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Let $A,B,C$ be collinear points in 3d. Show that their projections $a,b,c$ respectively, onto 2d image plane are also collinear. If $A=[x_1,y_1,z_1], B=[x_2,y_2,z_2], C=[x_3,y_3,z_3]$ then $a=[x_1/z_1, y_1/z_1,1], b=[x_2/z_2, y_2/z_2,1], c=[x_3/z_3, y_3/z_3,1]$. Is a correct approach here to find the projected line $l'$ and show that $l'.a=0, l'.b=0, l'.c=0$ meaning that each projected point belongs on the projected line? If so how can this be shown?

Sebastiano
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jroc
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  • Welcome to MSE! <> The natural strategy may depend on how you've seen projective coordinates. To take a step back, is it clear (geometrically or otherwise) that a line projects to a line (or point)? – Andrew D. Hwang Sep 04 '23 at 21:04
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    No, i'm having trouble understanding why that is the case – jroc Sep 04 '23 at 21:06
  • I see; in that case, since mathematical notation and terminology are not universal, could you please edit the question to clarify: 0. Is this a question about affine geometry, or projective geometry? 1. Do square brackets denote Cartesian coordinates in three-space, or homogeneous coordinates in projective space, or something else? 2. Is "projection" defined by the formula $[x, y, z] \mapsto [x/z, y/z, 1]$? 3. What is the working definition of a line? <> Those items should help passersby give an answer fitting into your framework. :) – Andrew D. Hwang Sep 04 '23 at 21:21

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