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I need help understanding the formula for intersection between 2 planes

Where is $\alpha$ actually, its not clear in the diagram. It appears that I could just interpret it as the below?

Where $\beta = \theta$?

UPDATE

Now given the question:

What I did

Whats wrong? Correct answer $60.7^{\circ}$. Did I "expanded" the equation for plane whose equation was in Cartesian form correctly?

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The diagram is somewhat confusing, but: remember that two sets of angles are formed when two non-perpendicular lines cross. You have one set of obtuse "vertical angles", and you have one set of acute "vertical angles". The acute and obtuse angles are complementary; that is, letting $\theta$ be the acute angle and $\alpha$ be the obtuse one, $\alpha+\theta=\pi$. –  J. M. Nov 14 '11 at 2:46

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Maybe it's best to ignore the diagram from your book. Just consider two planes with normal vectors ${\bf n}_1$ and ${\bf n}_2$. Let's suppose that the angle between these vectors is $\alpha$ (which is acute). Then that's the angle between the two planes.

Now what if our first plane had had the normal vector $-{\bf n}_1$? Then the angle between our two normals would have been $\theta$. This angle is obtuse. So it should be replaced by $\alpha = \pi - \theta$.

Anytime you are computing the angle between two planes (which is done by computing the angle between their normal vectors), you should make sure your answer is an acute angle. If your initial answer is obtuse, you've used normals which point out of "incompatible" sides of the planes. Swapping one of the normals for its negative will change the angle to its complement and then give you an acute angle (the desired answer).

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