Find the area of the triangle with vertices A(1,1,2)B(2,3,5) and C(1,5,5).

Explanation

The vertices of triangle ABC are given as A (1,1,2) , B (2,3,5) and C (1,5,5).
The adjacent sides `vec(AB) and vec(BC) of triangleABC` are given as
`vec(AB)=(2-1)hati+(3-1)hatj+(5-2)hatk=hati+2hatj + 3hatk`
`vec(BC)=(1-2)hati+(5+3)hatj+(5-5)hatk=-hati+2hatj`
`Area of triangleABC=1/2|vec(AB)xxvec(BC)|`
`vec(AB)xxvec(BC)=|{:(hati,hatj,hatk),(1,2,3),(-1,2,0):}|=hati(-6)-hatj(3)+hatk(2+2)=-6hati-3hatj+4hatk`
`|vec(AB)xxvec(BC)|=sqrt((-6)^(2)+(-3)^(2)+4^(2))=sqrt(36+9+16)=sqrt61`
Hence , the area of `triangle is sqrt61/2` square units.