The condition `(a.b)^(2)=a^(2)b^(2)` is satisfied when

Options

A. a is parallel to b
B. `aneb`
C. `a.b=1`
D. `abotb`

Explanation

To solve the problem, we need to analyze the condition \((\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2\) and determine when it holds true.

### Step-by-step Solution:

1. **Understanding the Dot Product**:
The dot product of two vectors \(\mathbf{a}\) and \(\mathbf{b}\) is given by:
\[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta \]
where \(\theta\) is the angle between the two vectors.

2. **Substituting the Dot Product**:
We can substitute the expression for the dot product into the given condition:
\[ (\mathbf{a} \cdot \mathbf{b})^2 = (|\mathbf{a}| |\mathbf{b}| \cos \theta)^2 \]
This simplifies to:
\[ (\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2 \cos^2 \theta \]

3. **Setting Up the Equation**:
Now we set the left-hand side equal to the right-hand side:
\[ a^2 b^2 \cos^2 \theta = a^2 b^2 \]

4. **Dividing by \(a^2 b^2\)**:
Assuming \(a \neq 0\) and \(b \neq 0\), we can divide both sides by \(a^2 b^2\):
\[ \cos^2 \theta = 1 \]

5. **Solving for \(\theta\)**:
The equation \(\cos^2 \theta = 1\) implies:
\[ \cos \theta = \pm 1 \]
This occurs when \(\theta = 0^\circ\) or \(\theta = 180^\circ\). In both cases, the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are either parallel or anti-parallel.

6. **Conclusion**:
Therefore, the condition \((\mathbf{a} \cdot \mathbf{b})^2 = a^2 b^2\) is satisfied when the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are parallel.

### Final Answer:
The condition is satisfied when **\(\mathbf{a}\) is parallel to \(\mathbf{b}\)**.

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