How to prove that three different points in $RP^2$, $[x_1;y_1;z_1]$ $[x_2;y_2;z_2]$ $[x_3;y_3;z_3]$ Are on the same projective line $RP^2$ if and only if the det of the row's matrix of the three points (above) equals 0. Can you help in this, and by the way suggest me a good book on the topic of Projective space and geometry.


Using homogeneous coordinates, the lines in $\Bbb R^3$ through the origin correspond to the points of $\Bbb RP^2$, and theplanes through the origin correspond to the lines of $\Bbb RP^2$.

So, the 3 given points are collinear in $\Bbb RP^2$ iff the 3 given vectors are in a common plane (iff they don't span the whole space $\Bbb R^3$ iff they are linearly dependent) iff their determinant is $0$.

  • $\begingroup$ Thanks! , Is there a way to prove it with calculating? ,I guess i got it but not totally. $\endgroup$ – Maths1999_ Sep 30 '20 at 14:44
  • $\begingroup$ Why are the lines and planes in $R^3$ through the origin? $\endgroup$ – Maths1999_ Sep 30 '20 at 14:49
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    $\begingroup$ One way to visualize it is to put the ordinary plane on the affine plane $S:z=1$ in $\Bbb R^3$. Then any nonzero vector determines a unique point in the projectivized plane of $S$ by taking the intersection of the line of $v$ (through the origin) with $S$. If the vector is parallel to $S$, i.e. lies on the $x,y$-plane, then it determines a 'point in infinity'. $\endgroup$ – Berci Sep 30 '20 at 15:36

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