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1)

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


A) cos1(12)

B) cos1(16)

C) cos1(13)

D) cos1(14)

Answer:

Option C

Explanation:

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 Vector perpendicular  to face OAB=  n1

= OA×OB

 =  (ˆi2ˆj+ˆk)×(2ˆi+ˆj+ˆk)

=(21)ˆi(21)ˆj+(14)ˆk)

= 3ˆi3ˆj3ˆk

 Vector perpendicular to face ABC=  n2.

= AB×AC

= (3ˆi+3ˆj)×(ˆj+ˆk)

=   3ˆi+3ˆj3ˆk

 Since,  angle between faces is equal to angle between their normals.

  cosθ=n1.n2|n1||n2|

 =(3)(3)+(3)(3)+(3)(3)9+9+99+9+9

=99+92727=13

θ=cos1(13)