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.


1 Answer 1


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

  • $\begingroup$ Thanks@David Quinn for your help $\endgroup$
    – A. Khan
    Sep 11, 2016 at 4:11
  • 1
    $\begingroup$ You can accept the answer and close the question if this answer is what you are looking for. $\endgroup$ Sep 11, 2016 at 6:34

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