Find the distance from $(1, 1, 1)$ to the subspace span ${(1, 1, 0), (0, 1, 1)}$ of $\mathbb R^3$.

Question

Consider $\mathbb R ^3$ with the usual inner product. If d is the distance from $Q=(1, 1, 1)$ to the subspace $P$ span ${(1, 1, 0), (0, 1, 1)}$of $\mathbb R ^3$ , then $ 3d ^2 $ = ?

Can anyone give me a hint? I can not understand what to do with the inner product. I know that I will get an orthogonal vector of $(1, 1, 1)$ in the subspace span ${(1, 1, 0), (0, 1, 1)}$.

Answer 1

Note that the vector $(1, -1, 1)$ is orthogonal to both of these vectors. Now $$ (1, 1, 1) = \frac{2}{3}(1, 1, 0) + \frac{2}{3}(0, 1, 1) + \frac{1}{3}(1, -1, 1). $$ Hence the orthogonal projection is $\frac{1}{3}(1, -1, 1)$ and so $d$ is the length of this vector. Then $d^2 = 1/3$ and $3d^2 = 1$.

Answer 2

Project the vector onto the subspace. Subtract the result from $(1,1,1)$. Then compute the norm.

If you're familiar with Gram-Schmidt, it's basically the same.

First I need to get an orthogonal basis for $P$: $(1,1,0)-\frac12(0,1,1)=(1,\frac12,-\frac12)$.

So, we get $(1,1,1)-(0,1,1)-\frac23(1,\frac12, -\frac12) =(\frac13,-\frac13,\frac13) $.

Now the norm is $d=\sqrt{\frac13}$.

So $3d^2=1 $.