This is my code to find the coordinate of projection of the point pA = {2, 3, 5} on the plane -284 + 14 x + 2 y + 5 z == 0.
Clear["Global`*"] pA = {2, 3, 5}; myP = -284 + 14 x + 2 y + 5 z; {x, y, z} /. Solve[{x == pA[[1]] + Coefficient[myP, x] t, y == pA[[2]] + Coefficient[myP, y] t, z == pA[[3]] + Coefficient[myP, z] t, myP == 0}, {x, y, z, t}, Reals] {{16, 5, 10}}
What is simpler way to find the projection of a point on a plane?
RegionNearest[ImplicitRegion[myP == 0, {x, y, z}]]@pA
{16, 5, 10}
Clear["Global`*"] pA = {2, 3, 5}; myP = -284 + 14 x + 2 y + 5 z; The normal vector to the plane is:
Coefficient[myP, {x, y, z}] {14, 2, 5}
The line that passes through the point pA and is parallel to the normal vector is given by:
line = Coefficient[myP, {x, y, z}] t + pA {2 + 14 t, 3 + 2 t, 5 + 5 t}
For Mathematica processing, convert to rules:
pline = Thread[{x, y, z} -> line] {x -> 2 + 14 t, y -> 3 + 2 t, z -> 5 + 5 t}
To find where the parametric line intersects the plane (i.e., closest point), solve for t:
sol = First@(myP == 0 /. pline // Solve[#, t] &) {t -> 1} res = line /. sol {16, 5, 10}