0
$\begingroup$

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,

$\endgroup$
1
$\begingroup$

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, $$45=7k^2$$ $$\frac{45}7 = k^2$$ $$k = \pm\sqrt{\frac{45}7} = \pm3\sqrt{\frac{5}7}$$

$\endgroup$
3
  • $\begingroup$ 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
2
$\begingroup$

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

Squaring both sides

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.