If vector `a` is a variable vector, whose magnitude is constant, then choose the statement which is always true.

Options

A. `(a)(da)/(dt)` is perpendicular to `a`
B. `(b)(da)/(dt)` is parallel to `a`
C. `(c)(da)/(dt)` is a constant vector
D. (d)None of these

Answer

To solve the problem, we need to analyze the properties of a variable vector \( \mathbf{a} \) that has a constant magnitude.

### Step-by-Step Solution:

1. **Understanding the Vector**:
We have a vector \( \mathbf{a} \) whose magnitude is constant. This means that while the vector can change in direction, its length (magnitude) remains the same.

2. **Variable Vector**:
Since \( \mathbf{a} \) is a variable vector, its direction can change over time. We can denote the magnitude of \( \mathbf{a} \) as \( |\mathbf{a}| = c \), where \( c \) is a constant.

3. **Differentiating the Vector**:
To understand how \( \mathbf{a} \) changes over time, we differentiate it with respect to time \( t \):
\[ \frac{d\mathbf{a}}{dt} \]
This derivative represents the rate of change of the vector \( \mathbf{a} \).

4. **Magnitude is Constant**:
Since the magnitude of \( \mathbf{a} \) is constant, we can use the property of differentiation:
\[ \frac{d}{dt}(|\mathbf{a}|^2) = 0 \]
This implies:
\[ \frac{d}{dt}(\mathbf{a} \cdot \mathbf{a}) = 0 \]
Using the product rule, we get:
\[ 2\mathbf{a} \cdot \frac{d\mathbf{a}}{dt} = 0 \]
This indicates that the dot product of \( \mathbf{a} \) and \( \frac{d\mathbf{a}}{dt} \) is zero.

5. **Conclusion**:
The result \( \mathbf{a} \cdot \frac{d\mathbf{a}}{dt} = 0 \) implies that \( \frac{d\mathbf{a}}{dt} \) is perpendicular to \( \mathbf{a} \). Therefore, the statement that is always true is:
\[ \frac{d\mathbf{a}}{dt} \text{ is perpendicular to } \mathbf{a} \]

### Final Answer:
The correct option is: \( \frac{d\mathbf{a}}{dt} \) is perpendicular to \( \mathbf{a} \).

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