If the position vectors of the vertices a, B and C of a `Triangle ABC` be `(1, 2, 3), (-1, 0, 0)` and `(0, 1, 2)` respectively then find `angleABC`.

Answer

`cos^(-1) (10/sqrt(102))`

Explanation

Let `angle ABC= theta`.
Now, `vec(BA)=("p.v. of A")-("p.v. of B")=(2 hat(i)+2hat(j)+3hat(k))`
and `vec(BC)=("p.v. of C")-("p.v. of B")=(hat(i)+hat(j)+2 hat(k))`.
`cos theta =(vec(BA). Vec(BC))/(|vec(BA)||vec(BC)|)= ((2xx1+2xx1+3xx2))/({sqrt(2^(2)+2^(2)+3^(3))}{sqrt(1^(2)+1^(2)+2^(2))})`
`=10/(sqrt(17)xxsqrt(6))=10/sqrt(102)`
`implies theta = cos^(-1) (10/sqrt(102))`.