Find the angle between the vector $x=(2,2,1,1)$ and the space formed by the linear combination of the vectors $a_1 = (3,4,-4,-1)$ and $a_2=(0,1,-1,2)$.
I found the vector $y=(3,1,3,1)$ that is perpendicular to both $a_1$ and $a_2$. Then found the angle between $x$ and $y$: $$\varphi = \arccos\frac{\langle x,y \rangle}{|x||y|}=\arccos\frac{3\sqrt 2}{5}$$ As the answer I took $\pi/2 - \varphi$. Is this correct?
Update: I found this duplicate but could you please help me understand the difference between my solution and that in the link I provided? Why should we find an orthonormal basis as the accepted answer states? It would be appreciated.
EDIT (my original answer was incorrect)
To see why your solution is not correct, I will consider a simpler example. Let $a_1 , a_2$ be such that $span\{ a_1 , a_2 \} = span \{ (1,0,0,0) , (0,1,0,0) \}$. Let $x=(0,0,1,0)$. Then any vector that is perpendicular to combinations of $a_1 , a_2$ is of the form $y = (0,0, y_3 , y_4)$. I will show that your solution can lead to different answers:
If you found that $y = (0,0,1,0)$, then $\cos \varphi = \langle y,x\rangle /(\|y \| \| x \|)=1$
If you found that $y = (0,0,1,1)$, then $\cos \varphi = \langle y,x\rangle /(\|y \| \| x \|)=1/\sqrt{2}$
To see why the linked answer is correct, notice that they simply compute the orthogonal projection of $x$ onto $span \{ a_1 , a_2 \}$. Denote this projection $y$. Then the angle between $x$ and $y$ is exactly what you are looking for. Note that in order to compute the projection, you need an orthonormal basis (this is how projection is defined)