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Question:

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)