Find the dihedral angle

Geometry Level pending

As shown above, in a 3D-rectangular coordinate system system:

A ( 0 , 0 , 0 ) , B ( 1 , 0 , 0 ) , C ( 1 , 2 , 0 ) , D ( 0 , 1 , 0 ) , E ( 0 , 0 , 2 ) , F ( 1 , 2 , 1 ) A(0,0,0), B(1,0,0), C(1,2,0), D(0,1,0), E(0,0,2), F(1,2,1) .

Then find the dihedral angle E B D F E-BD-F . Submit the angle in degrees and round to the hundredth.


The answer is 74.21.

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1 solution

One of the simplest problems that you post :P

B D = j ^ i ^ , B E = 2 k ^ i ^ , B F = 2 j ^ + k ^ \vec {BD}=\hat j-\hat i, \vec {BE}=2\hat k-\hat i, \vec {BF}=2\hat j+\hat k .

The required dihedral angle is

cos 1 ( ( B D × B E ) . ( B D × B F ) B D × B E B D × B F ) = cos 1 ( 2 2 + 2 3 × 6 ) = cos 1 ( 2 3 6 ) 74.20683 ° \cos^{-1} \left (\dfrac {(\vec {BD}\times \vec {BE}).(\vec {BD}\times \vec {BF})}{|\vec {BD}\times \vec {BE}||\vec {BD}\times \vec {BF}|}\right ) =\cos^{-1}\left (\dfrac{2-2+2}{3\times \sqrt 6}\right ) =\cos^{-1}\left (\dfrac{2}{3\sqrt 6}\right ) \approx \boxed {74.20683\degree} .

Yeah. I sometime post simple problems, but my formula leads me to the trouble of deciding whether it is acute angle or obtuse angle. (My formula is find their normal vectors seperately, then find the angle of the two normal vectors :) )

Alice Smith - 1 year, 1 month ago

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