For three vectors `veca, vecb and vecc`, If `|veca|=2, |vecb|=1, vecaxxvecb=vecc` and `vecbxxvecc=veca`, then the value of `[(veca+vecb,vecb+vecc,vecc+veca)]` is equal to

Answer

To solve the problem, we need to find the value of the scalar triple product \([( \vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a} )]\). Given the conditions of the vectors, we can follow these steps:

### Step 1: Understand the given conditions
We have:
- \(|\vec{a}| = 2\)
- \(|\vec{b}| = 1\)
- \(\vec{a} \times \vec{b} = \vec{c}\)
- \(\vec{b} \times \vec{c} = \vec{a}\)

### Step 2: Calculate the magnitude of \(\vec{c}\)
Using the formula for the magnitude of the cross product:
\[ |\vec{c}| = |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Since \(\vec{a}\) and \(\vec{b}\) are perpendicular (as implied by the cross product), \(\sin \theta = 1\). Thus:
\[ |\vec{c}| = |\vec{a}| \cdot |\vec{b}| = 2 \cdot 1 = 2 \]

### Step 3: Confirm the orthogonality of the vectors
From the relationships given:
- \(\vec{c} = \vec{a} \times \vec{b}\) implies \(\vec{c}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).
- \(\vec{a} = \vec{b} \times \vec{c}\) implies \(\vec{a}\) is perpendicular to both \(\vec{b}\) and \(\vec{c}\).

Thus, all three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular.

### Step 4: Calculate the scalar triple product
The scalar triple product can be expressed as:
\[ [\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w}) \]
For our case:
\[ [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] \]
We can expand this using the distributive property of the dot and cross products.

### Step 5: Expand the expression
Let:
- \(\vec{u} = \vec{a} + \vec{b}\)
- \(\vec{v} = \vec{b} + \vec{c}\)
- \(\vec{w} = \vec{c} + \vec{a}\)

We need to compute:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) \]

### Step 6: Calculate \(\vec{v} \times \vec{w}\)
Using the distributive property:
\[ \vec{v} \times \vec{w} = (\vec{b} + \vec{c}) \times (\vec{c} + \vec{a}) = \vec{b} \times \vec{c} + \vec{b} \times \vec{a} + \vec{c} \times \vec{c} + \vec{c} \times \vec{a} \]
Since \(\vec{c} \times \vec{c} = \vec{0}\), we have:
\[ \vec{v} \times \vec{w} = \vec{a} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} \]

### Step 7: Substitute back into the scalar triple product
Now substituting this back into the scalar triple product:
\[ [\vec{u}, \vec{v}, \vec{w}] = (\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) \]

### Step 8: Evaluate the dot products
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular:
- \(\vec{a} \cdot \vec{b} = 0\)
- \(\vec{a} \cdot \vec{c} = 0\)
- \(\vec{b} \cdot \vec{c} = 0\)

Thus, we can simplify:
\[ [\vec{u}, \vec{v}, \vec{w}] = |\vec{a}|^2 |\vec{b}| |\vec{c}| = 2^2 \cdot 1 \cdot 2 = 8 \]

### Final Answer
The value of \([( \vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a} )]\) is equal to 8.