Question

If \( \vec{a} \) and \( \vec{b} \) are unit vectors and the angle between them is \(60^\circ\), find \( |\vec{a}-\vec{b}| \).

• A. \(1\)
• B. \(\sqrt{2}\)
• C. \(\sqrt{3}\)
• D. \(2\)

Hint:

For unit vectors, \( \vec{a}\cdot\vec{b} = \cos\theta \). This makes magnitude formulas like \( |\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 -2\vec{a}\cdot\vec{b} \) much easier to evaluate.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two unit vectors, meaning their magnitudes are 1. The angle between them is \(60^\circ\). We need to find the magnitude of their difference.
Step 2: Key Formula or Approach:
We use the property that the square of the magnitude of a vector is the dot product of the vector with itself.
\[ |\vec{a} - \vec{b}|^2 = (\vec{a} - \vec{b}) \cdot (\vec{a} - \vec{b}) \] Expanding this gives:
\[ |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2(\vec{a} \cdot \vec{b}) + |\vec{b}|^2 \] We also need the formula for the dot product: \( \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta \).
Step 3: Detailed Explanation:
Given:
\( |\vec{a}| = 1 \)
\( |\vec{b}| = 1 \)
\( \theta = 60^\circ \)
First, calculate the dot product \( \vec{a} \cdot \vec{b} \):
\[ \vec{a} \cdot \vec{b} = (1)(1)\cos(60^\circ) = 1 \times \frac{1}{2} = \frac{1}{2} \] Now, substitute this into the magnitude squared formula:
\[ |\vec{a} - \vec{b}|^2 = (1)^2 - 2\left(\frac{1}{2}\right) + (1)^2 \] \[ |\vec{a} - \vec{b}|^2 = 1 - 1 + 1 = 1 \] Finally, take the square root to find the magnitude:
\[ |\vec{a} - \vec{b}| = \sqrt{1} = 1 \] Step 4: Final Answer:
The magnitude \( |\vec{a} - \vec{b}| \) is 1.