Orthogonal complement of subspace $W = span(5,1+t)$ I have this subspace of $P_2(\mathbb R)$ and I need to find its orthogonal complemente, using the inner product defined as $$<p(t),q(t)> = \int_o^1 p(t)q(t) dt$$
So I'm assuming the vector $$v = a+bt+ct^2$$ as being the vector such that $$<v,5> = 0\\<v, 1+t> = 0$$
So I did:
$$\int_0^1 (a+bt+ct^2)5dt = 0 \implies \frac{5}{6}(6 a+3 b+2 c) = 0\\\int_0^1 (a+bt+ct^2)(1+t)dt = 0 \implies \frac{1}{12} (18 a+10 b+7 c) = 0$$
So I have two equations:
$$6 a+3 b+2 c=0\\18a+10 b+7 c=0$$
Does it means that my subspace is spanned by what? Do I have to choose $c$ to be a free variable? How do I represent my orthogonal complemente as a subspace? All the $t$'s are gone D: 
Thak you so much!
Update:
By fixing c, I've found:
$$a = a, b = -6a$$
Then my polynomial should be:
$$a-6at+ct^2$$
But I don't know what to do for $c$. Could somebody help me?
Update 2:
I've managed to solve for $c$ once I knew $a$ and $b$, so I got: $c=6a$. Then my polynomial should be:
$$a-6at+6at^2$$
But my answer is:
$$-bt^2/6+bt-b$$

Your computations are correct. Now you want to find one nonzero solution of
$$
\begin{cases}
6a+3b+2c=0\\
18a+10b+7c=0
\end{cases}
$$
Multiply the first equation by $3$ and subtract it from the second, getting
$$
b+c=0
$$
so $b=-c$; then $6a-3c+2c=0$ or $6a=c$. Thus you get a nonzero solution by taking $c=1$ (or any nonzero number). The polynomial you're looking for is $\dfrac{1}{6}-t+t^2$ (or any scalar multiple thereof).
Why just one? The subspace you want the orthogonal complement of has dimension $2$, so the orthogonal complement has dimension $1$; hence a single nonzero vector spans it.
