If ` vec a\ ` is a non zero vecrtor iof modulus ` vec a\ a n d\ m` is a non zero scalar such that `m a` is a unit vector, then`

Options

A. `m = pm 1 `
B. `a = |m|`
C. `a = 1//|m|`
D. `a = 1/m`

Explanation

To solve the problem, we need to analyze the given information step by step.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
We are given that \( \vec{a} \) is a non-zero vector with a modulus (magnitude) denoted as \( |\vec{a}| \). We also have a non-zero scalar \( m \) such that \( m\vec{a} \) is a unit vector.

2. **Definition of a Unit Vector:**
A unit vector has a magnitude of 1. Therefore, we can express this as:
\[ |m\vec{a}| = 1 \]

3. **Using the Property of Magnitudes:**
The magnitude of a scalar multiplied by a vector can be expressed as:
\[ |m\vec{a}| = |m| \cdot |\vec{a}| \]
Substituting this into our previous equation gives:
\[ |m| \cdot |\vec{a}| = 1 \]

4. **Rearranging the Equation:**
From the equation \( |m| \cdot |\vec{a}| = 1 \), we can isolate \( |\vec{a}| \):
\[ |\vec{a}| = \frac{1}{|m|} \]

5. **Conclusion:**
This means the modulus of the vector \( \vec{a} \) is the reciprocal of the absolute value of the scalar \( m \). Therefore, we conclude that:
\[ |\vec{a}| = \frac{1}{|m|} \]

### Final Answer:
The correct conclusion is that \( |\vec{a}| = \frac{1}{|m|} \).