# In which case three projetive points exist on the same projetive line $RP^2$.

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 $$0$$.

• Thanks! , Is there a way to prove it with calculating? ,I guess i got it but not totally. – Maths1999_ Sep 30 '20 at 14:44
• Why are the lines and planes in $R^3$ through the origin? – Maths1999_ Sep 30 '20 at 14:49
• 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'. – Berci Sep 30 '20 at 15:36