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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 1 Maths Textbook Solution.

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Answer:  proved

Hint:  you must know the rules of solving vectors.

Given:  \hat{a} and \hat{b} are unit vector in dined at angle \theta then prove  \cos \frac{\theta }{2}=\frac{1}{2}\mid \hat{a}+\hat{b}\mid

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

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

We have,

 \begin{aligned} &|\hat{a}+\hat{b}|^{2}=\mid \widehat{a \mid}^{2}+\mid \widehat{b \mid}^{2}+2 \hat{a} \cdot \hat{b}\\ \end{aligned}

\begin{aligned} &=1+1+2|\hat{a}||\hat{b}| \cos \theta\\ &=2+2(1)(1) \cos \theta\\ \end{aligned}

                                                            \begin{aligned} &|\hat{a}+\hat{b}|^{2}=2+2 \cos \theta\\ \end{aligned}

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

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

\begin{aligned} &=(1)^{2}+(1)^{2}-2|\hat{a}||\hat{b}| \cos \theta\\ \end{aligned}

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

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

Now,

\begin{aligned} &\cos \theta=\sqrt{\frac{1+\cos \theta }{2}} \end{aligned}

\begin{aligned} &=\sqrt{\frac{1+\frac{|\hat{a}+\hat{b}|^{2}-2}{2}}{2}} \\ &=\sqrt{\frac{2+|\hat{a}+\hat{b}|^{2}-2}{4}} \\ &=\sqrt{\frac{|\hat{a}+\hat{b}|^{2}}{4}} \\ &\cos \frac{\vartheta}{2} \Rightarrow \frac{1}{2}|\hat{a}+\hat{b}| \quad=\text { Proved } \end{aligned}

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