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If $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ then prove that: i. $\cos \frac{\theta}{2}=\frac{1}{2}|\hat{\mathrm{a}}+\hat{\mathrm{b}}|$ ii. $\tan \frac{\theta}{2}=\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$
If $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ then prove that:
i. $\cos \frac{\theta}{2}=\frac{1}{2}|\hat{\mathrm{a}}+\hat{\mathrm{b}}|$
ii. $\tan \frac{\theta}{2}=\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$
CBSE | Class 12 | Maths | RS Aggarwal | Scalar, or Dot, Product of Vectors
If $\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$ then prove that:
i. $\cos \frac{\theta}{2}=\frac{1}{2}|\hat{\mathrm{a}}+\hat{\mathrm{b}}|$
ii. $\tan \frac{\theta}{2}=\frac{|\hat{a}-\hat{b}|}{|\hat{a}+\hat{b}|}$

Solution:

$\hat{a}$ and $\hat{b}$ are unit vectors inclined at an angle $\theta$
i) $\begin{aligned} &(\hat{a}+\hat{b})(\hat{a}+\hat{b})=\mathrm{I} \hat{\mathrm{a}} \mathrm{I}_{2}+\mathrm{I} \hat{\mathrm{b}} \mathrm{I}_{2}+2 \hat{a} \hat{b} \\ \mathrm{I} \hat{a}+\hat{b}_{2} &=1+1+2 \mathrm{I} \hat{\mathrm{I}} \mathrm{I} \hat{b} \mathrm{I} \cos \theta \\ \mathrm{I} \hat{a}+\hat{b}_{2} &=2+2 \cos \theta=2(1+\cos \theta) \\ \mathrm{I} \hat{a}+\hat{b}_{2} &=2 \cdot 2 \cos ^{2} \frac{\theta}{2} \\ \mathrm{I} \hat{a}+\hat{b}_{2} &=4 \cos ^{2} \frac{\theta}{2} \\ \frac{1 \hat{a}+\hat{b}}{2} &=\cos \frac{\theta}{2} \\ \text { ii) } \quad \mathrm{I} \hat{a}-\hat{b} \mathrm{I}_{2}=\mathrm{I} \hat{\mathrm{I}}_{2}+\mathrm{I} \hat{b} \mathrm{I}_{2}-2 \hat{a} \hat{b} \end{aligned}$ $\mathrm{I} \hat{a}-\hat{b} \mathrm{I}_{2}=1+1-2 \cos \theta$
$\mathrm{I} \hat{a}-\hat{b} \mathrm{I}_{2}=2(1-\cos \theta)$
$\mathrm{I} \hat{a}-\hat{b} \mathrm{I}_{2}=2.2 \sin ^{2} \frac{\theta}{2}$
$\mathrm{I} \hat{a}-\hat{\mathrm{b}} \mathrm{I}=2 \sin \frac{\theta}{2}$
Clearly $\frac{1}{1 \hat{a}+b_{1}}=\frac{2 \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}$

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