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If unit vectors veca and vecb are inclin...

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

A

` [ 0,pi/6) cup ( 5pi //6,pi ]`

B

` [ 0,pi]`

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
we have
` |veca -vecb|^(2) = |veca |^(2) +|vecb|^(2) =-2 (veca.vecb)`
` Rigtharrow |veca -vecb|^(2) = |veca|^(2) +|vecb|^(2) -2|veca||vecb|cos 2 theta`
` Rightarrow |veca -vecb|^(2) =2-2 cos 2 theta " " [ |veca| =|vecb|=1]`
` Rightarrow |veca -vecb|^(2) =4 sin ^(2) theta`
` Rightarrow |veca -vecb| =2 |sin theta |`
now,
` |veca -vecb|lt 1`
` Rightarrow 2|sin theta| lt 1 Rightarrow |sin theta| lt 1/2 Rightarrow theta in [ 0, pi//6) cup ( 5pi//6,pi]`
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