If direction cosines of a line are `(1/a, 1/a, 1/a)` then

Options

A. `agt0`
B. `0lt alt 1`
C. `a = pm sqrt(3)`
D. `a gt 2`

Answer

To solve the problem, we need to find the value of \( a \) given that the direction cosines of a line are \( \left(\frac{1}{a}, \frac{1}{a}, \frac{1}{a}\right) \).

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
The direction cosines of a line are denoted as \( \cos \alpha, \cos \beta, \cos \gamma \). For the given direction cosines, we have:
\[ \cos \alpha = \frac{1}{a}, \quad \cos \beta = \frac{1}{a}, \quad \cos \gamma = \frac{1}{a} \]

2. **Using the Direction Cosine Relation**:
The fundamental property of direction cosines states that:
\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]
Substituting the given values into this equation:
\[ \left(\frac{1}{a}\right)^2 + \left(\frac{1}{a}\right)^2 + \left(\frac{1}{a}\right)^2 = 1 \]

3. **Simplifying the Equation**:
This simplifies to:
\[ \frac{1}{a^2} + \frac{1}{a^2} + \frac{1}{a^2} = 1 \]
Which can be written as:
\[ \frac{3}{a^2} = 1 \]

4. **Finding \( a^2 \)**:
To isolate \( a^2 \), we can take the reciprocal of both sides:
\[ 3 = a^2 \]
Thus, we have:
\[ a^2 = 3 \]

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

6. **Conclusion**:
Therefore, the values of \( a \) are \( \sqrt{3} \) and \( -\sqrt{3} \).

### Final Answer:
The correct options for \( a \) are \( a = \sqrt{3} \) and \( a = -\sqrt{3} \).

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