Question

Let S be the subspace of x+y+z=0 in R^3. Let V be (1,0,1). Find distance of v to s.

Answer 1

Let (x,y,1-x-y) be a point in the plane x+y+z=0
The distance squared from that point to (1,0,1) is
\[
f(x,y)=1 - x)^2 + y^2 + (1 - (1 - x - y))^2=1 - 2 x + 2 x^2 + 2 x y + 2 y^2\\
\]

Answer 2

You can show that this function has a minimum at x=2/3 and y=-1/3.
Hence the point (2/3,-1/3,-1/3) is the closest to your point.
What is the distance between these two points?

Answer 3

You compute the distance and you find it equal to \(\sqrt 2\)