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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1 Answers
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}}