let a,b, and c be three unit vectors such that a is perpendicular to the plane off b and c. if the angle betweenn b annd c is `(pi)/(3)`, then `|axxb-axxc|` is equal to

Options

A. `1//3`
B. `1//2`
C. `1`
D. 2

Answer

To solve the problem, we need to find the value of \( | \mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c} | \) given that \( \mathbf{a}, \mathbf{b}, \) and \( \mathbf{c} \) are unit vectors, \( \mathbf{a} \) is perpendicular to the plane of \( \mathbf{b} \) and \( \mathbf{c} \), and the angle between \( \mathbf{b} \) and \( \mathbf{c} \) is \( \frac{\pi}{3} \).

### Step 1: Understand the Cross Product
The expression \( \mathbf{a} \times \mathbf{b} \) gives a vector that is perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). Similarly, \( \mathbf{a} \times \mathbf{c} \) is perpendicular to both \( \mathbf{a} \) and \( \mathbf{c} \).

### Step 2: Use the Magnitude of the Cross Product
The magnitude of the cross product \( | \mathbf{a} \times \mathbf{b} | \) can be calculated using the formula:
\[ | \mathbf{a} \times \mathbf{b} | = | \mathbf{a} | | \mathbf{b} | \sin(\theta) \]
where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \). Since \( \mathbf{a} \) is perpendicular to \( \mathbf{b} \), \( \theta = \frac{\pi}{2} \) and thus:
\[ | \mathbf{a} \times \mathbf{b} | = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 3: Calculate \( | \mathbf{a} \times \mathbf{c} | \)
Similarly, since \( \mathbf{a} \) is also perpendicular to \( \mathbf{c} \):
\[ | \mathbf{a} \times \mathbf{c} | = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{2}\right) = 1 \]

### Step 4: Find the Dot Product
Next, we need to find \( | \mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c} | \). We can use the property of the magnitude of the difference of two vectors:
\[ | \mathbf{u} - \mathbf{v} |^2 = | \mathbf{u} |^2 + | \mathbf{v} |^2 - 2 \mathbf{u} \cdot \mathbf{v} \]
Let \( \mathbf{u} = \mathbf{a} \times \mathbf{b} \) and \( \mathbf{v} = \mathbf{a} \times \mathbf{c} \).

### Step 5: Calculate \( \mathbf{u} \cdot \mathbf{v} \)
Using the quadruple product identity:
\[ \mathbf{u} \cdot \mathbf{v} = (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{c}) = \mathbf{a} \cdot \mathbf{a} (\mathbf{b} \cdot \mathbf{c}) - (\mathbf{a} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{b}) \]
Since \( \mathbf{a} \) is a unit vector, \( \mathbf{a} \cdot \mathbf{a} = 1 \). Also, \( \mathbf{a} \cdot \mathbf{c} = 0 \) and \( \mathbf{a} \cdot \mathbf{b} = 0 \) because \( \mathbf{a} \) is perpendicular to both \( \mathbf{b} \) and \( \mathbf{c} \).

The angle between \( \mathbf{b} \) and \( \mathbf{c} \) is \( \frac{\pi}{3} \), so:
\[ \mathbf{b} \cdot \mathbf{c} = | \mathbf{b} | | \mathbf{c} | \cos\left(\frac{\pi}{3}\right) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2} \]
Thus,
\[ \mathbf{u} \cdot \mathbf{v} = 1 \cdot \frac{1}{2} - 0 = \frac{1}{2} \]

### Step 6: Calculate \( | \mathbf{u} - \mathbf{v} |^2 \)
Now we can substitute back into the equation:
\[ | \mathbf{u} - \mathbf{v} |^2 = | \mathbf{u} |^2 + | \mathbf{v} |^2 - 2 \mathbf{u} \cdot \mathbf{v} \]
Substituting the values:
\[ | \mathbf{u} - \mathbf{v} |^2 = 1^2 + 1^2 - 2 \cdot \frac{1}{2} = 1 + 1 - 1 = 1 \]

### Step 7: Take the Square Root
Finally, taking the square root gives:
\[ | \mathbf{u} - \mathbf{v} | = \sqrt{1} = 1 \]

Thus, the final answer is:
\[ | \mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c} | = 1 \]