Adding 2D Measurements to 3D coordinates

Question

Apologies if this is the wrong place, or if the question is stupid, I get flustered easily with math.

I have a set of 3D coordinates, which describe four positions of a object within a room taken by a camera. Imagine I am holign a piece of paper vertically in front of a camera, I have the four corner points of this object. (please see diagram for details).

Diagram of setup

The four points describe the four corners of an object in the room, but the object also possesses other features, mainly dips and valleys in it's design. I would like to get the 3D positions of these features, for graphing purposes.

Example of shape in room

Given I have the two 3D points, and all the 2D data of the shape (its height, width etc in centimetres), is there an easy way to use this data to get more 3D points, such as where the dips are?

Edit: These are the four points I have:

(432.8458196, 28.52526426, -119.3253036)

(430.6633017, 29.41752562, -80.5229784)

(404.8092209, 126.9603963, -83.04548288)

(406.0723883, 128.6560902, -119.3584913)

I also have a few other points in the world, such as the positions of a robot arm which was interacting with the object, and as close to paralele with the object as possible:

(460.6903766, 52.57003202, -78.05766964)

(446.4389674, 50.82420874, -81.01552504)

(443.7663845, 49.770538, -81.0743616)

(441.5178063, 49.50744226, -83.29886566)

Additionally, the 2D data I know is in mm, for example, 56mm from the one edge of the object, the height of the object reduces by 6mm, it then alters again after 350mm etc. Given this information, I can easily convert it into 2D coordinates.

Answer 1

Suppose you know the 3D coordinates of three points of your 2D shape, labeled $A$, $B$, $C$ in figure below. To find the coordinates of point $P$ you must draw two axes $AC$ and $AB$ (not necessarily perpendicular) and compute numbers $t$ and $u$ as in figure below: $$ t={AD\over AB},\quad u={AE\over AC}, $$ where $PD$ is parallel to $AC$ and $PE$ is parallel to $AB$. Note that $AD$ and $AE$ are signed distances, that is $AD>0$ if $D$ lies on ray $AB$, otherwise $AD<0$, and the same for $AE$ on ray $AC$.

The 3D coordinates of $P$ are then given by $$ P=(1-t-u)A+tB+uC. $$