A tetrahedron has vertices O (0,0,0), A(1,2,1,), B(2,1,3) and C(-1,1,2), the angle between faces OAB and ABC will be

Options

A. `90^(@)`
B. `cos^(-1)(19/35)`
C. `cos^(-1)(17/31)`
D. `30^(@)`

Answer

B

Explanation

Let `vecn_(1)` and `vecn_(2)` be the vectors normal to the faces `OAB` and `ABC`. Then `
`vecn_(1)=vec(OA)xxvec(OB)=|(hati,hatj,hatk),(1,2,1),(2,1,3)|=5hati-hatj-3hatk`
and
`vecn_(2)=vec(AB)xxvec(AC)=|(hati,hatj,hatk),(1,-1,2),(-2,-1,1)|=hati-5hatj-3hatk`
If `theta` is the angle between the faces OAB and ABC, then
`cos theta=(vecn_(1).vecn_(2))/(|vecn_(1)||vecn_(2)|)`
`impliescos theta=(5+5+9)/(sqrt(251+9)sqrt(1+25+9))=19/35impliestheta=cos^(-1)(19/35)`