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A tetrahedron has vertices O (0,0,0), A(...

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

A

`cos^(-1).((7)/(31))`

B

`cos^(-1).((9)/(35))`

C

`cos^(-1).((19)/(35))`

D

`cos^(-1).((17)/(31))`

Text Solution

Verified by Experts

The correct Answer is:
C
The given vertices of tetrahedron PQRO are P(1,2,1)
`Q (2,1,3) ,R (-1,1,2) " and " O(0,0,0) `
The normal vector to the face OPQ
`= OP xx OQ =(hat(i) + 2hat(i) + hat(k) )xx (2hat(i) + hat(j) + 3hat(k))`
`= |{:(hat(i) ,,hat(j) ,,hat(k) ),(1,,2,,1),(-2,,-1,,1):}|`
`=hat(i) (-1+2) -hat(j) (1+4) + hat(k)`
Now the angle between the faces OPQ and PQR is the angle between their normals
`= " cos"^(-1) (|5+5+9|)/(sqrt(25+1+9)sqrt(1+25+9)) = "cos "^(-1) ((19)/(35))`
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