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This is more of a follow-up question to the previous: Recreating a line after projective transformation

Referring to the below diagram (adapted from the wiki page on vanishing points): enter image description here

Say if I have a line $L$ that:

  • Lies exactly on the plane together with $L_1, L_2$,
  • Bounded by $L_1, L_2$,
  • Parallel to both $L_1, L_2$,
  • contains a point $P$.

Assuming I know the perpendicular distances from $P$ to $P_1$ and $P_2$ (which sits on $L_1, L_2$), as $d_1, d_2$ respectively.

Given the projection of $P$ on the image plane as $P'$, can I always assert that:

$\frac{\|{P_1' - P'}\|}{\|P_2' - P'\|}$ = $\frac {d_1}{d_2} $

where $\|P_1' - P'\|$ is the euclidean distance (in image coordinates) between $P_1'$ and $P'$ ?

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  • $\begingroup$ Is the line through $P_1'$ and $P_2'$ parallel to the line through $P_1$ and $P_2$? If yes, then the answer to your question is yes, by the similarity of the triangles $O P_1' P_2'$ and $O P_1 P_2$. $\endgroup$ – flawr Mar 18 '18 at 8:37
  • $\begingroup$ @flawr it is unsure. All that is know of the line is that it passes through all the points, $P_1', P_2', P'$. In other words, I am not sure whether a line segment, $L_0$ perpendicular to and lies on the same plane as with $L_1, L_2$ and its projection $L_0'$ will be necessarily parallel or not. $\endgroup$ – John Tan Mar 18 '18 at 8:45
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If you don't know whether this line through $P_1, P_2$ is parallel to the image plane, then you cannot cannot infer the relationship of the two projected distances: To illustrate this, consider the extreme case where $L=V$ and where $\Vert P_1' - P'\Vert = \Vert P_2' - P \Vert$. Then we could have the situation where $P_1 = P_1'$ and $P_2$ is a lot further away from $P_2'$. That means

$$\frac{\Vert P_1' -P'\Vert}{\Vert P_2' - P'\Vert} = 1$$

but

$$\frac{d_1}{d_2} << 1.$$

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  • $\begingroup$ To clarify it is known that the line through $P_1, P_2$ is parallel to the ground plane (in fact it lies exactly on the ground plane). Did u mean line through $P_1', P_2'$ on the image plane instead? $\endgroup$ – John Tan Mar 18 '18 at 9:08
  • $\begingroup$ I edited my answer: I mean the question is whether $P_1, P_2$ is parallel to the image plane. $\endgroup$ – flawr Mar 18 '18 at 9:12

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