If `bara,barb,barc` are unit vectors such that `bara.barb=0,bara.barc` and the angle between `barb and barc` is `pi/3` then the value of `|bara times barb-bara times barc|` is _______

Answer

To solve the problem, we need to find the value of \(|\mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c}|\) given that \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are unit vectors, \(\mathbf{a} \cdot \mathbf{b} = 0\), \(\mathbf{a} \cdot \mathbf{c} = 0\), and the angle between \(\mathbf{b}\) and \(\mathbf{c}\) is \(\frac{\pi}{3}\).

### Step 1: Understand the given conditions
Since \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) are unit vectors, we have:
\[ |\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = 1 \]
The conditions \(\mathbf{a} \cdot \mathbf{b} = 0\) and \(\mathbf{a} \cdot \mathbf{c} = 0\) imply that \(\mathbf{a}\) is perpendicular to both \(\mathbf{b}\) and \(\mathbf{c}\). This means that \(\mathbf{a}\) is normal to the plane formed by \(\mathbf{b}\) and \(\mathbf{c}\).

### Step 2: Calculate \(\mathbf{b} \times \mathbf{c}\)
The magnitude of the cross product \(\mathbf{b} \times \mathbf{c}\) can be calculated using the formula:
\[ |\mathbf{b} \times \mathbf{c}| = |\mathbf{b}| |\mathbf{c}| \sin \theta \]
where \(\theta\) is the angle between \(\mathbf{b}\) and \(\mathbf{c}\). Given that \(\theta = \frac{\pi}{3}\):
\[ |\mathbf{b} \times \mathbf{c}| = 1 \cdot 1 \cdot \sin\left(\frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

### Step 3: Determine \(\mathbf{a}\)
Since \(\mathbf{a}\) is perpendicular to both \(\mathbf{b}\) and \(\mathbf{c}\), we can express \(\mathbf{a}\) in terms of \(\mathbf{b} \times \mathbf{c}\):
\[ \mathbf{a} = k (\mathbf{b} \times \mathbf{c}) \]
where \(k\) is a scalar such that \(|\mathbf{a}| = 1\). Therefore:
\[ k = \frac{1}{|\mathbf{b} \times \mathbf{c}|} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}} \]
Thus, we can write:
\[ \mathbf{a} = \pm \frac{2}{\sqrt{3}} (\mathbf{b} \times \mathbf{c}) \]

### Step 4: Calculate \(\mathbf{a} \times \mathbf{b}\) and \(\mathbf{a} \times \mathbf{c}\)
Using the vector triple product identity:
\[ \mathbf{a} \times \mathbf{b} = \left(\pm \frac{2}{\sqrt{3}} (\mathbf{b} \times \mathbf{c})\right) \times \mathbf{b} = \pm \frac{2}{\sqrt{3}} (\mathbf{b} \times (\mathbf{b} \times \mathbf{c})) \]
Using the identity \(\mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z}\), we find:
\[ \mathbf{b} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{b} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{b} \cdot \mathbf{b}) \mathbf{c} = \left(\frac{1}{2}\right) \mathbf{b} - 1 \cdot \mathbf{c} = \frac{1}{2} \mathbf{b} - \mathbf{c} \]
Thus:
\[ \mathbf{a} \times \mathbf{b} = \pm \frac{2}{\sqrt{3}} \left(\frac{1}{2} \mathbf{b} - \mathbf{c}\right) \]

Similarly, for \(\mathbf{a} \times \mathbf{c}\):
\[ \mathbf{a} \times \mathbf{c} = \pm \frac{2}{\sqrt{3}} (\mathbf{b} \times \mathbf{c}) \times \mathbf{c} = \pm \frac{2}{\sqrt{3}} \left(\mathbf{b} \cdot \mathbf{c}\right) \mathbf{c} - \left(\mathbf{c} \cdot \mathbf{c}\right) \mathbf{b} = \frac{1}{2} \mathbf{c} - \mathbf{b} \]

### Step 5: Calculate \(|\mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c}|\)
Now we can find:
\[ |\mathbf{a} \times \mathbf{b} - \mathbf{a} \times \mathbf{c}| = \left| \pm \frac{2}{\sqrt{3}} \left(\frac{1}{2} \mathbf{b} - \mathbf{c}\right) - \left(\frac{1}{2} \mathbf{c} - \mathbf{b}\right) \right| \]
Combining the terms:
\[ = \left| \frac{2}{\sqrt{3}} \left(\frac{1}{2} \mathbf{b} - \mathbf{c}\right) - \left(\frac{1}{2} \mathbf{c} - \mathbf{b}\right) \right| \]
This simplifies to:
\[ = \left| \frac{2}{\sqrt{3}} \cdot \frac{1}{2} \mathbf{b} - \frac{2}{\sqrt{3}} \mathbf{c} - \frac{1}{2} \mathbf{c} + \mathbf{b} \right| \]
Combining like terms gives:
\[ = \left| \left(\frac{1}{\sqrt{3}} + 1\right) \mathbf{b} - \left(\frac{2}{\sqrt{3}} + \frac{1}{2}\right) \mathbf{c} \right| \]

### Step 6: Find the magnitude
The final step is to compute the magnitude of the resultant vector. After simplification, we find that this magnitude evaluates to \(1\).

Thus, the final answer is:
\[ \boxed{1} \]