Your basis is incorrect. The subspace $A$ has dimension $2$, not $1$, so you need to start with two basis vectors, not one.
In addition, your vector $B$ is not even in the subspace: if you plug in $x_1=1$, $x_2=-4$, $x_3=-1$ into $x_1-4x_2-x_3$, you do not get $0$.
So, first write out the solutions to $x_1-4x_2-x_3=0$, e.g., in parametric form. This will give you the two basis vectors. Then apply Gram-Schmidt to the basis you got.
(In the special case where the vector space is $1$-dimensional, the orthogonalization part of Gram-Schmidt is empty, so all you need to do is normalize a single basis vector).
The choice of inner product defines the notion of orthogonality.
The usual notion of being "perpendicular" depends on the notion of "angle" which turns out to depend on the notion of "dot product".
If you change the way we measure the "dot product" to give a more general inner product then we change what we mean by "angle", and so have a new notion of being "perpendicular", which in general we call orthogonality.
So when you apply the Gram-Schmidt procedure to these vectors you will NOT necessarily get vectors that are perpendicular in the usual sense (their dot product might not be $0$).
Let's apply the procedure.
It says that to get an orthogonal basis we start with one of the vectors, say $u_1 = (-1,1,0)$ as the first element of our new basis.
Then we do the following calculation to get the second vector in our new basis:
$u_2 = v_2 - \frac{\langle v_2, u_1\rangle}{\langle u_1, u_1\rangle} u_1$
where $v_2 = (-1,1,2)$.
Now $\langle v_2, u_1\rangle = 3$ and $\langle u_1, u_1\rangle = 3$ so that we are given:
$u_2 = v_2 - u_1 = (0,0,2)$.
So your basis is correct. Let's check that these vectors are indeed orthogonal. Remember, this is with respect to our new inner product. We find that:
$\langle u_1, u_2\rangle = 3(-1)(0) + (1)(0) + 2(0)(2) = 0$
(here we also happened to get a basis that is perpendicular in the traditional sense, this was lucky).
Now is the basis orthonormal? (in other words, are these unit vectors?). No they arent, so to get an orthonormal basis we must divide each by its length. Now this is not the length in the usual sense of the word, because yet again this is something that depends on the inner product you use. The usual Pythagorean way of finding the length of a vector is:
$||x||=\sqrt{x_1^2 + ... + x_n^2} = \sqrt{x . x}$
It is just the square root of the dot product with itself. So with more general inner products we can define a "length" via:
$||x|| = \sqrt{\langle x,x\rangle}$.
With this length we see that:
$||u_1|| = \sqrt{2(-1)(-1) + (1)(1) + 3(0)(0)} = \sqrt{3}$
$||u_2|| = \sqrt{2(0)(0) + (0)(0) + 3(2)(2)} = 2\sqrt{3}$
(notice how these are different to what you would usually get using the Pythagorean way).
Thus an orthonormal basis is given by:
$\{\frac{u_1}{||u_1||}, \frac{u_2}{||u_2||}\} = \{(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 0), (0,0,\frac{1}{\sqrt{3}})\}$
Best Answer
You have a hidden inner product in your phrase "best approximation". Presumably your best approximation is one that minimizes $\int_0^1(f(x)-p(x))^2w(x)dx$ where $w(x)$ is a weight function. Each choice of $w(x)$ gives a different set of orthogonal polynomials. If you choose $w(x)=1$ you get the Legendre polynomials. On the interval $[-1,1]$ the first three are $1,x,\frac 12(3x^2-1)$