Achieve a perspective projection on the z = 0 plane of the unit cube, demonstrated in Figure (l) from the cop at E (0, 0, 10) upon the z-axis.
![564_Achieve a Perspective Projection on the Plane of Unit Cube.png](https://www.expertsmind.com/CMSImages/564_Achieve%20a%20Perspective%20Projection%20on%20the%20Plane%20of%20Unit%20Cube.png)
Figure: I
01: currently center of projection
E = (0, 0, -d) = (0, 0, 10).
∴ d = - 10
We identify here (from equation - 1), the particular point perspective transformation of the projection along with z = 0, plane; here cop is at (0, 0, -d) is provided by:
![134_Achieve a Perspective Projection on the Plane of Unit Cube 1.png](https://www.expertsmind.com/CMSImages/134_Achieve%20a%20Perspective%20Projection%20on%20the%20Plane%20of%20Unit%20Cube%201.png)
--------------------------(I)
Pn' = P. Pper, z --------------- (II)
Hence the perspective transformation of a provided cubes v = [ABCDEFGH] to V' = [A'B'C'D'E'F'G'H'] with d = - 10 is provided by:
[V'] = [V] . [Pper, z]
![1957_Achieve a Perspective Projection on the Plane of Unit Cube 2.png](https://www.expertsmind.com/CMSImages/1957_Achieve%20a%20Perspective%20Projection%20on%20the%20Plane%20of%20Unit%20Cube%202.png)
Hence the projected points of a provided cube V = [ABCDEFGH] are as:
A' =(0, 0, 0), B' = (1.11, 0, 0), C' = (1.11, 1.11, 0), D' = (0, 1.11, 0), E' = (0, 0, 0) F' = (1, 0, 0), G' = (1, 1, 0) and H' = (0, 1, 0).