# Find the angle between the vectors \vec{a}=\hat{i}-\hat{j}+2\hat{k} \,and \,\vec{b}=3\hat{i}+2\hat{j}+4\hat{k}.

Category: QuestionsFind the angle between the vectors \vec{a}=\hat{i}-\hat{j}+2\hat{k} \,and \,\vec{b}=3\hat{i}+2\hat{j}+4\hat{k}.
Editor">Editor Staff asked 11 months ago

Find the angle between the vectors \vec{a}=\hat{i}-\hat{j}+2\hat{k} \,and \,\vec{b}=3\hat{i}+2\hat{j}+4\hat{k}.

(a) cos^{-1}⁡\sqrt{\frac{58}{3}}

(b) cos^{-1}⁡\frac{\sqrt{58}}{3}

(c) cos^{-1}\frac{⁡58}{3\sqrt{3}}

(d) cos^{-1}⁡\frac{\sqrt{58}}{3\sqrt{3}}

This question was addressed to me in a national level competition.

This interesting question is from Product of Two Vectors-1 in division Vector Algebra of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options
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Editor">Editor Staff answered 11 months ago

The correct option is (d) cos^{-1}⁡\frac{\sqrt{58}}{3\sqrt{3}}

To explain: The angle between the two vectors is given by

cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}

|\vec{a}|=\sqrt{1^2+(-1)^2+2^2}=\sqrt{1+1+4}=\sqrt{6}

|\vec{b}|=\sqrt{3^2+2^2+(-4)^2}=\sqrt{9+4+16}=\sqrt{29}

\vec{a}.\vec{b}=1(3)-1(2)+2(4)=9

∴cos⁡θ=\frac{\sqrt{6}.\sqrt{29}}{9}=\frac{\sqrt{58}}{3\sqrt{3}}

∴θ=cos^{-1}\frac{⁡\sqrt{58}}{3\sqrt{3}}