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Say the place is parametrized as $x_1 = y_1 + y_2$ , $x_2=y_1-y_2 , x_3 = y_1+y_2$ and a point in $\mathbb{R}^3$ $x_1 = 2 , x_2 = 1 , x_3 = 3$

What would be the most efficient way to find the projection without using a normal vector (noticing that the plane is characterized by $(1,1,1)$ and $(1,-1,1)$, then finding that the normal vector is $(1,0,-1)$)? where then we have $(2,1,3) - t(1,0,-1) = (x_1^0 , x_2^0 , x_3^0) + y_1(1,1,1) + y_2(1,-1,1)$ and solving for the scalars etc.

I understand there is a "Normal equation" I could use, but I am not familiar with the concept, any hints are much appreciated

rannoudanames
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  • Using a normal to the plane is quite efficient. A normal vector can be computed via a single cross product, after which you project onto the normal and subtract. – amd May 13 '19 at 20:06
  • You’re using $x_1$ &c to mean two different things in your question. – amd May 13 '19 at 20:18
  • @amd I have $x_1$ which is a coordinate and $x_1^0$ which is the $x_1$ coordinate of a point on the plane. Would there be a way I could solve this problem using least squares? Thank you for your insight! – rannoudanames May 13 '19 at 20:21
  • In your very first sentence, you have $x_1$ as a coordinate of a point on the plane and also as a coordinate of the fixed point not on the plane. – amd May 13 '19 at 20:23
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    Yes, this can be solved as a least-squares fit, but that’s not the most efficient way to solve this specific problem. – amd May 13 '19 at 20:23

1 Answers1

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Once you have a normal to the plane, you can compute the orthogonal projection onto the plane directly using a standard formula: the projection of $\mathbf v$ onto another vector $\mathbf n$ is given by $$\mathbf\pi_{\mathbf n}\mathbf v = {\mathbf v\cdot\mathbf n\over\mathbf n\cdot\mathbf n}\mathbf n$$ and the projection of $\mathbf v$ onto the plane through the origin perpendicular to $\mathbf n$ is simply $\mathbf v - \mathbf\pi_{\mathbf n}\mathbf v$, that is, it’s what’s left over after you remove the component of $\mathbf v$ orthogonal to $\mathbf n$.

Alternatively, find the values of $y_1$ and $y_2$ that minimize the distance between that point on the plane and $(2,1,3)$, i.e., that minimize the function $(y_1+y_2-2)^2+(y_1-y_2-1)^2+(y_1+y_2-3)^2$. This can be done without using calculus, although that’s the simplest method to solve this minimization problem, I think.

In a comment you ask about solving this using least-squares methods. That’s certainly possible, too, though it’s not nearly as efficient as the above methods. Let $$A=\begin{bmatrix}1&1\\1&-1\\1&1\end{bmatrix}, \mathbf b=\begin{bmatrix}2\\1\\3\end{bmatrix}$$ so that you’re looking for the least-squares solution to $A\mathbf y=\mathbf b$. As usual, multiply both sides by $A^T$ to get $A^TA\mathbf y=A^T\mathbf b$, from which the solution is $\mathbf y = (A^TA)^{-1}A\mathbf b$ and so the projection onto the plane is $A(A^TA)^{-1}A^T\mathbf b$. This is in fact a standard formula for projection onto the column space of $A$ when $A$ has full column rank.

amd
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  • This makes sense, thank you! The last solution is awfully similar to linear regression correct? – rannoudanames May 13 '19 at 20:33
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    @rannoudanames I forgot to mention that the point-plane intersection that you proposed can be computed directly using the line’s Plücker matrix. See https://math.stackexchange.com/a/2434194/265466 for an example. – amd May 13 '19 at 23:02