I am trying to find a vector $\alpha$. I have the angle between the vector $\alpha$ and $\beta$ which is $\beta = \begin{pmatrix}0 \\1 \\0\end{pmatrix}$.
To find the angle between them I use this formula:
$\begin{equation*} \cos\theta = \frac{\beta\cdot\alpha}{\left|\beta \right|\left|\alpha \right|} \end{equation*}$
Both the magnitude of the vector $\beta$ and $\alpha$ is 1, so the formula is simplified to
$\begin{equation*} \cos\theta = \beta\cdot\alpha \end{equation*}$
Lets take the vector $\alpha = \begin{pmatrix}\alpha_x \\ \alpha_y \\ \alpha_z\end{pmatrix}$ and $\beta= \begin{pmatrix}\beta_x \\ \beta_y \\ \beta_z\end{pmatrix}$. I have found the $\alpha_y$ by using this formula
$\begin{equation*}\cos\theta = \alpha_x\beta_x + \alpha_y\beta_y + \alpha_z\beta_z\end{equation*}$
$\begin{equation*}\cos\theta = \alpha_x(0) + \alpha_y(1) + \alpha_z(0)\end{equation*}$
$\begin{equation*}\cos\theta = \alpha_y\end{equation*}$
I know the angle between them is 60 degrees which $\theta = 60^{\circ}$. So
$\begin{equation*}\cos(60)= \alpha_y = \frac{1}{2}\end{equation*}$
I tried to find $\alpha_x$ and $\alpha_z$ by using this formula
$\begin{equation*}\sqrt{\alpha_x^2+\alpha_y^2+\alpha_z^2} = 1\end{equation*}$
$\begin{equation*}\alpha_x^2+\alpha_y^2+\alpha_z^2 = 1\end{equation*}$
$\begin{equation*}\alpha_x^2+(\frac{1}{2})^2+\alpha_z^2 = 1\end{equation*}$
$\begin{equation*}\alpha_x^2+\frac{1}{4}+\alpha_z^2 = 1\end{equation*}$
$\begin{equation*}(\alpha_x+\alpha_z)^2 = 1 - \frac{1}{4}\end{equation*}$
$\begin{equation*}\alpha_x+\alpha_z = \frac{\sqrt{3}}{2}\end{equation*}$
If you simplify it again and again. In the end you are left with an unknown and I cant seem to figure out $\alpha_x$.
can I find the $\alpha_x$ and $\alpha_z$ component or is it impossible to find them?
I have found the answer. $\alpha = \begin{pmatrix}0.866 \\ 0.5 \\ 0\end{pmatrix}$. $\theta$ is between the vector $\alpha$ and the y axis. So to calculate the $\alpha_x$ is simply $\sin\theta$ which in turm makes the $\alpha_z = 0$