Question

Let $ \vec{a} $ and $ \vec{b} $ be two equal vectors inclined at an angle $ θ $ , then $ a\,sin\, \frac{\theta}{2} $ is equal to

• A. $ \frac{\left|\vec{a}-\vec{b}\right|}{2} $
• B. $ \frac{\left|\vec{a}+\vec{b}\right|}{2} $
• C. $ \left|\vec{a}-\vec{b}\right| $
• D. $ \left|\vec{a}+\vec{b}\right| $

Answer 1

Answer: (A)

$\left|\vec{a}-\vec{b}\right|=\sqrt{\vec{a}^{2}+\vec{b}^{2}-2\left|\vec{a}\right|\cdot\left|\vec{b}\right|\,cos\,\theta}$
Here, $\vec{a}=\vec{b}$
$\therefore \left|\vec{a}-\vec{b}\right|=\sqrt{\vec{a}^{2}+\vec{a}^{2}-2\vec{a}^{2}\,cos\,\theta}$
$=\sqrt{2\vec{a}^{2}\,\left(1-cos\,\theta\right)}=\sqrt{2a^{2}\cdot2\,sin^{2}\, \frac{\theta}{2}}$
$=2\,a\,sin\, \frac{\theta}{2}$
$\Rightarrow \left|\vec{a}-\vec{b}\right|=2\,a\,sin\, \frac{\theta}{2}$
$\Rightarrow a\,sin\, \frac{\theta}{2}=\frac{\left|\vec{a}-\vec{b}\right|}{2}$