orthogonal projection of a vector onto a plane

Question

I am trying to find the orthogonal projection of a vector $\vec u= (1,-1,2)$ onto a plane which has three points $\vec a=(1,0,0)$ ,$\vec b=(1,1,1)$, and $\vec c=(0,0,1)$. I started by projecting $\vec u$ onto $\vec a$ and projecting the same vector onto $\vec c$ and finally adding both projections ,but I am not getting what I am expecting. Can someone please help?

Answer 1

It is easy to guess the equation of the plane as $x-y+z=1$ from which you see that the normal is $(1,-1,1)$. To find the projection choose $\alpha$ such that $(1,-1,2)+\alpha (1,-1,1)$ lies in the plane $x-y+z=1$. This gives $\alpha =-1$ and the projection is nothing but $\vec c=(0,0,1)$.

Answer 2

You can work out the normal to the plane by computing $$\vec n=(\vec a-\vec b)\times (\vec b- \vec c)$$where $\times$ is the vector cross product. Then project your vector $\vec u$ onto this normal to get $\vec u_\parallel$. Then the required projection onto the plane is $$\vec u_\perp=\vec u-\vec u_\parallel+\vec a$$ where the $\vec a$ is added on to ensure the vector lies on the plane, rather than lying parallel to the plane, but starting at the origin.