Assertion : A vector can have zero magnitude if one of its rectangular components is not zero.
Reason : Scalar product of two vectors cannot be a negative quantity.

Options

A. `[epsilon_(0)]=[M^(-1)L^(-3)T^(2)I]`
B. `[epsilon_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`
C. `[mu_(0)]=[MLT^(-2)I^(-2)]`
D. `[mu_(0)]=[ML^(2)T^(-1)I]`

Explanation

To solve the given question, we need to analyze both the assertion and the reason separately.

### Step-by-Step Solution:

1. **Understanding the Assertion**:
- The assertion states: "A vector can have zero magnitude if one of its rectangular components is not zero."
- A vector's magnitude is calculated using the formula:
\[ |\mathbf{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \]
where \(A_x\), \(A_y\), and \(A_z\) are the rectangular components of the vector along the x, y, and z axes, respectively.
- If the magnitude of the vector is zero, then:
\[ \sqrt{A_x^2 + A_y^2 + A_z^2} = 0 \]
This implies that:
\[ A_x^2 + A_y^2 + A_z^2 = 0 \]
Since squares of real numbers are non-negative, the only solution is:
\[ A_x = 0, A_y = 0, A_z = 0 \]
- Therefore, if the magnitude of the vector is zero, all its components must also be zero. Thus, the assertion is **false**.

2. **Understanding the Reason**:
- The reason states: "Scalar product of two vectors cannot be a negative quantity."
- The scalar product (or 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.
- The cosine of an angle can take values from -1 to 1. Therefore, if \(\theta\) is between 90° and 270°, \(\cos(\theta)\) will be negative.
- This means the scalar product can indeed be negative if the angle between the two vectors is obtuse (greater than 90°). Thus, the reason is also **false**.

3. **Conclusion**:
- Both the assertion and the reason are false.

### Final Answer:
Both the assertion and the reason are false.