If `veca,vecb` are two unit vectors such that `veca+(veca xx vecb)=vecc`, where `|vecc|=2`, then value of `[vecavecbvecc]` is

Options

A. 0
B. `+-1`
C. `-3`
D. 3

Explanation

To solve the problem step by step, we start with the given information and work through the necessary calculations.

### Step 1: Understand the given information
We have two unit vectors \(\vec{a}\) and \(\vec{b}\) such that:
\[ \vec{a} + (\vec{a} \times \vec{b}) = \vec{c} \]
and the magnitude of \(\vec{c}\) is given as:
\[ |\vec{c}| = 2 \]

### Step 2: Take the dot product of the equation with \(\vec{a}\)
Taking the dot product of both sides of the equation with \(\vec{a}\):
\[ \vec{a} \cdot \vec{a} + \vec{a} \cdot (\vec{a} \times \vec{b}) = \vec{a} \cdot \vec{c} \]
Since \(\vec{a}\) is a unit vector, \(|\vec{a}|^2 = 1\) and \(\vec{a} \cdot (\vec{a} \times \vec{b}) = 0\) (because \(\vec{a} \times \vec{b}\) is perpendicular to \(\vec{a}\)). Thus, we have:
\[ 1 + 0 = \vec{a} \cdot \vec{c} \]
This simplifies to:
\[ \vec{a} \cdot \vec{c} = 1 \]

### Step 3: Take the dot product of the original equation with \(\vec{c}\)
Now, we take the dot product of the original equation with \(\vec{c}\):
\[ \vec{c} \cdot \vec{c} = \vec{a} \cdot \vec{c} + (\vec{a} \times \vec{b}) \cdot \vec{c} \]
We know that \(|\vec{c}|^2 = 4\) (since \(|\vec{c}| = 2\)), so:
\[ 4 = \vec{a} \cdot \vec{c} + (\vec{a} \times \vec{b}) \cdot \vec{c} \]
Substituting \(\vec{a} \cdot \vec{c} = 1\):
\[ 4 = 1 + (\vec{a} \times \vec{b}) \cdot \vec{c} \]
This simplifies to:
\[ (\vec{a} \times \vec{b}) \cdot \vec{c} = 3 \]

### Step 4: Conclusion
The value of the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\) is:
\[ [\vec{a}, \vec{b}, \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c} = 3 \]

Thus, the final answer is:
\[ \boxed{3} \]