# 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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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}}.