A vector A is rotated by a small angle Δθ radians (Δθ<<1) to get a new vector B. In that case |B−A| is
|A|Δθ
|B|Δθ−|A|
|A|(1−Δθ²/2)
0
Solution:
Δθ is very small we can say that |A|=|B| ∴|A−B|=√A²+B²−2ABcosΔθ=√A²+A²−2A²cosΔθ=2A√(1−cosΔθ)/2=2A√sin²(Δθ/2)=2Asin(Δθ/2) as Δθ/2<<1 we also can say sinΔθ/2≈Δθ/2 ∴|A−B|=2A×Δθ/2 ∴|A−B|=(AΔθ) Thus option (C)