Finding a point on a plane closest to another point I have the point $(1,1,1)$ and the plane $2x+2y+z = 0$. I want to find a point that is closest to my point on the plane.
In other words, I want to find a point along the line $(1,1,1)+t(2,2,1)$ but on my plane. Notice that the vector $(2,2,1)$ is my normal vector and therefore I want to find the point parallel to this vector, but from my original point.
I need a nudge to complete this problem! Thank you.

Hint:
The distance between point $\;(a,b,c)\in\Bbb R^3\;$ and the plane $\;Ax+By+Cz+D=0\;$ is given by
$$\frac{|Aa+Bb+Cc+D|}{\sqrt{A^2+B^2+C^2}}$$
After you do the above (Result: $\;\sqrt3\;$), now just substitute in the euclidean distance formula with a general point  on the plane .
