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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors Exercise 22.9 Question 7 Sub Question 2 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.9 Question 7 Sub Question 2 Maths Textbook Solution.

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Answer: \frac{\pi }{2},\frac{\pi }{4},\frac{3\pi }{4}

Given: \hat{j}-\hat{k}

Hint: Find  \cos \alpha ,\cos \beta ,\cos \gamma

Explanation: Let \vec{u}=\hat{j}-\hat{k} be the given vector

The magnitude of vector \mu  is |\mu|=\sqrt{0^{2}+(1)^{2}+(-1)^{2}}=\sqrt{2}

Let the direction angle of the vector are \alpha ,\beta ,\gamma

We have \cos \alpha=\frac{\mu i}{|\mu|}=\frac{0}{\sqrt{2}}=0 \Rightarrow \alpha=\cos ^{-1}(0)=\frac{\pi}{2}\left[\because \cos \frac{\pi}{2}=0\right]

We have \cos \beta=\frac{\mu \cdot j}{|\mu|}=\frac{1}{\sqrt{2}} \Rightarrow \beta=\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]

We have \cos \gamma=\frac{\mu . k}{|\mu|}=\frac{-1}{\sqrt{2}} \Rightarrow \gamma=\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4}\left[\because \cos \frac{3 \pi}{4}=\frac{-1}{\sqrt{2}}\right]

\therefore The angles of the given vector are \frac{\pi }{2},\frac{\pi }{4},\frac{3\pi }{4}

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