If `hat(i)+hat(j), hat(j)+hat(k), hat(i)+hat(k)` are the position vectors of the vertices of a `Delta ABC` taken in order, then `angleA` is equal to

Options

A. `pi/2`
B. `pi/5`
C. `pi/6`
D. `pi/3`

Answer

D

Explanation

Let position vector of the vertices are
`OA = hati +hatj , OB = hatj + hatk` and `OC = hati + hatk`
Now, `AB = - hati + hatk` and `AC = hatk - hati`
`:.cos theta = ((AB).(AC))/(|AB||AC|) = ((-hati + hatk).(hatk -hatj))/(sqrt(1^(2) + 1^(2))sqrt(1^(2) +1^(2))) = (1)/(sqrt(2)sqrt(2)) = 1/2`
`rArr theta= (pi)/(3)`