Let $V=P(t)$ be the space of all polynomials in t with degree 2 or less, and having real numbers as coefficients. Define an inner product on $V$ by $$\langle f,g\rangle = \int_{0}^{1} f(t)g(t) dt$$ Give a basis for the subspace of $V$ orthogonal to $h(t)=2t+1$ under this inner product.
I am completely lost on how to do this.
We can find such vectors by taking an arbitrary vector $p(t) = at^2 + bt + c \in W$. What conditions does this impose on $a, b, c$? Well, \begin{align*} 0 &= \int_0^1 p(t)(2t + 1) \,\mathrm{d}t = \int_0^1(at^2 + bt + c)(2t + 1) \,\mathrm{d}t \\ &= \int_0^1 2at^3 + (2b + a)t^2 + (2c+b)t + c \,\mathrm{d}t \\ &= \frac{a}{2} + \frac{2b + a}{3} + \frac{2c+b}{2} + c \\ &= \frac{5}{6}a + \frac{7}{6}b + 2c \\ \end{align*} Solving for $c$, we get, $$c = -\frac{5}{12}a - \frac{7}{12}b,$$ so our vector takes the form, $$p(t) = at^2 + bt - \frac{5}{12}a - \frac{7}{12}b = a\left(t^2 - \frac{5}{12}\right) + b\left(t - \frac{7}{12}\right).$$ So, as we've shown, $W$ is contained in the span of two polynomials: $t^2 - \frac{5}{12}$ and $t - \frac{7}{12}$. If you show that both of these polynomials belong to $W$, then you have a spanning set of $W$. It's also clear that these are linearly independent, as they are not multiples of each other (since they're of different degrees). Hence, they form a basis for $W$.