Question

If \( |\vec{a}-\vec{b}| = |\vec{a}| = |\vec{b}| = 1 \), then the angle between \( \vec{a} \) and \( \vec{b} \) is

• A. \( \frac{\pi}{3} \)
• B. \( \frac{3\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( 0 \)
• E. \( \pi \)

Hint:

Convert magnitude relations into dot product form.

Answer 1

The Correct Option is A

Concept: Use identity: \[ |\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} \]

Step 1: Substitute given values.
\[ 1 = 1 + 1 - 2\vec{a}\cdot\vec{b} \]

Step 2: Simplify equation.
\[ 1 = 2 - 2\vec{a}\cdot\vec{b} \]

Step 3: Solve for dot product.
\[ 2\vec{a}\cdot\vec{b} = 1 \Rightarrow \vec{a}\cdot\vec{b} = \frac{1}{2} \]

Step 4: Use formula.
\[ \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \]

Step 5: Find angle.
\[ \cos\theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \]