If the vertices A,B, C of a triangle ABC are (1,2,3),(-1, 0,0), (0, 1,2), respectively, then find `angle ABC`.

Explanation

The vertices of `triangleABC` are given as A(1,2,3) , B(-1,0,0), and C ( 0,1,2)
Also, it is given that `angle ABC` is the angle between the vectors `vec(BA) and vec(BC)` thus
`vec(BA)={1-(-1)}hati+(2-0)hatj+ (3-0)hatk`
`2hati+ 2hatj + 3hatk`
`vec(BC)={0-(-1)}hati+(1-0)hatj + (2-0)hatk`
`= hati + hatj + 2hatk`
`vec(BA).vec(BC)=(2hati+ 2hatj + 3hatk). (hati + hatj + 2hatk)`
`2xx1+2xx1+3xx2`
`= 2 + 2 +6 =10`
`|vec(BA)|=sqrt(2^(2)+2^(2)+3^(2))=sqrt(4+4+9)=sqrt17`
`|vec(BA)|=sqrt(1+1+2^(2))=sqrt6`
Now, it is known that :
`vec(BA). vec(BC)= |vec(BA)||vec(BC)|cos(angleABC)`
`10 =sqrt17 xx sqrt6 cos(angleABC)`
`or cos(angleABC) = 10/(sqrt17xxsqrt6)`
`or angleABC = cos^(-1)(10/sqrt102)`