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Let $ax+by+cz=0$ be a line in projective space. Let the line be satisfied by two points $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$. Then we have $$a_1x+a_2y+a_3z=0$$ $$b_1x+b_2y+b_3z=0$$ This implies that $\begin{vmatrix} a_1&a_2\\b_1&b_2\end{vmatrix}=\begin{vmatrix} a_2&a_3\\b_2&b_3\end{vmatrix}=\begin{vmatrix} a_1&a_3\\b_1&b_3\end{vmatrix}=0$.

Now my textbook says that the equation of this line is $\begin{vmatrix} a_1&a_2\\b_1&b_2\end{vmatrix}x_1+\begin{vmatrix} a_2&a_3\\b_2&b_3\end{vmatrix}x_2+\begin{vmatrix} a_1&a_3\\b_1&b_3\end{vmatrix}x_3=0$.

I don't understand how this is. This relation should be satisfied by ANY $(x_1,x_2,x_3)$, and not just the point on the line, as the coefficients (the determinants) are all 0!! So you get $0.x_1+0.x_2+0.x_3=0$ for every $(x_1,x_2,x_3)$.

Am I correct?

algebraically_speaking
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  • Yes you're right. But maybe you're misleading the notation of the book. Could you please say the name of the book so I can check it myself? – user40276 Aug 15 '14 at 08:40

1 Answers1

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Your claim that the three $2\times 2$ minors being all equal zero are simply not true. Since you claim $ax+by+cz=0$ to be a line, I assume you are talking about the projective plane with homogeneous coordinate $(x,y,z)$.

The two equations you gave is supposed to solve the line coordinates $(x,y,z)$, or rather the $(a,b,c)$ in your line equation $ax+by+cz=0$. The minors are not necessarily all $0$. Basically, you are looking for a vector that is perpendicular to two other vectors $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$, and therefore the results is given by cross product. The three minors pop up in the computation of the cross product.

Troy Woo
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  • I think the OP is assuming that $(a_1: a_2: a_3) = (b_1: b_2: b_3)$ and in this case the minor determinants vanish. – user40276 Aug 16 '14 at 06:09