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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 1 Maths Textbook Solution.

Need Solution for R.D .Sharma Maths Class 12 Chapter 22 Algebra of Vectors  Exercise 22.9 Question 7 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer: \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)

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

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

Explanation: Let \mu =\hat{i}-\hat{j}+\hat{k} be the given vector

Then magnitude of vector \mu is |\vec{\mu}|=\sqrt{1^{2}+(-1)^{2}+(1)^{2}}=\sqrt{3}

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

We have \begin{aligned} &\cos \alpha=\frac{\mu i}{|\vec{\mu}|}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \\ \end{aligned}

We have \begin{aligned} &\cos \beta=\frac{\mu \cdot j}{|\vec{\mu}|}=\frac{-1}{\sqrt{3}} \Rightarrow \beta=\cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right) \\ \end{aligned}

We have \begin{aligned} &\cos \gamma=\frac{\mu \cdot k}{|\vec{\mu}|}=\frac{1}{\sqrt{3}} \Rightarrow \gamma=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \end{aligned}

\therefore The angles of the given vector  are \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{-1}{\sqrt{3}}\right), \cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)

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