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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

Answer»

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

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

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



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