Find the angle between \(\vec{a} \,and \,\vec{b}\) if \(|\vec{a}|=2,|\vec{b}|=\frac{1}{2√3}\) and \(\vec{a}×\vec{b}=\frac{1}{2}\).

Category: QuestionsFind the angle between \(\vec{a} \,and \,\vec{b}\) if \(|\vec{a}|=2,|\vec{b}|=\frac{1}{2√3}\) and \(\vec{a}×\vec{b}=\frac{1}{2}\).
Editor">Editor Staff asked 11 months ago

Find the angle between \(\vec{a} \,and \,\vec{b}\) if \(|\vec{a}|=2,|\vec{b}|=\frac{1}{2√3}\) and \(\vec{a}×\vec{b}=\frac{1}{2}\).
 
(a) \(\frac{2π}{3}\)
 
(b) \(\frac{4π}{5}\)
 
(c) \(\frac{π}{3}\)
 
(d) \(\frac{π}{2}\)
 
I got this question during an interview for a job.
 
My doubt is from Product of Two Vectors-2 topic in chapter Vector Algebra of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

The correct option is (c) \frac{π}{3}
 
To explain: Given that, |\vec{a}|=2, \,|\vec{b}|=\frac{1}{2√3} and \vec{a}×\vec{b}=\frac{1}{2}
 
We know that, \vec{a}×\vec{b}=\vec{a}.\vec{b} \,sin⁡θ
 
∴ sin⁡θ=\frac{(\vec{a}×\vec{b})}{|\vec{a}|.|\vec{b}|}
 
sin⁡θ=\frac{\frac{1}{2}}{2×\frac{1}{2√3}}=\frac{\sqrt{3}}{2}
 
θ=sin^{-1}⁡\frac{\sqrt{3}}{2}=\frac{π}{3}