If` vec(a)` is a non-zero vector and m is a non-zero scalar, then m `vec(a)` is a unit vector if

Options

A. `m = +-1`
B. `m = |a|`
C. `m = +- | vec(a)|`
D. `m = +- (1)/( |vec(a) |)`

Answer

To determine the conditions under which \( m \vec{a} \) is a unit vector, we start with the definition of a unit vector. A vector is considered a unit vector if its magnitude is equal to 1.

### Step-by-step Solution:

1. **Understanding the Magnitude of the Vector**:
The magnitude of the vector \( \vec{a} \) is denoted as \( |\vec{a}| \) or \( \text{mod } \vec{a} \). Since \( \vec{a} \) is a non-zero vector, \( |\vec{a}| \) is a positive scalar.

2. **Magnitude of the Scaled Vector**:
The magnitude of the vector \( m \vec{a} \) (where \( m \) is a non-zero scalar) can be expressed as:
\[ |m \vec{a}| = |m| \cdot |\vec{a}| \]

3. **Setting the Condition for a Unit Vector**:
For \( m \vec{a} \) to be a unit vector, we need:
\[ |m \vec{a}| = 1 \]
Substituting the expression from step 2, we get:
\[ |m| \cdot |\vec{a}| = 1 \]

4. **Solving for \( |m| \)**:
Rearranging the equation gives:
\[ |m| = \frac{1}{|\vec{a}|} \]

5. **Considering the Sign of \( m \)**:
Since \( m \) can be either positive or negative, we conclude:
\[ m = \pm \frac{1}{|\vec{a}|} \]

### Final Conclusion:
Thus, \( m \vec{a} \) is a unit vector if:
\[ m = \pm \frac{1}{|\vec{a}|} \]