Equation of the 5D coordinates of the projection of a point

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The discussion centers on finding the coordinates of the projection of a point above a plane onto that plane. Participants emphasize the importance of identifying the nearest point on the plane to the projected point, as this will represent the projection itself. A request is made for a generic expression for a point in the plane to facilitate this process. Clarifications are exchanged to ensure understanding of the projection concept. The focus remains on deriving the correct equation for the projection coordinates.
user199
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Homework Statement
I have no idea how to calculate the equation of the 5D coordinates of the projection onto the plane as stated in the part A of the attached problem. Help in this regards would be greatly appreciated.
Relevant Equations
I only know the solution in 3D but in higher dimensions I have no idea how to calculate the equation of the coordinates of the projection onto the plane. My solution for 3D problem is:

A given point A(x0, y0, z0) and its projection A′ determine a line of which the direction vector s
coincides with the normal vector N of the projection plane P. As the point A′ lies at the same time on the line AA′ and the plane P, the coordinates of the radius (position) vector of a variable point of the line written in the parametric form
x = x0 + a · t,
y = y0 + b· t
and
z = z0+ c· t,
These variable coordinates of a point of the line plugged into the equation of the plane will determine the value of the parameter t such that this point will be, at the same time, on the line and the plane.

Example: Find the orthogonal projection of the point A(5, -6, 3) onto the plane 3x -2y + z -2 = 0.
Solution: The direction vector of the line AA′
is s = N = 3i -2 j + k, so the parametric equation of
the line which is perpendicular to the plane and passes through the given point A
these coordinates of the radius vector of the point A′ must satisfy the equation of the given plane that is
3 · (3t+ 5) -2 · (-2t -6)+ 1 · (t+ 3) -2 = 0 => t = -2
therefore, the coordinates of the point A′ are,
x =3t+ 5 =3 · (-2)+ 5 = -1, y = -2t-6 = -2· (-2)-6 = -2
and
z = t+ 3 = -2+ 3 = 1,
thus the orthogonal projection of the point A onto the given plane is A′(-1,-2,1).
PP.jpg
 
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The hint implies to me finding the point in the plane nearest to the projected point.
Can you write a generic expression for a point in the plane?
 
haruspex said:
The hint implies to me finding the point in the plane nearest to the projected point.
Can you write a generic expression for a point in the plane?
Thanks for your comment.
NO, my means is that if we have a points above the plane as described in the statement of the image, how can we find the equation of the coordinates of the projections of these points on the plane. In other words, how one would find coordinates of the projection point?
 
user199 said:
Thanks for your comment.
NO, my means is that if we have a points above the plane as described in the statement of the image, how can we find the equation of the coordinates of the projections of these points on the plane. In other words, how one would find coordinates of the projection point?
Yes, I understand what you are trying to do. As a first step, I am asking you to write a generic expression for a point in the given plane.
Maybe you misinterpreted what I wrote. I mean the hint implies to me finding the point in the plane nearest to the point being projected. That nearest point will be the projection of the point being projected.
 

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