Lets say I have the point $(x, y, z)$ and the plane with normal $(a, b, c)$ with the point $(d, e, f)$. I am trying to use this in $3D$ programming. Thank you!
How do I find the projection of a point onto a plane
4 Answers
You want to find $t$ such that $(x+ta,y+tb,z+tc)$, $(x,y,z)$, and $(d,e,f)$ form a right angled triangle, with the first of these (the point you are looking for) being the right angle. You can do this with dot products, and this will give you
$t = \frac{ad-ax+be-by+cf-cz}{a^2+b^2+c^2}.$
Substitute this into $(x+ta,y+tb,z+tc)$ and you have your result.
Take the displacement vector from the point in the plane to the given point: $$ {\bf v}=(x-d , y-e, z-f) $$ and let ${\bf w}$ be the normal vector to the plane.
We can describe ${\bf v}$ as a sum of two vectors; one that is perpendicular to the normal vector ${\bf w}$ (denoted by ${\bf v}_\perp$), and another that is parallel to the normal vector ${\bf w}$ (denoted by ${\bf v}_\parallel$).
$$ {\bf v} = {\bf v}_\perp + {\bf v}_\parallel $$
${\bf v}_\parallel$ is given by $$ {\bf v}_\parallel = {{\bf v}\cdot{\bf w}\over\Vert{\bf w}\Vert^2} {\bf w} $$ Then $$ {\bf v}_\perp = {\bf v} - {{\bf v}\cdot{\bf w}\over\Vert{\bf w}\Vert^2} {\bf w} $$
From this, the required point is $(d,e,f)+{\bf v}_\perp$.
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0I know it's been a while, but why would w⊥v⊥ hold? Shouldn't that be w⊥(v⊥-(d,e,f))? – 2018-12-31
Let us denote your point as $(x_0,y_0,z_0)$ instead of $(x,y,z)$ and projection as (x'_0,y'_0,z'_0)
Parametric equation of the line that passes through point and its projection is given by :
x'_0=x_0+a\cdot t
y'_0=y_0+b\cdot t
z'_0=z_0+c\cdot t
Equation of the plane is :
$a \cdot(x-d)+b\cdot(y-e)+c\cdot(z-f)=0$
Now , since point (x'_0,y'_0,z'_0) belongs to the plane you have to substitute its coordinates into equation of the plane instead of $x,y,z$ and calculate parameter $t$ .
You have to take the parallel line to the normal vector in the point $(x,y,z)$ and the projection will be the intersection of this line whit the plane.