Question

Let $\vec{a}$ and $\vec{b}$ be two unit vectors. If the angle between them is $\theta$, then $\cos(\theta/2) =$

• A. $\frac{1}{2}|\vec{a} + \vec{b}|$
• B. $\frac{1}{2}|\vec{a} - \vec{b}|$
• C. $|\vec{a} + \vec{b}|$
• D. $|\vec{a} - \vec{b}|$

Hint:

For unit vectors, remember the standard forms: $|\vec{a}+\vec{b}| = 2\cos(\theta/2)$ and $|\vec{a}-\vec{b}| = 2\sin(\theta/2)$.

Answer 1

The Correct Option is A

Step 1: Use the sum-of-vectors length rule.
For any two vectors, $|\vec a + \vec b|^2 = |\vec a|^2 + |\vec b|^2 + 2\,\vec a\cdot\vec b$.

Step 2: Put in unit vectors.
Both have length 1, and $\vec a\cdot\vec b = \cos\theta$. So \[ |\vec a + \vec b|^2 = 1 + 1 + 2\cos\theta \]

Step 3: Group it.
\[ |\vec a + \vec b|^2 = 2(1 + \cos\theta) \]

Step 4: Use a half-angle identity.
Since $1 + \cos\theta = 2\cos^2\frac{\theta}{2}$: \[ |\vec a + \vec b|^2 = 4\cos^2\frac{\theta}{2} \]

Step 5: Take square roots.
\[ |\vec a + \vec b| = 2\cos\frac{\theta}{2} \]

Step 6: Solve for the cosine.
\[ \cos\frac{\theta}{2} = \frac{1}{2}|\vec a + \vec b| \] \[ \boxed{ \cos\frac{\theta}{2} = \tfrac12|\vec a + \vec b| } \]