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If \(\vec a\) and \(\vec b\) are unit vectors and θ is the angle between them, then prove that \(sin\frac\theta 2=\frac12|\vec a - \vec b|.\) |
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Answer» \(\vec a .\vec b= |\vec a| |\vec b| cos \theta\) ⇒ \(\vec a .\vec b= cos \theta\) \((|\vec a|= |\vec b|=1)\) Now, \(|\vec a - \vec b|^2 = (\vec a - \vec b).(\vec a - \vec b)\) \(= \vec a .\vec a - 2\vec a.\vec b + \vec b. \vec b \) \(=|a|^2 - 2cos \theta + |\vec b|^2\) \(= 1 - 2cos \theta + 1 \) \((\because |\vec a| = |\vec b| = 1)\) \(= 2 (1 - cos \theta )\) \(= 2 (1 - (1 - 2sin^2 \frac\theta 2))\) \(= 4 sin^2\frac\theta 2\) ⇒ \(sin^2 \frac\theta2 = \frac14 |\vec a - \vec b|^2\) ⇒ \(sin \frac\theta2 = \frac12 |\vec a - \vec b|\) |
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