Find the distance between (1,0,0,-2) in Euclidean space $\displaystyle R^4$ and a surface L:$\displaystyle \left\{\begin{array}{cc}x+y+z-w=0&\mbox{}
\\2x-y+w=0 & \mbox{}\end{array}\right.$
Thanks to anyone able to solve it.
Find the distance between (1,0,0,-2) in Euclidean space $\displaystyle R^4$ and a surface L:$\displaystyle \left\{\begin{array}{cc}x+y+z-w=0&\mbox{}
\\2x-y+w=0 & \mbox{}\end{array}\right.$
Thanks to anyone able to solve it.
Your "surface" is a (two-dimensional) vectorial plane. Find an orthonormal basis $\displaystyle (u,v)$ of $\displaystyle L$ (i.e. find two non-colinear vectors in $\displaystyle L$ and orthonormalize them), then the projection of $\displaystyle x=(1,0,0,-2)$ on $\displaystyle L$ is $\displaystyle \pi_L(x)=(x,u)u+(x,v)v$ (dot products) and thus by "Pythagoras theorem" $\displaystyle {\rm dist}(x,L)^2=\|x\|^2-\|\pi_L(x)\|^2=\|x\|^2-(x,u)^2-(x,v)^2$. I find $\displaystyle 3/2$ but I may very well be wrong.