Two unit vectors $\mathbf{x}$ and $\mathbf{y}$ in $\Bbb R^n$ satisfy $\mathbf{x}\cdot\mathbf{y}=\frac{\sqrt{2}}{2}$ in radians. How would I go about finding the angle between $\mathbf{x}$ and $\mathbf{y}$?
As I don't know the $\mathbf{x}$ and $\mathbf{y}$ unit vectors, would the unit circle be useful here? For instance, using $\frac{\sqrt{2}}{2}$ and plugging those values into $\dfrac{\mathbf{x}\cdot\mathbf{y}}{\mathbf{\|x\| \|y\|}}$ to find the angle?
On
The dotproduct between two vectors in euclidean space can be defined as $$\textbf{x}\cdot\textbf{y} = \cos\theta||\textbf{x}||||\textbf{y}||$$ Since you have unit vectors: $||\textbf{x}||,||\textbf{y}||=1$. The angle $\theta$ is then given by $$\theta = \arccos(\textbf{x}\cdot\textbf{y})=\arccos(\sqrt{2}/2)=\pi/4$$
No, I believe the unit circle is not really involved here.
It is simple. You already know the $cosinus$ of the angle $\theta$ between the two vectors. It is this expression:
$$cos(\theta) = \dfrac{\mathbf{x}\cdot\mathbf{y}}{\mathbf{||x||\cdot ||y||}}$$
Just plug in the numbers in this formula. Thus you get:
$$cos(\theta) = \frac{\sqrt{2}/2}{1.1}$$
And once you know that $a = cos(\theta) = \sqrt{2}/2$,
find $\theta = \arccos(a) = \arccos(\sqrt{2}/2) = \pi / 4$