Respuesta :
Answer:
Step-by-step explanation:
Given is a point (2,4,5) in 3 dimension.
a) The projection of the point on xy plane would be with coordinate z=0 and same x,y coordinates
i.e. (2,4,0)
Similarly projection xz plane would be (2,0,5)
and projection on yz plane is (0,4,5)
If (2,4,5) is opposite vertex and faces are parallel to the coordinate planes
we would have the sides as xy plane yz plane and zx plane with extreme vertex as (2,4,5)
So all the 6 vertices would be
(0,0,0) (0,4,5) (2,0,5) (2,4,0) (0,4,0) (0,0,5) and (2,0,0),(2,4,0)
We want to project a point on different planes. We will get the answers:
- Projection on the xy-plane: (2, 4, 0)
- Projection on the yz-plane: (0, 4, 5)
- Projection on the xz-plane: (2, 0, 5).
How to project a point?
So we have the point (2, 4, 5), first, we want to project it along the xy-plane.
You know that the xy-plane is that one where we have always z = 0, so the projection of this point on that plane just makes the z-component equal to zero, thus the projection is: (2, 4, 0)
Now we want to do the same thing but with the yz-plane, this time the x-component becomes zero: (0, 4, 5)
Finally, we do the same thing with the xz-plane, this time the y-component becomes zero: (2, 0, 5)
Now we want to find the vertices needed to draw a rectangular box with the vertices known are:
(0, 0, 0), (2, 4, 5), (2, 4, 0), (0, 4, 5), (2, 0, 5)
But a box has actually 8 vertices, so we need to find the other 3 ones, these are given by projecting again the projected points along a perpendicular plane, such that only one component is different than zero, this is trivial, so we will get other 3 vertices:
(2, 0, 0), (0, 4, 0) and (0, 0, 5)
So the needed vertices are:
(0, 0, 0), (2, 4, 5), (2, 4, 0), (0, 4, 5), (2, 0, 5), (2, 0, 0), (0, 4, 0), and (0, 0, 5).
If you want to learn more abot projections, you can read:
https://brainly.com/question/14789885
