If the directions cosines of a straight line are `ltk,k,kgt` then

Options

A. `kgt0`
B. `0ltklt1`
C. `k=1`
D. `k=1/sqrt3" or " -1/sqrt3`

Explanation

To solve the problem, we need to find the value of \( k \) given that the direction cosines of a straight line are \( l = k, m = k, n = k \).

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
The direction cosines of a line are defined as the cosines of the angles that the line makes with the positive x, y, and z axes. They are represented as \( l = \cos \alpha \), \( m = \cos \beta \), and \( n = \cos \gamma \).

2. **Setting Up the Equation**:
Since we are given that \( l = k, m = k, n = k \), we can express the relationship between the direction cosines:
\[ l^2 + m^2 + n^2 = 1 \]

3. **Substituting Values**:
Substitute \( l, m, n \) with \( k \):
\[ k^2 + k^2 + k^2 = 1 \]
This simplifies to:
\[ 3k^2 = 1 \]

4. **Solving for \( k^2 \)**:
Divide both sides by 3:
\[ k^2 = \frac{1}{3} \]

5. **Finding \( k \)**:
Taking the square root of both sides gives:
\[ k = \pm \sqrt{\frac{1}{3}} = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3} \]

6. **Conclusion**:
The values of \( k \) are:
\[ k = \frac{1}{\sqrt{3}} \quad \text{and} \quad k = -\frac{1}{\sqrt{3}} \]

### Final Answer:
The possible values of \( k \) are \( \frac{1}{\sqrt{3}} \) and \( -\frac{1}{\sqrt{3}} \).

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