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Explain Solution R.D. Sharma Class 12 Chapter 22 Algebra of Vectors  Exercise 22 .9 Question 7 Sub Question 3 Maths Textbook Solution.

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Answer: \cos ^{-1}\left(\frac{4}{9}\right), \cos ^{-1}\left(\frac{8}{9}\right), \cos ^{-1}\left(\frac{1}{9}\right)

Given: 4\hat{i}+8\hat{j}+\hat{k}

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

Explanation: Let \vec{u}=4\hat{i}+8\hat{j}+\hat{k} be the given vector

The magnitude of vector \mu  is |\mu|=\sqrt{4^{2}+(8)^{2}+(1)^{2}}=\sqrt{81}=9

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

We have \begin{aligned} &\cos \alpha=\frac{\mu i}{|\mu|}=\frac{4}{9} \Rightarrow \alpha=\cos ^{-1}\left(\frac{4}{9}\right) \\ \end{aligned}

We have \begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\mu|}=\frac{8}{9} \Rightarrow \beta=\cos ^{-1}\left(\frac{8}{9}\right) \\ \end{aligned}

We have \begin{aligned} &\cos \gamma=\frac{\mu \cdot j}{|\mu|}=\frac{1}{9} \Rightarrow \gamma=\cos ^{-1}\left(\frac{1}{9}\right) \end{aligned}

\therefore The angles of the given vector are \cos ^{-1}\left(\frac{4}{9}\right), \cos ^{-1}\left(\frac{8}{9}\right), \cos ^{-1}\left(\frac{1}{9}\right)

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