Consider the real Euclidean space $P(\mathbb R)$ the vectorial space of all polynomials with real coefficients with the following inner product: $$\langle p, q\rangle = \int_{-1}^{1}(1 + t^2)p(t)q(t)dt$$ Let's be sets: $$A = \{p ∈ P(R) | p(x) = 1 + ax, a ∈ \mathbb R\}, B = \{p ∈ P(R) | p(x) = −1 + x + bx^2 , b ∈ \mathbb R\}$$ Determine the polynomials pA ∈ A and pB ∈ B such that the distance between pA and pB is minimal.
My frst attempt was to do $\langle pA - pB, pA - pB\rangle = \int_{-1}^1(1+t^2)(2+(a-1)t-bt^2)^2dt = \frac{8}{105}(14a^2-28a+9b^2-56b+154)$
So to minimize this we need to take the parcial derivatives with respect to a and to b.
But I tried other way: I did the projection of pB in pA - 1 = pA' so that $||pB-pA|| = ||pB-1 - (pA-1)||$. So I found a orthonormal basis to pA', in this case $\frac{\sqrt{15}}{4}x$ such that $||pB-pA|| = \sqrt{||pB-1||^2 - ||proj_{pA'}(pB-1)||}$ And I found $$||pB-pA|| = \sqrt{\frac{24b^2}{35} - \frac{64b}{15}+\frac{176}{15}-\frac{4\sqrt{15}}{15}}$$
Why the two methods given different results? I think that they would be the same, right?
Any hint or suggestion will be apreciated.