Find the angle between the vectors \(\vec{a}=-\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}\)

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

Find the angle between the vectors \(\vec{a}=-\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}\)
 
(a) \(cos^{-1}⁡-\frac{\sqrt{3}}{2}\)
 
(b) \(cos^{-1}⁡-\frac{2}{\sqrt{3}}\)
 
(c) \(cos^{-1}⁡-\sqrt{2}\)
 
(d) \(cos^{-1}⁡-\sqrt{\frac{3}{2}}\)
 
I have been asked this question in final exam.
 
My query is from Product of Two Vectors-1 in chapter 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) \(cos^{-1}⁡-\sqrt{\frac{3}{2}}\)
 
To explain: The angle between two vectors is given by
 
\(cos⁡θ=\frac{|\vec{a}|.|\vec{b}|}{\vec{a}.\vec{b}}\)
 
\(|\vec{a}|=\sqrt{(-1)^2+(1)^2+(-1)^2}=\sqrt{3}\)
 
\(|\vec{b}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}\)
 
\(\vec{a}.\vec{b}\)=(-1)(1)+1(-1)+0=-2
 
\(cos⁡θ=\frac{\sqrt{3}.\sqrt{2}}{-2}=-\sqrt{\frac{3}{2}}\)
 
∴\(θ=cos^{-1}⁡-\sqrt{\frac{3}{2}}\)