If `overline(a)` is a non-zero vector and k is a scalar such that `|koverline(a)|=1,` then k=

Options

A. `|overline(a)|`
B. `1`
C. `(1)/(|overline(a)|)`
D. `pm(1)/(|overline(a)|)`

Answer

To solve the problem, we need to find the value of the scalar \( k \) given that \( |k \overline{a}| = 1 \) where \( \overline{a} \) is a non-zero vector.

### Step-by-Step Solution:

1. **Understanding the Modulus of a Vector**:
The modulus (or magnitude) of a vector \( \overline{a} \) is denoted as \( |\overline{a}| \). Given that \( \overline{a} \) is a non-zero vector, we know \( |\overline{a}| > 0 \).

2. **Applying the Modulus Property**:
The modulus of the scalar multiplication of a vector can be expressed as:
\[ |k \overline{a}| = |k| \cdot |\overline{a}| \]
This means that the magnitude of the product of a scalar \( k \) and a vector \( \overline{a} \) is equal to the product of the absolute value of \( k \) and the magnitude of \( \overline{a} \).

3. **Setting Up the Equation**:
According to the problem, we have:
\[ |k \overline{a}| = 1 \]
Therefore, substituting from the previous step, we get:
\[ |k| \cdot |\overline{a}| = 1 \]

4. **Isolating \( |k| \)**:
To find \( |k| \), we can rearrange the equation:
\[ |k| = \frac{1}{|\overline{a}|} \]

5. **Determining the Possible Values of \( k \)**:
Since \( |k| \) represents the absolute value of \( k \), it can take on two values:
\[ k = \frac{1}{|\overline{a}|} \quad \text{or} \quad k = -\frac{1}{|\overline{a}|} \]
Thus, we can conclude:
\[ k = \pm \frac{1}{|\overline{a}|} \]

### Final Answer:
The value of \( k \) is:
\[ k = \pm \frac{1}{|\overline{a}|} \]