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If unit vectors ` vec a` and ` vec b` are inclined at angle `2theta` such that `| vec a- vec b|<1a n d0lt=thetalt=pi ,t h e n` `theta` lies in interval a.`[0,pi//6)` b. `(5pi//6,pi]` c. `[pi//6,pi//2]` d. `[pi//2,5pi//6]`

A

`[0, pi//6)`

B

`(5 pi//6, pi]`

C

`[pi//6, pi//2]`

D

`(pi//2, 5pi//6]`

Text Solution

Verified by Experts

The correct Answer is:
a,b
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]`
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