Point on a plane closest to the origin Consider the plane with equation $x-y+z=1.$ Find the point on the plane closest to the origin. 

I think I need to find a function for the distance between the plane and the origin and then find it's derivative and set it equal to zero. I'm not sure exactly how to go about doing this though. Any ideas?
The perpendicular vector to the plane is in the direction of $v = (1,-1,1)$. Start at the origin and travel in a straight line in the direction of $v$. You will hit the plane at the point that minimises the distance.
The point on the plane $ax+by+cz=d$ nearest the origin is $$\biggl(\frac{da}{\sqrt{a^2+b^2+c^2}}, \frac{db}{\sqrt{a^2+b^2+c^2}}, \frac{dc}{\sqrt{a^2+b^2+c^2}}\biggr)$$

Here, the point on the plane $x-y+z=1$ nearest the origin is $$\Biggl(\frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times-1}{\sqrt{1^2+(-1)^2+1^2}}, \frac{1\times1}{\sqrt{1^2+(-1)^2+1^2}}\Biggr) = \biggl(\frac{1}{\sqrt3}, \frac{-1}{\sqrt3}, \frac{1}{\sqrt3}\biggr)$$



This is an extension into three dimensions of the two dimensional nearest point to the origin on the line $ax+by=c$ is $\Bigl(\frac{ca}{\sqrt{a^2+b^2}}, \frac{cb}{\sqrt{a^2+b^2}}\Bigr)$, which is the point of intersection of $ax+by=c$ and $bx-ay=0$, perpendicular and passing through the origin.