The position vectors of the points A, B, and C are `hati + hatj + hatk.hati+ 5hatj-hatk` and `2hati + 3hatj + 5hatk`, respectively . The greatest angle of triangle ABC is -

Options

A. `120^(@)`
B. `90^(@)`
C. `cos^(-1)(3//4)`
D. None of these

Answer

2

Explanation

`{:("A"(veca = hati + hatj + hatk)),("B"(vecb = hati + 5hatj +hatk)),("C"(vecc = 2hati+ 3hatj+ hatk)):}]`Position vectors

`rArr a= BC = |vec(BC)|`
`= |vec(OC)- vec(OB)|=sqrt(41)`
`rArr b= CA = |vec(CA)| = |vec(OA)-vec(OC)|= sqrt(21)`
`rArr = AB=|vec(AB)| = |vec(OB)-vec(OA)|=sqrt(20)`
`cosA = (b^(2) + c^(2)-a^(2))/(2ab)=(21+ 20-41)/(2sqrt(21)sqrt(20))=40`
`therefore /_A=(pi)/(2)`