If unit vectors a→ and b→ are inclined at an angle 2θ such |a→−b→|<1 and 0≤θ≤π, then θ lies in the interval
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a
[0,π/6)
b
(5π/6,π]
c
[π/6,π/2)
d
(π/2,5π/6]
answer is A.
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Detailed Solution
we have |a→−b→|2=|a→|2+|b→|2−2(a→⋅b→)or |a→−b→|2=|a→|2+|b→|2−2|a→||b→|cos2θor |a→−b→|2=2−2cos2θ (∵|a→|=|b→|=1) =4sin2θ⇒ |a→−b→|=2|sinθ|now |a→−b→|<1⇒ 2|sinθ|<1or |sinθ|<12⇒θ∈[0,π/6) or θ∈(5π/6,π]