Solving for k in 3D Plane Intersection at 60 Degrees

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To find the value of k for the plane x + ky + 2z - 9 = 0, which intersects at a 60-degree angle with the plane 2x + 2y - z = 0, one must first determine the normal vectors of both planes. The angle between the planes corresponds to the angle between these normal vectors, which can be calculated using the dot product. The cosine of the angle can be derived from the dot product formula, but it may be necessary to prove that the angle between the planes is indeed the angle between their normals. Additionally, there is concern about whether solutions are being prematurely provided without confirmation from the original poster. Understanding these concepts is crucial for solving the problem effectively.
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Homework Statement


Given the plane x + ky + 2z - 9 = 0, find k if the plane makes an angle of 60 degrees with the plane 2x + 2y - z = 0.


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The Attempt at a Solution


I'm stumped on this question. No idea of even how to start this one.
 
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can you find the angle between two vectors?
 
First find the normal vector for each plane- that should be easy. Then use the dot product to find the cosine of those two vectors.
 
as hallsofIvy mentioned, the angle between two planes is the angle between their normals. Using the dot product will give you the answer,

BUT

your teacher might require you to actually prove that the angle between two planes is the angle between their normals. to prove this, just study the quadrilateral formed by the two planes and the two normals.
 
Just wondering... Are we now just giving out the solutions? We haven't even heard back from the OP.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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