Given two planes find the value of k such that the angle between two planes is ${\frac\pi3}$

Question

How can we determine the coefficients of the components of a plane when given the angle?

I am given two planes $$\Pi_1: 2x-y+2z=5$$
$$\Pi_2: x+2y+kz=3$$

and asked to find the value of k such that the angle between the planes is
${\frac\pi3}$.

I know how to find the angle when given planes (with all of the coefficients) but I can't find any examples of how to find k when given the angle.

I think I'm supposed to use the equation $cos\theta=\frac{\vec{n_1}\cdot\vec{n_2}}{||{\vec{n_1}||||\vec{n_2}}||}$

$$\theta={\frac\pi3}$$

The normal vectors are $\vec{n_1}=<2,-1,2>$, $\vec{n_2}=<1,2,k>$

The dot product between the two normal vectors is ${\vec{n_1}\cdot\vec{n_2}}=2k$

The magnitudes are

$||\vec{n_1}||=3$

$||\vec{n_2}||=\sqrt{5+k^2}$

So if I put all of it together I have $$\frac\pi3=\cos^{-1}\frac{2k}{3\sqrt{5+k^2}}$$
And I'm stuck here, I'm not even sure that the work above is what I was supposed to do, but if it is correct where should I go from here?

Any advice would be greatly appreciated.

Thank you,

Answer 1

$$\frac{2k}{3\sqrt{5+k^2}}=\cos\frac\pi3=\frac{1}{2}$$

Squaring both sides