Finding value of $m$ using vector dot product Given the acute angle between the vectors  $\textbf{a}=\binom{m}{0},\:\textbf{b}=\binom{1}{m}$ is thirty degrees, find the possible value of $m$ if $m$ is real.
A simple dot product gives
$$\frac{\sqrt{3}}{2}=\frac{1}{m^2+1}$$
which I solved to be $\pm \frac{1}{\sqrt{3}}$. By inspection, the negative solution is impossible since there is no acute angle. Then what is the other positive value of $m$? It seems too be the reciprocal, but I don't get how this pops out algebraically.
My intuition tells me it have something to do with the complex conjugate theorem...
EDIT: The other supposed root is $\sqrt{3}$
 A: we have $\vec{a}.b=||\vec{a}||*||\vec{b}|| \cos{\theta}$
so $\cos{\theta}=\frac{\vec{a}.\vec{b}}{||\vec{a}||*||\vec{b}||}$
$\vec{a}.\vec{b}=m*1+0*m=m$
$||\vec{a}||=m  , ||\vec{b}||=\sqrt{m^2+1}$
then $\cos{30^{\circ}}=\frac{\sqrt3}{2}=\frac{m}{m\sqrt{m^2+1}}=\frac{1}{\sqrt{m^2+1}}$
$\frac{1}{\sqrt{m^2+1}}=\frac{\sqrt3}{2}=\frac{1}{\sqrt\frac{4}{3}}$
so $m^2+1=\frac{4}{3}$
so $m=\pm \frac{1}{\sqrt3}$
so your solution is right and as you mentioned there is really  problem with the negative value because the angle will not be 30 degrees...but we don't have another positive solution because we obtain two solutions and exclude the negative one because it doesn't satisfy angle 30 degrees :


Also you said that that $\sqrt3$ is solution but it is not because:
if $m=\sqrt3$ then $\vec{a}=\binom{\sqrt3}{0}$ and $\vec{b}=\binom{1}{\sqrt3}$
then $\cos{\theta}=\frac{\vec{a}.\vec{b}}{||\vec{a}||*||\vec{b}||}=\frac{\sqrt3*1+0*\sqrt3}{(\sqrt{(3+0)})\sqrt{(1+3)}}=\frac{\sqrt3}{(\sqrt3)(\sqrt4)}=\frac{1}{2}$
so $\theta =60^{\circ}$ not $30^\circ$
A: Let's look at this geometrically. Here is a graph of the Cartesian coordinate plane with some relevant vectors plotted on it.
Also note the line $x = 1$ parallel to the $y$-axis.

The vector $\mathbf a = \begin{pmatrix}m \\ 0\end{pmatrix}$
is a vector parallel to the $x$ axis, so it must have the same direction as one of the red vectors in the figure.
The vector $\mathbf b = \begin{pmatrix}1 \\ m\end{pmatrix}$ must have the same direction and length as the direction and distance from the origin to one of the points on the line $x = 1.$
In the case where $m < 0$ the vector $\mathbf a$ must point in the same direction as the vector $A_2,$ and the only possible vectors that make $30$-degree
angles to $\mathbf a$ are the vectors marked $B_3$ and $B_4$ and vectors that are positive scalar multiples of those vectors.
But there is no positive scalar multiple of either vector that will give you the direction and distance from the origin to a point on the line $x = 1.$
So neither of these two vectors is suitable. There is no solution for $m < 0.$
In the case where $m > 0$ the vector $\mathbf a$ must point in the same direction as the vector $A_1,$ and the only possible vectors that make $30$-degree
angles to $\mathbf a$ are the vectors marked $B_1$ and $B_2$ and vectors that are positive scalar multiples of those vectors.
Either vector can be scaled to give the direction and distance from the origin to a point on the line $x = 1,$ but in the case of $B_2$ we see that the vector's
$y$-coordinate is negative, whereas the $y$-coordinate of $\mathrm b$ is $m,$ which in this case is positive.
So no solution can put $\mathrm b$ in the direction of $B_2$.
This leaves $B_1$ as the only possible solution.
And indeed if we scale $B_1$ so that we get the distance from the origin to
the line $x = 1$ in that direction, we get the vector
$\mathbf b = \begin{pmatrix}1 \\ 1/\sqrt{3}\end{pmatrix}$.
There is no other solution.
The vector $\begin{pmatrix}1 \\ \sqrt{3}\end{pmatrix}$ that you would obtain from the supposed "other root" $m = \sqrt3$ is the vector $V$ in the graph.
It makes a $30$-degree angle with the $y$-axis, not with any vector parallel to the $x$-axis. I don't know who came up with that "solution" or how, but it is not a valid solution to the problem as asked.
