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: Concept
The magnitude of the sum of two vectors is given by the relation $|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}$.

Step 2: Meaning
Since $\vec{a}$ and $\vec{b}$ are unit vectors, we have $|\vec{a}| = 1$ and $|\vec{b}| = 1$. The dot product is $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta = \cos\theta$.

Step 3: Analysis
Substituting the values: \[ |\vec{a} + \vec{b}|^2 = 1^2 + 1^2 + 2\cos\theta \] \[ |\vec{a} + \vec{b}|^2 = 2 + 2\cos\theta = 2(1 + \cos\theta) \] Using the trigonometric half-angle formula $1 + \cos\theta = 2\cos^2(\theta/2)$: \[ |\vec{a} + \vec{b}|^2 = 2\left(2\cos^2\frac{\theta}{2}\right) = 4\cos^2\frac{\theta}{2} \] Taking the square root on both sides: \[ |\vec{a} + \vec{b}| = 2\cos\frac{\theta}{2} \implies \cos\frac{\theta}{2} = \frac{1}{2}|\vec{a} + \vec{b}| \]

Step 4: Conclusion
The value of $\cos(\theta/2)$ is equal to $\frac{1}{2}|\vec{a} + \vec{b}|$. Final Answer: (A)