Q80 / 180 · KCET 2020 · Mathematics
If $\mathbf{a}$ and $\mathbf{b}$ are unit vectors and $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$, then $\sin \frac{\theta}{2}$ is equal to
- A$|\mathbf{a}+\mathbf{b}|$
- B$\frac{|\mathbf{a}+\mathbf{b}|}{2}$
- Correct answerC$\frac{|\mathbf{a}-\mathbf{b}|}{2}$
- D$|\mathbf{a}-\mathbf{b}|$
Solution
Given $\mathbf{a}$ and $\mathbf{b}$ are unit vectors
$\therefore|\mathbf{a}|=1 \text { and }|\mathbf{b}|=1$
We know that, $|\mathbf{a}-\mathbf{b}|^{2}=(\mathbf{a}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{b})$
$=|\mathbf{a}|^{2}-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^{2}$
$=|\mathbf{a}|^{2}-2|\mathbf{a}||\mathbf{b}| \cos \theta+|\mathbf{b}|^{2}$
$=1-2 \cdot 1 \cdot 1 \cdot \cos \theta+1$
$=2-2 \cos \theta$
$\Rightarrow \quad|\mathbf{a}-\mathbf{b}|^{2}=2(1-\cos \theta)$
$=2\left(2 \sin ^{2} \theta / 2\right)$
$=4 \sin ^{2} \theta / 2$
$\begin{array}{ll}\Rightarrow & \sin ^{2} \frac{\theta}{2}=\frac{1}{4}|\mathbf{a}-\mathbf{b}| \\ \Rightarrow & \sin \frac{\theta}{2}=\frac{1}{2}|\mathbf{a}-\mathbf{b}|\end{array}$
Hence, the correct answer is option: 'C'.