Find the angle between the two vectors \vec{a} and \vec{b} with magnitude \sqrt{3} and \sqrt{2} respectively and \vec{a.} \,\vec{b}=3\sqrt{2}.

Category: QuestionsFind the angle between the two vectors \vec{a} and \vec{b} with magnitude \sqrt{3} and \sqrt{2} respectively and \vec{a.} \,\vec{b}=3\sqrt{2}.
Editor">Editor Staff asked 11 months ago

Find the angle between the two vectors \vec{a} and \vec{b} with magnitude \sqrt{3} and \sqrt{2} respectively and \vec{a.} \,\vec{b}=3\sqrt{2}.
 
(a) cos^{-1}⁡\frac{1}{\sqrt{3}}
 
(b) cos^{-1}⁡\sqrt{3}
 
(c) cos^{-1}⁡\frac{3}{\sqrt{2}}
 
(d) cos^{-1}⁡\frac{2}{\sqrt{3}}
 
I got this question during an internship interview.
 
Asked question is from Product of Two Vectors-1 in chapter Vector Algebra of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Correct answer is (a) cos^{-1}⁡\frac{1}{\sqrt{3}}
 
Explanation: Given that, |\vec{a}|=\sqrt{3} \,and \,|\vec{b}|=\sqrt{2}
 
Also, \vec{a.} \vec{b}=3\sqrt{2}
 
The angle between two vectors is given by
 
cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}
 
∴cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{3\sqrt{2}}=\frac{1}{\sqrt{3}}
 
∴θ=cos^{-1}⁡\frac{1}{\sqrt{3}}.