Question

If \(\vec{a}\) and \(\vec{b}\) are unit vectors, then the angle between \(\vec{a}\) and \(\vec{b}\) for \( \sqrt{3}\vec{a} - \vec{b} \) to be a unit vector is given by :

• A. \(\pi/6\)
• B. \(\pi/4\)
• C. \(\pi/3\)
• D. \(2\pi/3\)

Hint:

When dealing with unit vectors and magnitudes, always square the vector equation to convert magnitudes into dot products. This leverages the properties \( |\vec{v}|^2 = \vec{v} \cdot \vec{v} \) and \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \) for unit vectors.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem involves two unit vectors \(\vec{a}\) and \(\vec{b}\). We are given that a linear combination of these vectors, \( \sqrt{3}\vec{a} - \vec{b} \), is also a unit vector. We need to find the angle between \(\vec{a}\) and \(\vec{b}\).

Step 2: Key Formula or Approach:
1. If \(\vec{v}\) is a unit vector, then \( |\vec{v}|^2 = 1 \).
2. For any vector \(\vec{v}\), \( |\vec{v}|^2 = \vec{v} \cdot \vec{v} \).
3. The dot product of two vectors \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta \).
4. Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \( |\vec{a}| = 1 \) and \( |\vec{b}| = 1 \).

Step 3: Detailed Explanation:
Let \(\vec{v} = \sqrt{3}\vec{a} - \vec{b}\).
Since \(\vec{v}\) is a unit vector, \( |\vec{v}|^2 = 1 \).
\[ |\sqrt{3}\vec{a} - \vec{b}|^2 = 1 \]
Expand the square of the magnitude using the dot product property:
\[ (\sqrt{3}\vec{a} - \vec{b}) \cdot (\sqrt{3}\vec{a} - \vec{b}) = 1 \]
\[ (\sqrt{3}\vec{a}) \cdot (\sqrt{3}\vec{a}) - 2(\sqrt{3}\vec{a} \cdot \vec{b}) + \vec{b} \cdot \vec{b} = 1 \]
\[ (\sqrt{3})^2 |\vec{a}|^2 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) + |\vec{b}|^2 = 1 \]
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors, \( |\vec{a}|^2 = 1 \) and \( |\vec{b}|^2 = 1 \).
\[ 3(1) - 2\sqrt{3} (\vec{a} \cdot \vec{b}) + 1 = 1 \]
\[ 4 - 2\sqrt{3} (\vec{a} \cdot \vec{b}) = 1 \]
\[ 3 = 2\sqrt{3} (\vec{a} \cdot \vec{b}) \]
Now, substitute \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta = (1)(1)\cos\theta = \cos\theta \):
\[ 3 = 2\sqrt{3} \cos\theta \]
Solve for \(\cos\theta\):
\[ \cos\theta = \frac{3}{2\sqrt{3}} = \frac{3\sqrt{3}}{2 \times 3} = \frac{\sqrt{3}}{2} \]
The angle \(\theta\) whose cosine is \( \frac{\sqrt{3}}{2} \) is \( 30^{\circ} \) or \( \pi/6 \) radians.

Step 4: Final Answer:
The angle between \(\vec{a}\) and \(\vec{b}\) is \(\pi/6\). This corresponds to option (a).