Let $P_2[x]$ be the space of polynomials of degree less than or equal to 2. If $W = \{p ∈ P_2[x] \mid p(2) = p(1)\}$, then find a basis for $ W^⊥$ where $P_2[x]$ is equipped with an inner product given by $\langle p, q \rangle= \int^1_0p(x)q(x)dx$
My solution: If $ax^2+bx+c \in P$ with $a,b,c \in \mathbb{R}$ I have calculated the integrals \begin{align*} \frac{-11a}{20}+\frac{-3b}{4}+\frac{-7c}{6} &= 0\\ \frac{a}{3}+\frac{b}{4}+c &= 0 \end{align*}
Then what should I do?
Hint: Start by calculating a basis for $W$. I assume you can do this so lets just take $p_1$ and $p_2$ to be the basis elements that you've found (yes, there will be two of them).
Now take the polynomial $f = ax^2 + bx + c$. You want to figure out conditions on $a, b, c$ such that $f \in W^\perp$. This will allow you to describe $W^\perp$ as a subspace and then you can find a basis for it just like you found a basis for $W$.
So, since $p_1$ and $p_2$ are a basis for $W$, we have $f \in W^\perp$ if and only if $\langle f, p_1\rangle = 0$ and $\langle f, p_2\rangle = 0$. So compute those two integrals and set them equal to zero. This gives you a system of two equations in the variables $a, b, c$. Solve those equations to get the conditions that give $f \in W^\perp$.