A regular tetrahedron has vertices #A, B, C# and #D# with co-ordinates #(0,0,0,), (0,1,1), (1,1,0), "and" (1,0,1)# respectively. Show the angle between any two faces of the tetrahedron is #arccos(1/3)#?

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This is in a chapter of my textbook about the scalar product of vectors, and the angle between planes.

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Answer 1 · 2018-07-06T17:57:46

Please refer to The Explanation.

Recall that angle btwn. two planes #pi_1 and pi_2# is the

angle btwn. their resp. normals #n_1 and n_2#.

The tetrahedron #A-BCD# has #4# faces, namely,

#ABC, ABD, ACD and BCD#.

Let us find the angle #theta in (0,pi/2)# btwn. # pi_1=ABC and pi_2=BCD#.

Recall that, the normal #vecn_1# of #pi_1# is given by,

#vecn_1=vec(AB)xxvec(AC)#,

#=|(i,j,k),(0-0,1-0,1-0),(1-0,1-0,0-0)|#,

#=|(i,j,k),(0,1,1),(1,1,0)|#,

#=i(0-1)-j(0-1)+k(0-1)#.

# vecn_1=(-1,1,-1)#.

On the similar lines, we have,

#vecn_2=vec(BC)xxvec(BD)=(-1,-1,-1)#,

#:. costheta=|vecn_1*vecn_2|/{||vecn_1||*||vecn_2||}#,

#=|(-1)(-1)+(1)(-1)+(-1)(-1)|/[sqrt{(-1)^2+1^2+(-1)^2}sqrt(1+1+1)]=1/3#.

# theta=/_(pi_1,pi_2)=arccos(1/3)#, as desired!

The other angles can similarly be shown to be #arccos(1/3)#.

Enjoy Maths.!