppose I have two planes $\alpha \leftrightarrow (\alpha_0,\alpha_1,\alpha_2,\alpha_3)$ and $\beta \leftrightarrow (\beta_0,\beta_1,\beta_2,\beta_3)$. I want to determine the intersection line of $\alpha$ and $\beta$.

I tried the same approach as in linear geometry, that is, the equation of the intersection line has to satisfy two equation, i.e
$\left \{\begin{array}{ll} \alpha_0x_0+\alpha_1x_1+\alpha_2x_2+\alpha_3x_3 \\ \beta_0x_0+\beta_1x_1+\beta_2x_2+\beta_3x_3 \end{array} \right.$

Now I choose $x_0$ and $x_1$ free as parameters, let $x_0=t$ and $x_1=l$ and I treat them as constant numbers. Eventually I get the following result for the intersection line:

$x_2=\displaystyle \frac{(-\alpha_0\alpha_2\beta_3+\alpha_0\beta_2\alpha_3-\alpha_3\alpha_0\beta_2+\alpha_3\beta_0)t+(-\alpha_1\alpha_2\beta_3+\alpha_1\beta_2\alpha_3-\alpha_3\alpha_1\beta_2+\alpha_3\beta_1)l}{\alpha_2(\alpha_2\beta_3-\alpha_3\beta_2)}$
$x_3 = \displaystyle \frac{\alpha_0\beta_2-\beta_0)t+(\alpha_1\beta_2-\beta_1)l}{\alpha_2\beta_3-\beta_2\alpha_3}$

So, I was wondering. Is this the correct way? Is there a better/easier and more beautiful way?