Let `vec a` and `vec b` be two vectors such that `|vec b|=1`and `|vec b timesvec a|=2` .Then `|(vec b timesvec a)-vec b|^(2)` is equal to

Options

A. 3
B. 4
C. 1
D. 5

Answer

To solve the problem, we need to find the value of \(|\vec{b} \times \vec{a} - \vec{b}|^2\) given that \(|\vec{b}| = 1\) and \(|\vec{b} \times \vec{a}| = 2\).

### Step-by-step solution:

1. **Understanding the given conditions**:
- We know that \(|\vec{b}| = 1\), which means that \(\vec{b}\) is a unit vector.
- We also know that \(|\vec{b} \times \vec{a}| = 2\).

2. **Using the properties of cross product**:
- The magnitude of the cross product \(|\vec{b} \times \vec{a}|\) can be expressed as:
\[ |\vec{b} \times \vec{a}| = |\vec{b}| |\vec{a}| \sin \theta \]
where \(\theta\) is the angle between vectors \(\vec{b}\) and \(\vec{a}\).
- Given that \(|\vec{b}| = 1\), we have:
\[ |\vec{b} \times \vec{a}| = |\vec{a}| \sin \theta = 2 \]

3. **Finding \(|\vec{b} \times \vec{a} - \vec{b}|^2\)**:
- We can express \(|\vec{b} \times \vec{a} - \vec{b}|^2\) using the formula:
\[ |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 - 2\vec{u} \cdot \vec{v} \]
where \(\vec{u} = \vec{b} \times \vec{a}\) and \(\vec{v} = \vec{b}\).

4. **Calculating \(|\vec{b} \times \vec{a}|^2\)**:
- Since \(|\vec{b} \times \vec{a}| = 2\), we have:
\[ |\vec{b} \times \vec{a}|^2 = 2^2 = 4 \]

5. **Calculating \(|\vec{b}|^2\)**:
- Since \(|\vec{b}| = 1\), we have:
\[ |\vec{b}|^2 = 1^2 = 1 \]

6. **Calculating the dot product \(\vec{u} \cdot \vec{v}\)**:
- The dot product \(\vec{b} \times \vec{a} \cdot \vec{b}\) is equal to 0 because the cross product is orthogonal to both vectors involved:
\[ \vec{b} \times \vec{a} \cdot \vec{b} = 0 \]

7. **Putting it all together**:
- Now substituting these values into the equation:
\[ |\vec{b} \times \vec{a} - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2 - 2(\vec{b} \times \vec{a} \cdot \vec{b}) \]
\[ = 4 + 1 - 2 \cdot 0 = 4 + 1 = 5 \]

### Final Answer:
\[ |\vec{b} \times \vec{a} - \vec{b}|^2 = 5 \]