Dtjytyk

I want to find the orthogonal projection of the vector $vec y$ onto a plane.

I have $vec y = (1, -1, 2)$ and a plane that goes through the points
begin{align*}u_1 = (1, 0, 0) \ u_2 = (1, 1, 1) \ u_3 = (0, 0, 1)
end{align*}
I started by finding the equation for the plane by calculating: $vec{PQ} = u_2 - u_1$ and $vec{PR} = u_3 - u_1$.

I then took the cross product between $vec{PQ}$ and $vec{PR}$ and got $(1, -1, 1)$. I used the cross product as coefficients $a, b, c$ in the equation:
$$a(x - x_0) + b(y-y_0) + c(z-z_0) = 0$$

With this I got the plane equation to become $x - y + z = 1$.

Building on this, I went on to calculate the point where the vector $vec y$ intersects the plane. I used $(0, 0, 0)$ as a starting point and $(1, -1, 2)$ as the endpoint.

$$r(t) = {x_0, y_0, z_0} + t{x_1-x_0, y_1-y_0, z_1-z_0} = (t, -t, 2t)$$

I inserted these parameters into the plane equation and got $t = 1/4$.

So the vector intersects the plane in $(1/4, -1/4, 1/2)$.

Now my task is to find the projection of the vector y onto the plane. My idea was to use the point of intersection together with the cross product to find a vector that is perpendicular to the plane. By using the point of intersection as the starting point and the cross product as the endpoint.

I could subtract y with this perpendicular vector and get the endpoint for the projection of y onto the plane, while also here using the point of intersection as the starting point.

Howevever, the resulting projection is not correct. Apparently, both the starting point and the end point has to be calculated differently.

I also tried using the Gram-Schmidt process to transform the base vectors $u_1$, $u_2$, $u_3$ into an ortogonal base. With this I tried to use the equation $$vec y' = frac{vec y·u_1}{u_1·u_1}cdot u_1 + frac{vec y·u_2}{u_2·u_2}cdot u_2 + frac{vec y·u_3}{u_3·u_3}cdot u_3$$ to find the projection but a bit surprisingly arrived back at the original vector y when doing this.

Tremendously grateful for any tips.

Image of my problem:

I think some confusion might have come from the way Mathematica creates a 2D plane out of the two given points.

By using that a vector that passes through a plane (y) can be broken down into the sum of a vector (normal) orthogonal to the plane (n) and a vector that runs parallell to the plane and is a projection (x).

y = n + x

(1, -1, 2) = (1, -1, 1) + (a, b, c)

Projection = Vektor - VektorNormal

Projection = (0, 0, 1)

I used this and calculated where the VektorNormal intersects the plane and used that as the starting point for the projection.

It seems reasonable that it could work and it looks like it might do the trick.

Although far from as elegant as Haris Gusic's calculations.

It looks like you’ve corrected the fundamental conceptual error that you were making in trying to find where $vec y$ (really the line segment from the origin to $vec y$) intersects the plane. That line of attack is suspect since there’s no a priori reason to believe that this intersection even exists.

However, once you’ve found an equation for the plane, the orthogonal projection of $vec y$ onto this plane can be computed directly: it’s simply the foot of the perpendicular from $vec y$ to the plane. A simple way to compute this point is to substitute $vec y+tvec n$, where $vec n$ is normal to the plane, into the equation of the plane, and then solve for $t$: $$(1+t)-(-1-t)+(2+t)-1 = 3t+3 = 0,$$ so $t=-1$ and the orthogonal projection of $vec y$ onto the plane is $(0,0,1)$.

You can instead compute the projection without finding an implicit Cartesian equation for the plane or even computing its normal by using the fact that the orthogonal projection of $vec y$ onto the plane is the nearest point on the plane to $vec y$. The plane can be parameterized by the affine combination $$vec r(alpha,beta)=alpha u_1+beta u_2+(1-alpha-beta)u_3 = (alpha+beta,beta,1-alpha).$$ Minimizing the distance between $vec y$ and $vec r$ is equivalent to minimizing the square of the distance, namely $$(alpha+beta-1)^2+(beta+1)^2+(1-alpha-2)^2 = 2alpha^2+2alphabeta+2beta^2+3=frac12(alpha-beta)^2+frac32(alpha+beta)^2+3,$$ from which it’s obvious that the minimum is attained when $alpha=beta=0$, i.e., that the closest point to $vec y$ is $(0,0,1)$.