question_answer

If \[\hat{a}\] and \[\hat{b}\] are unit vectors inclined at an angle \[\theta ,\] then prove that \[\sin \frac{\theta }{2}=\frac{1}{2}|\hat{a}-\hat{b}|.\]

Answer:

To prove, \[\sin \frac{\theta }{2}=\frac{1}{2}|\hat{a}-\hat{b}|\]             We know that,             \[|\hat{a}-\hat{b}{{|}^{2}}=(\hat{a}-\hat{b})\cdot (\hat{a}-\hat{b})\]                         \[=\hat{a}\cdot \hat{a}\,-\hat{a}\cdot \hat{b}\,-\hat{b}\cdot \hat{a}\,+\hat{b}\cdot \hat{b}\]                         \[=\,\,|\hat{a}{{|}^{2}}-2\,\hat{a}\cdot \hat{b}+|\hat{b}{{|}^{2}}\] [\[\because \] dot product is commutative]                         \[=1-2\hat{a}\cdot \hat{b}+1\]   \[[\because \,\,\,|\hat{a}|\,\,=\,\,|\hat{b}|\,\,=1]\]             \[\Rightarrow \]   \[|\hat{a}-\hat{b}{{|}^{2}}=2-2\cos \theta \] \[\begin{align}   & [\text{if}\,\,\hat{a}\cdot \hat{b}=\,\,|\hat{a}||\hat{b}|\,\,\cos \theta \,\,\text{and}\,\,|\hat{a}|\,\,=\,\,|\hat{b}|\,\,=1, \\  & then\,\,\hat{a}\cdot \hat{b}=\cos \theta ] \\ \end{align}\]                         \[=2\,(1-\cos \theta )=2\left( 2{{\sin }^{2}}\frac{\theta }{2} \right)=4{{\sin }^{2}}\frac{\theta }{2}\]                                                 \[\left[ \because \,\,\,1-\cos \theta =2{{\sin }^{2}}\frac{\theta }{2} \right]\]             \[\Rightarrow \]   \[{{\sin }^{2}}\frac{\theta }{2}=\frac{1}{4}|\hat{a}-\hat{b}{{|}^{2}}\]             \[\therefore \]      \[\sin \frac{\theta }{2}=\frac{1}{2}|\hat{a}-\hat{b}|\] [Taking positive square root as modulus cannot be negative]   Hence proved.