`[a " " b " "axxb] +[a.b]^(2) =`

Options

A. `(a+b)^(2)`
B. `|a|^(2)|b|^(2)`
C. `|a|^(2) + |b|^(2)`
D. 2

Answer

To solve the expression \([ \mathbf{a} \times \mathbf{b} ] + [ \mathbf{a} \cdot \mathbf{b} ]^2\), we will break it down into two parts: the cross product \(\mathbf{a} \times \mathbf{b}\) and the dot product \([\mathbf{a} \cdot \mathbf{b}]^2\).

### Step-by-Step Solution:

1. **Calculate the Cross Product \(\mathbf{a} \times \mathbf{b}\)**:
- The cross product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:
\[ \mathbf{a} \times \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \sin \theta \, \mathbf{n} \]
where \(\theta\) is the angle between the two vectors and \(\mathbf{n}\) is the unit vector perpendicular to the plane formed by \(\mathbf{a}\) and \(\mathbf{b}\).

2. **Calculate the Dot Product \(\mathbf{a} \cdot \mathbf{b}\)**:
- The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]

3. **Square the Dot Product**:
- Now, we need to square the result of the dot product:
\[ [\mathbf{a} \cdot \mathbf{b}]^2 = (|\mathbf{a}| |\mathbf{b}| \cos \theta)^2 = |\mathbf{a}|^2 |\mathbf{b}|^2 \cos^2 \theta \]

4. **Combine the Results**:
- Now we combine the results of the cross product and the squared dot product:
\[ [\mathbf{a} \times \mathbf{b}] + [\mathbf{a} \cdot \mathbf{b}]^2 \]
- This can be expressed as:
\[ |\mathbf{a}| |\mathbf{b}| \sin \theta \, \mathbf{n} + |\mathbf{a}|^2 |\mathbf{b}|^2 \cos^2 \theta \]

5. **Final Expression**:
- The final expression combines both components, but since they are fundamentally different (one is a vector and the other is a scalar), they do not combine further. Thus, the expression remains:
\[ \mathbf{a} \times \mathbf{b} + |\mathbf{a}|^2 |\mathbf{b}|^2 \cos^2 \theta \]