# Prove that the equation of two planes inclined at an angle $\alpha$ to x-y plane

Prove that the equation of two planes inclined at an angle $\alpha$ to x-y plane and containing the line $y=0, z \cos\beta=x\sin\beta$ is $~~(x^2+y^2) \tan^2\beta+z^2-2zx~\tan\beta=y^2\tan^2\alpha$.

My approach: Let $l_1 x+m_1y+n_1z=d_1$ and $l_2 x+m_2y+n_2z=d_2$ be the two planes containing the lines formed by intersection of two planes then angle between direction ratios of planes and lines will be 90 degree.

Since the plane contains the line which passes through the origin, we can write the equation of the plane as $$Ax+By+z=0$$

It contains the line described, so $$A\cos\beta+\sin\beta=0$$

Meanwhile, using scalar product, the the angle $\alpha$ satisfies $$\cos\alpha=\frac{1}{\sqrt{A^2+B^2+1}}=\frac{1}{\sqrt{B^2+\sec^2\beta}}$$

Rearranging this gives $$B=\pm\sqrt{\sec^2\alpha-\sec^2\beta}$$

Substituting in the values of $A$ and $B$ into the plane equation gives

$$-x\tan\beta\pm y\sqrt{\sec^2\alpha-\sec^2\beta}+z=0$$

You can now rearrange this so that the square root gets squared, and required result follows almost immediately.

I hope this helps

• Thanks@David Quinn for your help Sep 11, 2016 at 4:11
• You can accept the answer and close the question if this answer is what you are looking for. Sep 11, 2016 at 6:34