# How do I find the orthogonal projection of a point onto a plane

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luc on 16 Mar 2015
Commented: NGOC TAM LAM on 23 Apr 2020
Say I have a plane spanned by two vectors A and B. I have a point C=[x,y,z], I want to find the orthogonal projection of this point unto the plane spanned by the two vectors. How do I do this?

Andrew Newell on 16 Mar 2015
That's a math question. If you tell us the formula, we can tell you how to implement it.
luc on 23 Mar 2015
True, unfortunately I know how to implement it, but not the math.

Torsten on 23 Mar 2015
min: (x0+lambda*a0+mu*b0-x)^2 + (y0+lambda*a1+mu*b1-y)^2 + (z0+lambda*a2+mu*b2-z)^2
gives the distance squared from the point (x,y,z) to the plane
w=(x0,y0,z0)+lambda*(a0,a1,a2)+mu*(b0,b1,b2).
Differentiate the distance squared with respect to lambda and mu, set the partial derivatives to 0 and solve for lambda and mu.
If the result is lambda^, mu^, then
(x0,y0,z0)+(lambda^)*(a0,a1,a2)+(mu^)*(b0,b1,b2)
is the orthogonal projection of (x,y,z) onto the plane.
Best wishes
Torsten.

#### 1 Comment

luc on 23 Mar 2015
Thanks Torsten!

Noah on 3 Oct 2019
This is an old post, but it deserves a simpler answer. Your plane is spanned by vectors A and B, but requires some point in the plane to be specified in 3D space. Call a point in the plane P. You can compute the normal (call it "n" and normalize it). Then the projection of C is given by translating C against the normal direction by an amount dot(C-P,n).
% compute the normal
n = cross(A, B) ;
n = n / sqrt(sum(n.^2)) ;
% project onto the plane
C_proj = C - dot(C - P, n) * n

#### 1 Comment

NGOC TAM LAM on 23 Apr 2020
How can you express P in the formula?