Question

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 1

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

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

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