Question

If \( \theta \) be the angle between two unit vectors \( \hat{a} \) and \( \hat{b} \), prove that \( \sin\frac{\theta}{2} = \frac{1}{2} |\hat{a} - \hat{b}| \).

Hint:

This is a standard and useful vector identity. A similar identity exists for the sum of two unit vectors: \( \cos\frac{\theta}{2} = \frac{1}{2} |\hat{a} + \hat{b}| \). Memorizing both can save time in exams. The key to proving them is to start with the squared magnitude and use the properties of the dot product.

Answer 1

Step 1: Understanding the Concept:
This problem requires proving a vector identity that relates the angle between two unit vectors to the magnitude of their difference. The proof will utilize the properties of the dot product and trigonometric half-angle identities.
Step 2: Key Formula or Approach:
We will use the following properties: 1. For any vector \(\vec{v}\), its squared magnitude is \(|\vec{v}|^2 = \vec{v} \cdot \vec{v}\). 2. The dot product of two vectors \(\vec{u}\) and \(\vec{v}\) is \(\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta\). 3. For unit vectors \( \hat{a} \) and \( \hat{b} \), \(|\hat{a}|=1\) and \(|\hat{b}|=1\). 4. The trigonometric half-angle identity: \(1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)\).
Step 3: Detailed Explanation:
Let's start by considering the expression \( |\hat{a} - \hat{b}|^2 \).
Using the dot product property, we can write: \[ |\hat{a} - \hat{b}|^2 = (\hat{a} - \hat{b}) \cdot (\hat{a} - \hat{b}) \] Expanding the dot product: \[ = \hat{a} \cdot \hat{a} - \hat{a} \cdot \hat{b} - \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b} \] Since \(\hat{a}\) and \(\hat{b}\) are unit vectors, \( \hat{a} \cdot \hat{a} = |\hat{a}|^2 = 1^2 = 1 \) and \( \hat{b} \cdot \hat{b} = |\hat{b}|^2 = 1^2 = 1 \).
The dot product is commutative, so \( \hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{a} \). The expression becomes: \[ |\hat{a} - \hat{b}|^2 = 1 - 2(\hat{a} \cdot \hat{b}) + 1 = 2 - 2(\hat{a} \cdot \hat{b}) \] Now, substitute the definition of the dot product: \( \hat{a} \cdot \hat{b} = |\hat{a}||\hat{b}|\cos\theta = (1)(1)\cos\theta = \cos\theta \). \[ |\hat{a} - \hat{b}|^2 = 2 - 2\cos\theta = 2(1 - \cos\theta) \] Using the half-angle identity \( 1 - \cos\theta = 2\sin^2(\frac{\theta}{2}) \): \[ |\hat{a} - \hat{b}|^2 = 2 \left( 2\sin^2\left(\frac{\theta}{2}\right) \right) = 4\sin^2\left(\frac{\theta}{2}\right) \] Taking the square root of both sides: \[ |\hat{a} - \hat{b}| = \sqrt{4\sin^2\left(\frac{\theta}{2}\right)} = 2\left|\sin\left(\frac{\theta}{2}\right)\right| \] The angle \(\theta\) between two vectors is usually taken in the range \( [0, \pi] \). Therefore, \( \frac{\theta}{2} \) is in the range \( [0, \frac{\pi}{2}] \), where the sine function is non-negative. So, \( \left|\sin\left(\frac{\theta}{2}\right)\right| = \sin\left(\frac{\theta}{2}\right) \). \[ |\hat{a} - \hat{b}| = 2\sin\left(\frac{\theta}{2}\right) \] Step 4: Final Answer:
Rearranging the equation, we get: \[ \sin\left(\frac{\theta}{2}\right) = \frac{1}{2} |\hat{a} - \hat{b}| \] This completes the proof.