Question

If \(\vec{a}\) and \(\vec{b}\) are unit vectors and θ is the angle between \(\vec{a}\) and \(\vec{b}\), then sin\(\frac{θ}{2}\) is

• A. \(|\vec{a}+\vec{b}|\)
• B. \(\frac{|\vec{a}+\vec{b}|}{2}\)
• C. \(\frac{|\vec{a}-\vec{b}|}{2}\)
• D. \(|\vec{a}-\vec{b}|\)

Answer 1

The Correct Option is C

If \(\vec{a}\) and \(\vec{b}\) are unit vectors and \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\), then we need to find \(\sin \frac{\theta}{2}\).

Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\).

We know that \(|\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{a} - 2 \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b}\).

\(|\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2 |\vec{a}| |\vec{b}| \cos \theta + |\vec{b}|^2\).

Since \(|\vec{a}| = 1\) and \(|\vec{b}| = 1\), we have:

\(|\vec{a} - \vec{b}|^2 = 1 - 2 \cos \theta + 1 = 2 - 2 \cos \theta\)

\(|\vec{a} - \vec{b}|^2 = 2(1 - \cos \theta)\)

We know that \(1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})\), so:

\(|\vec{a} - \vec{b}|^2 = 2(2 \sin^2(\frac{\theta}{2})) = 4 \sin^2(\frac{\theta}{2})\)

Taking the square root of both sides:

\(|\vec{a} - \vec{b}| = 2 \sin(\frac{\theta}{2})\)

\(\sin(\frac{\theta}{2}) = \frac{|\vec{a} - \vec{b}|}{2}\)

Therefore, the correct option is (C) \(\frac{|\vec{a} - \vec{b}|}{2}\).