Let a , b and c be three unit vectors out of which vectors b and c are non -parallel. If `alpha " and " beta` are the angles which vector a makes with vectors b and c respectively and `axx (b xx c) = (1)/(2) b,` Then `|alpha -beta|` is equal to

Options

A. `30^(@)`
B. `45^(@)`
C. `90^(@)`
D. `60^(@)`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Given Information
We have three unit vectors \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \). The angles \( \alpha \) and \( \beta \) are defined as the angles between vector \( \mathbf{a} \) and vectors \( \mathbf{b} \) and \( \mathbf{c} \) respectively. We also know that:
- \( |\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = 1 \) (since they are unit vectors)
- \( \mathbf{b} \) and \( \mathbf{c} \) are non-parallel.
- The condition given is \( \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \frac{1}{2} \mathbf{b} \).

### Step 2: Use the Vector Triple Product Identity
We can use the vector triple product identity:
\[ \mathbf{u} \times (\mathbf{v} \times \mathbf{w}) = (\mathbf{u} \cdot \mathbf{w}) \mathbf{v} - (\mathbf{u} \cdot \mathbf{v}) \mathbf{w} \]
Applying this to our situation:
\[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]
Setting this equal to \( \frac{1}{2} \mathbf{b} \):
\[ (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} = \frac{1}{2} \mathbf{b} \]

### Step 3: Compare Coefficients
Now we can compare coefficients from both sides:
1. For the coefficient of \( \mathbf{b} \):
\[ \mathbf{a} \cdot \mathbf{c} = \frac{1}{2} \]
2. For the coefficient of \( \mathbf{c} \):
\[ -(\mathbf{a} \cdot \mathbf{b}) = 0 \implies \mathbf{a} \cdot \mathbf{b} = 0 \]

### Step 4: Find Angles \( \alpha \) and \( \beta \)
From the dot product definitions:
- Since \( \mathbf{a} \cdot \mathbf{b} = 0 \), we have:
\[ |\mathbf{a}| |\mathbf{b}| \cos(\alpha) = 0 \implies \cos(\alpha) = 0 \implies \alpha = \frac{\pi}{2} \text{ (90 degrees)} \]
- For \( \mathbf{a} \cdot \mathbf{c} = \frac{1}{2} \):
\[ |\mathbf{a}| |\mathbf{c}| \cos(\beta) = \frac{1}{2} \implies \cos(\beta) = \frac{1}{2} \implies \beta = \frac{\pi}{3} \text{ (60 degrees)} \]

### Step 5: Calculate \( |\alpha - \beta| \)
Now we calculate:
\[ |\alpha - \beta| = \left| \frac{\pi}{2} - \frac{\pi}{3} \right| \]
Finding a common denominator (which is 6):
\[ |\alpha - \beta| = \left| \frac{3\pi}{6} - \frac{2\pi}{6} \right| = \left| \frac{\pi}{6} \right| = \frac{\pi}{6} \]

### Final Answer
Thus, the value of \( |\alpha - \beta| \) is \( \frac{\pi}{6} \).

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