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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 8 Sub Question 2 Maths Textbook Solution.

Provide Solution For R.D. Sharma Maths Class 12 Chapter 23 Scalar or Dot Products  Exercise 23.1 Question 8 Sub Question 2 Maths Textbook Solution.

Answers (1)

Answer:   Proved

Hint: you must know the property of solving vectors.

Given: if \hat{a} and \hat{b} are unit vectors inclined at angle \theta, prove that \tan \frac{\theta }{2}=\mid \frac{\hat{a}-\hat{b}}{\hat{a}+\hat{b}}\mid

Solution: given that \hat{a} and \hat{b}  are unit vectors

So \mid \hat{a}\mid =1,\mid \hat{b}\mid =1

We have,

\begin{aligned} &|\hat{a}-\hat{b}|^{2}=|\widehat{a}|^{2}+|\hat{b}|^{2}-2 \hat{a} \cdot \hat{b} \\ &=1+1-2|\hat{a}||\hat{b}| \cos \theta \\ &|\hat{a}-\hat{b}|^{2}=2-2 \cos \theta \\ &\cos \theta=\frac{2-|\hat{a}-\hat{b}|^{2}}{2} \end{aligned}

And \begin{aligned} &\sin \frac{\theta}{2}=\sqrt{\frac{1-\cos \theta}{2}} \\ \end{aligned}

\begin{aligned} &\sqrt{\frac{1-\frac{2-|\hat{a}-\hat{b}|^{2}}{2}}{2}} \\ &\Rightarrow \sqrt{\frac{2-2 + |\hat{a}-\hat{b}|^{2}}{4}} \end{aligned}

\begin{aligned} &\Rightarrow \sqrt{\frac{|\hat{a}-\hat{b}|^{2}}{4}} \\ &\sin \frac{\theta}{2}=\frac{1}{2}|\hat{a}-\hat{b}| \end{aligned}

And similarly\begin{aligned} &\cos \frac{\theta}{2}=\frac{1}{2}|\hat{a}+\hat{b}| \end{aligned}

\begin{aligned} &\therefore \tan \frac{\theta}{2}=\frac{\sin \theta / 2}{\cos \theta / 2}=\frac{\frac{1}{2}|\hat{a}-\hat{b}|}{\frac{1}{2}|\hat{a}+\hat{b}|} \\ &\Rightarrow \frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|} \end{aligned}

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