Orthogonal projection of a point to plane

Question

A point and a plane is given: point $P(-4, -9, -5)$ and plane defined with three points : $A(0, 1, 3)$, $B(-3, 2, 4)$ and $C(4, 1, -2)$. So far I've managed to calculate the equation of this plane $5x + 11y + 4z = 23$. How can I calculate the coordinates of the orthogonal projection of this point $P$ to the plane?

Answer 1

You asked for another way to do this, so here are a couple. The projection of $P$ is the intersection of the plane defined by the three points and the line through $P$ orthogonal to the plane—parallel to the plane’s normal. Since you’ve already found an equation of the plane, you can use that to compute this point directly in a couple of ways.

• Find the signed distance of $P$ from the plane and move toward it that distance along the normal: The signed distance of a point $(x,y,z)$ from the plane is $${5x+11y+4z-23 \over \sqrt{5^2+11^2+4^2}} = {(5,11,4)\cdot(x,y,z)-23 \over 9\sqrt2},$$ which comes out to $-9\sqrt2$ for $P$. From the equation that you derived, the corresponding unit normal is $${(5,11,4)\over\sqrt{5^2+11^2+4^2}} = \frac1{9\sqrt2}(5,11,4).$$ We want to move in the opposite direction, so the projection of $P$ onto the plane is $$(-4,-9,-5)-{-9\sqrt2 \over 9\sqrt2}(5,11,4) = (-4,-9,-5)+(5,11,4) = (1,2,-1).$$

• Move to homogeneous coordinates and use the Plücker matrix of the line: The line through points $\mathbf p$ and $\mathbf q$ can be represented by the matrix $\mathcal L = \mathbf p\mathbf q^T-\mathbf q\mathbf p^T$. The intersection of this line and a plane $\mathbf\pi$ is $\mathcal L\mathbf\pi = (\mathbf p\mathbf q^T-\mathbf q\mathbf p^T)\mathbf\pi = (\mathbf\pi^T\mathbf q)\mathbf p - (\mathbf\pi^T\mathbf p)\mathbf q.$ The quantities in parentheses in the final expression are just dot products of vectors. Applying this to the present problem, we have from the plane equation that you derived $\mathbf\pi=[5,11,4,-23]^T$. For $\mathbf p$ we can take the point $P$, i.e., $\mathbf p = [-4,-9,-5,1]^T$, and for $\mathbf q$ the direction vector of the plane normal $[5,11,4,0]^T$. Plugging these values into the above expression, we get $$\left([5,11,4,-23]\cdot[5,11,4,0]\right)[-4,-9,-5,1]-\left([5,11,4,-23]\cdot[-4,-9,-5,1]\right)[5,11,4,0] = 162[-4,-9,-5,1]+162[5,11,4,0] = [162,324,-162,162].$$ Dehomogenize this by dividing through by the last coordinate to get the point $(1,2,-1)$. You could of course set up a system of parametric or implicit Cartesian equations and solve them for the intersection of the line and plane, but this method allows you to compute it directly.

Answer 2

If $M$ is the projection of $P$ in that plane so $PM \perp AB$, $PM \perp AC$ and $PM \perp BC$ then you find a system of three equation with three unknowns which are the coordinates of the Point $M$, you can also use the equation of the plane instead of one of this equation, Can you take it from here?