Let `veca and vecb` be unit vectors that are perpendicular to each other l. then `[veca+ (veca xx vecb) vecb + (veca xx vecb) veca xx vecb]` will always be equal to

Options

A. 1
B. zero
C. `-1`
D. 3

Explanation

To solve the expression \(\vec{a} + (\vec{a} \times \vec{b}) \cdot \vec{b} + (\vec{a} \times \vec{b}) \times \vec{a} \times \vec{b}\), where \(\vec{a}\) and \(\vec{b}\) are unit vectors that are perpendicular to each other, we can follow these steps:

### Step 1: Understanding the given vectors
Since \(\vec{a}\) and \(\vec{b}\) are unit vectors and perpendicular, we have:
\[ |\vec{a}| = 1, \quad |\vec{b}| = 1, \quad \vec{a} \cdot \vec{b} = 0 \]

### Step 2: Simplifying the expression
The expression can be rewritten as:
\[ \vec{a} + (\vec{a} \times \vec{b}) \cdot \vec{b} + (\vec{a} \times \vec{b}) \times \vec{a} \times \vec{b} \]

### Step 3: Evaluating \((\vec{a} \times \vec{b}) \cdot \vec{b}\)
Since \(\vec{a}\) and \(\vec{b}\) are perpendicular, the cross product \(\vec{a} \times \vec{b}\) is a vector that is perpendicular to both \(\vec{a}\) and \(\vec{b}\). Therefore, the dot product of \((\vec{a} \times \vec{b})\) with \(\vec{b}\) is zero:
\[ (\vec{a} \times \vec{b}) \cdot \vec{b} = 0 \]

### Step 4: Simplifying the expression further
Now the expression reduces to:
\[ \vec{a} + 0 + (\vec{a} \times \vec{b}) \times \vec{a} \times \vec{b} \]
This simplifies to:
\[ \vec{a} + (\vec{a} \times \vec{b}) \times \vec{b} \]

### Step 5: Evaluating \((\vec{a} \times \vec{b}) \times \vec{b}\)
Using the vector triple product identity:
\[ \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \]
Let \(\vec{x} = \vec{a} \times \vec{b}\), \(\vec{y} = \vec{b}\), and \(\vec{z} = \vec{b}\):
\[ (\vec{a} \times \vec{b}) \times \vec{b} = (\vec{a} \cdot \vec{b}) \vec{b} - (\vec{b} \cdot \vec{b}) \vec{a} \]
Since \(\vec{a} \cdot \vec{b} = 0\) and \(|\vec{b}|^2 = 1\):
\[ (\vec{a} \times \vec{b}) \times \vec{b} = 0 \cdot \vec{b} - 1 \cdot \vec{a} = -\vec{a} \]

### Step 6: Final expression
Now substituting back, we have:
\[ \vec{a} - \vec{a} = \vec{0} \]

### Conclusion
Thus, the value of the expression \(\vec{a} + (\vec{a} \times \vec{b}) \cdot \vec{b} + (\vec{a} \times \vec{b}) \times \vec{a} \times \vec{b}\) is:
\[ \vec{0} \]