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A line which goes through the point $(0, 0, 0)$ and a plane which has the equation $ 2x - y + 3z = 1$ have the angle of $45^{\circ}$ between them. I am supposed to find the equation of the line.

I know that the normal vector of the plane is $n=(2, -1, 3)$ but I get stuck here and don't know how to continue. Any help is appreciated..

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    $\begingroup$ It appears that there's not only one single line that meets your demands. $\endgroup$ – Michael Hoppe Jul 5 '17 at 15:07
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You just need a vector $\textbf{w}$ such that the angle between $n=(2,-1,3), \textbf{w}$ is $45^o$. Letting $\textbf{w}$ be a unit vector we have,

\begin{align} (2,-1,3) \cdot \textbf{w} = \sqrt{14} \cos \theta_{n, \textbf{w}} &= \sqrt{14} \cos(45^o)\\ & = \sqrt{14} \cdot \frac{\sqrt{2}}{2} \end{align}

If you let $\textbf{w} = (x,y,z)$ then you have $ 2x-y+3c = \sqrt{7}$. It should be easy to get a $\textbf{w}$ from here. So once you do that, define $L$ to be the line parametrized by $\alpha(t) = \textbf{w}t$. As the other user pointed out, once you have one vector, rotation about the axis made by the normal line to the plane, gives all the others.

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