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Question

A tetrahedron has vertices at O(0,0,0),A(1,2,1),B(2,1,3)&C(1,1,2). Then the angle between the faces OAB&ABC will be?


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Solution

Finding the angle between the faces OAB&ABC :

Given that tetrahedron has vertices at O(0,0,0),A(1,2,1),B(2,1,3)&C(1,1,2).

Vector perpendicular to OAB,R1 =OAxOB

=ijk121213

R1=5ij3k

Vector perpendicular to ABCisR2=ABxAC

=ijk1-12-2-11

=i5j3k

angle between the faces OAB&ABC will be:

Cosθ=R1/R2/[|R1||R2|]=1935θ=cos-1(1935)

Hence, the angle between the faces OAB&ABC is θ=cos-1(1935).


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