If `hati, hatj, hatk` represent unit vectors along the X,Y, Z axes respectively, then the angle between the vectors `hati+hatj + hatk and hati+hatj` is equal to

Options

A. `90^(@)`
B. `sin^(-1)""(sqrt((2)/(3)))`
C. `cos^(-1)""((1)/(sqrt(3)))`
D. `sin^(-1)""((1)/(sqrt(3)))`

Answer

D

Explanation

Let `vec A = hati + hatj + hatk`
The magnitude of `hati + hatj + hatk = sqrt (1+1+1)= sqrt (3)" or "A= sqrt (3) and vec B = hati + hatj`
`therefore B = sqrt (1^(2)+1^(2)) = sqrt (2) and AB = sqrt (3) sqrt (2) and vec A cdot vec B = (hati + hatj + hatk) cdot (hati + hatj) =1+1=2`
The angle between `vec A and vec B` is given by `cos theta =(vec A cdot vec B)/(AB) =(2)/(sqrt (3) sqrt(2))=sqrt((2)/(3)) and sin theta = sqrt (1 - cos^(2) theta)= sqrt (1-(2)/(3))= sqrt ((1)/(3)) =(1)/(sqrt(3))`
`therefore theta = sin^(-1) ""((1)/(sqrt(3)))`