A tetrahedron has vertices `O(0,0,0),A(1,2,1),B(2,1,3),a n dC(-1,1,2),` then angle between face `O A Ba n dA B C` will be a. `cos^(-1)((17)/(31))` b. `30^0` c. `90^0` d. `cos^(-1)((19)/(35))`

Options

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

Answer

d

Explanation

Vector perpendicular to the face OAB is
`vec(OA)xxvec(OB)=(hati+2hatj+hatk)xx(2hati+hatj+3hatk)`
`=5hati-hatj-3hatk`
Vector perpendicular to face ABC is
`vec(AB)xxvecAC=(hati-hatj+2hatk)xx(-2hati-hatj+hatk)`
`=hati-5hatj-3hatk`
Since the angle between the face = angle between their normal, therefore
`costheta=(5+5+9)/(sqrt35sqrt35)=(19)/(35)ortheta=cos^(-1)((19)/(35))`