`[(veca,vecb,axxvecb)]+(veca.vecb)^(2)=`

Options

A. `|veca|^(2)|vecb|^(2)`
B. `|veca+vecb|^(2)`
C. `|veca|^(2)+|vecb|^(2)`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression \([( \vec{a}, \vec{b}, \vec{a} \times \vec{b})] + (\vec{a} \cdot \vec{b})^2\).

### Step 1: Understand the scalar triple product
The scalar triple product \(( \vec{a}, \vec{b}, \vec{c})\) is defined as \(\vec{a} \cdot (\vec{b} \times \vec{c})\). In our case, we need to calculate \(( \vec{a}, \vec{b}, \vec{a} \times \vec{b})\).

### Step 2: Apply the scalar triple product formula
Using the definition, we have:
\[ ( \vec{a}, \vec{b}, \vec{a} \times \vec{b}) = \vec{a} \cdot (\vec{b} \times (\vec{a} \times \vec{b})) \]

### Step 3: Use the vector triple product identity
We can simplify \(\vec{b} \times (\vec{a} \times \vec{b})\) using the vector triple product identity:
\[ \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \]
Applying this identity, we get:
\[ \vec{b} \times (\vec{a} \times \vec{b}) = (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \]

### Step 4: Substitute back into the scalar triple product
Now substituting back into our expression:
\[ ( \vec{a}, \vec{b}, \vec{a} \times \vec{b}) = \vec{a} \cdot \left( (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{b} \cdot \vec{a}) \vec{b} \right) \]
This expands to:
\[ = (\vec{b} \cdot \vec{b}) (\vec{a} \cdot \vec{a}) - (\vec{b} \cdot \vec{a}) (\vec{a} \cdot \vec{b}) \]

### Step 5: Simplify the expression
Notice that \(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}\), so we can rewrite:
\[ = |\vec{b}|^2 |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2 \]

### Step 6: Combine with the second part of the expression
Now we need to add \((\vec{a} \cdot \vec{b})^2\) to our result:
\[ |\vec{b}|^2 |\vec{a}|^2 - (\vec{a} \cdot \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 \]
The \((\vec{a} \cdot \vec{b})^2\) terms cancel out:
\[ = |\vec{b}|^2 |\vec{a}|^2 \]

### Final Result
Thus, the final value of the expression is:
\[ |\vec{a}|^2 |\vec{b}|^2 \]