If the unit vectors ` veca and vecb ` are inclined of an angle ` 2 theta` such that ` |veca -vecb| lt 1 and 0 le theta le pi` then ` theta` in the intervalA. `[0, pi//6)`B. `(5 pi//6, pi]`C. `[pi//6, pi//2]`D. `(pi//2, 5pi//6]`

If the unit vectors ` veca and vecb ` are inclined of an angle ` 2 the

1.	If the unit vectors ` veca and vecb ` are inclined of an angle ` 2 theta` such that ` \|veca -vecb\| lt 1 and 0 le theta le pi` then ` theta` in the intervalA. `[0, pi//6)`B. `(5 pi//6, pi]`C. `[pi//6, pi//2]`D. `(pi//2, 5pi//6]`
Answer» Correct Answer - 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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