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 $cos\theta=\frac{\vec{n_1}\cdot\vec{n_2}}{||{\vec{n_1}||||\vec{n_2}}||}$


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



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,


You're correct so far, all that is left is to solve for $k$. $$cos(\frac{\pi}3) = \frac{2k}{3\sqrt{5+k^2}}$$ $$\frac{1}2 = \frac{2k}{3\sqrt{5+k^2}}$$ Squaring both sides, $$\frac{1}4 = \frac{4k^2}{9(5+k^2)}$$ $$\frac{1}4 = \frac{4k^2}{45+9k^2}$$ Multiplying both sides by $45+9k^2$, $$\frac{45+9k^2}4 = 4k^2$$ Multiplying both sides by $4$, $$45+9k^2 = 16k^2$$ Simplifying, $$45=7k^2$$ $$\frac{45}7 = k^2$$ $$k = \pm\sqrt{\frac{45}7} = \pm3\sqrt{\frac{5}7}$$

  • $\begingroup$ The angle between two planes isn't in necessarily the angle between their normals... $\endgroup$ – Jean Marie May 8 '19 at 2:55
  • 1
    $\begingroup$ Any reason for rejecting the negative root? $\endgroup$ – Shubham Johri May 8 '19 at 2:59
  • $\begingroup$ Thanks, forgot to add the possibility of a negative $\endgroup$ – nsnave May 8 '19 at 14:14


Squaring both sides


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