Find a point on a plane in $\mathbb R^4$ closest to a point.

Question

Given a plane $P$ in $\mathbb R^4$ which includes the points $(0,0,0,0)$, $(1,0,−1,1),$ and $(0,1,1,2),$ find the point on the plane which is closest to the point $(−5,4,0,0). $

Getting the parametric equation of the plane in terms of vectors is pretty straightforward: $P(s,t) = <0,0,0,0> + s<1,0,-1,1> + t<0,1,1,2>.$ I'm having a hard time picturing a plane in $\mathbb R^4$ and how I can progress from this spot. Should I find a projection? I can't really find orthogonal vectors here, can I?

I'd appreciate some help!

Answer 1

HINT

$P(s,t) = <0,0,0,0> + s<1,0,-1,1> + t<0,1,1,2>=<s,t,t-s,2t+s>$

Now consider the function $$f(s,t)=||P(s,t)-<-5,4,0,0>||=||<s+5,t-4,t-s,2t+s>||$$ $$=\sqrt{(s+5)^2+(t-4)^2+(t-s)^2+(2t+s)^2}$$ where $||\bar{x}-\bar{y}||$ is the euclidean distance in $\mathbb{R}^4$

This function gives the distance of the plane from the point $<-5,4,0,0>$.

Can you minimize this function?

Answer 2

It is actually quite easy to find an orthornormal basis to your plane: Applying Gram-Schmidt to the given basis one obtains: $$u_1=\frac{1}{\sqrt{3}}(1,0,-1,1),\quad u_2=\frac{1}{\sqrt{51}}(-1,3,4,5)$$ The orthogonal projection of $(-5,4,0,0)$ onto the plane is therefore the vector $(-2,1,3,0)$ and the distance is $3\sqrt{3}$.