Computer Vision Geometry - Collinear Points In A Pinhole Camera Model

[jenny_shoars]

Homework Statement

Prove that a line in 3D space is imaged to a line on the image plane in a pinhole camera model.

Homework Equations

A 3D point give by (X,Y,Z) will be imaged on the image plane at
x = f(\frac{X}{Z})
and
y = f(\frac{Y}{Z})
where f is the focal point.

The Attempt at a Solution

My first thought was a more intuitive one. If you have a line in 3D space and the point which is the pinhole, you have a plane on which both the pinhole and line lie. Where this plane intersects the image plane it forms a line and this is where the 3D line is mapped to on the image plane. However, this seems like too much hand waving.

Instead I decided to try saying that for the line in 3D there must be a parametric equation given by
X = at + d
Y = bt + e
Z = ct + g
Then
x = f\frac{at+i}{ct+k}
y = f\frac{bt+j}{ct+k}
From here I know that for a line to exist on the image plane there must be a q and m such that
y = mx + q
Yet, this doesn't seem to lead me in the right direction.

Any suggestions? Thank you for your time!

 

[sunjin09]

Homework Statement

Prove that a line in 3D space is imaged to a line on the image plane in a pinhole camera model.

Homework Equations

A 3D point give by (X,Y,Z) will be imaged on the image plane at
x = f(\frac{X}{Z})
and
y = f(\frac{Y}{Z})
where f is the focal point.

The Attempt at a Solution

My first thought was a more intuitive one. If you have a line in 3D space and the point which is the pinhole, you have a plane on which both the pinhole and line lie. Where this plane intersects the image plane it forms a line and this is where the 3D line is mapped to on the image plane. However, this seems like too much hand waving.

Instead I decided to try saying that for the line in 3D there must be a parametric equation given by
X = at + d
Y = bt + e
Z = ct + g
Then
x = f\frac{at+i}{ct+k}
y = f\frac{bt+j}{ct+k}
From here I know that for a line to exist on the image plane there must be a q and m such that
y = mx + q
Yet, this doesn't seem to lead me in the right direction.

Any suggestions? Thank you for your time!

imaging a plane containing the line and the pinhole; this plane intersects with the image plane, the intersection
is obviously a line, isn't it just the image of the original line? I apologize for giving the answer directly, I simply can't think of any more disguised form ...

 

[jenny_shoars]

Like I said, that was my first thought, but it seemed to hand wavy. Maybe I'm just worrying to much and that's a fine answer.

Thank you.

 

[sunjin09]

Like I said, that was my first thought, but it seemed to hand wavy. Maybe I'm just worrying to much and that's a fine answer.

Thank you.

You're welcome, it's good to worry a bit more than others, as long as you're not obsessive :)