If unit vectors `veca and vecb` are inclined at an angle `2 theta` such that `|veca - vecb| lt 1 and 0 le theta le pi`, then `theta` lies in the interval

Options

A. `[0, pi//6)`
B. `(5 pi//6, pi]`
C. `[pi//6, pi//2]`
D. `(pi//2, 5pi//6]`

Answer

a,b

Explanation

we have `|veca -vecb|^(2) = |veca|^(2) + |vecb|^(2) -2 (veca. Vecb)`
`or |vec-vecb|^(2) = |veca |^(2) + |vecb|^(2) -2 |veca||vecb|cos2theta`
` |veca -vecb|^(2) = 2 - 2cos theta " " (|veca|=|vecb|=1)`
` = 4sin^(2) theta `
` or |veca -vecb| =2 |sin theta|`
Now, `|veca- vecb| gt 1`
` Rightarrow 2|sin theta|lt 1`
` or |sin theta| lt 1/2 `
` Rightarrrow theta in [ 0, pi//6) or theta in ( 5 pi//6, pi]`