I was doing a linear algebra assignment on vectors but I came across with some doubts. One of the problems says Find the module of a vector $\vec{w}$ if $\vec{c} =(-2;3)$ is perpendicular to $\vec{a}$ , $\langle \vec{a} +2 \vec{w}$ , $\vec{c}\rangle =2 \sqrt{13}$ and the angle formed by $\vec{w}$ and $\vec{c}$ is $60^\circ$
I don't know how to do it because it seems like there are many unknowns. I know that the product between two perpendicular vectors is equal to $0$. But how can I solve it?
Thanks
As you mentioned, the geometric angle between vectors (in Euclidean space, at least) can be related to their inner product.
The gist of the solution is to use the identities
\begin{align} \langle\vec{u}, \vec{v}\rangle &= uv\cos(\theta) \\ \langle\vec{u}, \vec{v}\rangle &= u_1 v_1 + u_2 v_2 \\ \langle\vec{u} + \vec{v}, \vec{w}\rangle &= \langle\vec{u},\vec{w}\rangle + \langle\vec{v},\vec{w}\rangle \end{align}
in order to turn the given information
\begin{align} \langle \vec{a},\vec{c}\rangle &= 0 \\ \langle \vec{a} + 2\vec{w}, \vec{c} \rangle &= 2\sqrt{13} \\ \langle \vec{w}, \vec{c} \rangle &= wc\cos(60^{\circ}) \end{align}
into an expression for $\vec{c}$ in terms of your basis for $\mathbb{R}^2$.