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

Concept: For vectors, the magnitude of the difference is given by \[ |\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} \] Also, \[ \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta \] where \( \theta \) is the angle between the vectors.
Step 1: {Use the given information.} Since \( \vec{a} \) and \( \vec{b} \) are unit vectors, \[ |\vec{a}| = 1, \qquad |\vec{b}| = 1 \] Angle between them: \[ \theta = 60^\circ \]
Step 2: {Find the dot product.} \[ \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos 60^\circ \] \[ = 1 \times 1 \times \frac{1}{2} \] \[ = \frac{1}{2} \]
Step 3: {Substitute in the magnitude formula.} \[ |\vec{a}-\vec{b}|^2 = 1^2 + 1^2 - 2\left(\frac{1}{2}\right) \] \[ = 1 + 1 - 1 \] \[ = 1 \]
Step 4: {Find the magnitude.} \[ |\vec{a}-\vec{b}| = \sqrt{1} = 1 \]