What is the value of n so that the angle between the lines having direction ratios (1,1,1) and (1,-1,n) is `60^@`?

Options

A. `sqrt3`
B. `sqrt6`
C. 3
D. None of these

Explanation

To find the value of \( n \) such that the angle between the lines with direction ratios \( (1, 1, 1) \) and \( (1, -1, n) \) is \( 60^\circ \), we can use the formula for the cosine of the angle between two lines given by their direction ratios.

### Step-by-Step Solution:

1. **Identify Direction Ratios:**
Let the direction ratios of the first line be \( (a_1, b_1, c_1) = (1, 1, 1) \) and for the second line be \( (a_2, b_2, c_2) = (1, -1, n) \).

2. **Use the Cosine Formula:**
The cosine of the angle \( \theta \) between two lines is given by:
\[ \cos \theta = \frac{a_1 a_2 + b_1 b_2 + c_1 c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \sqrt{a_2^2 + b_2^2 + c_2^2}} \]
Here, \( \theta = 60^\circ \), so \( \cos 60^\circ = \frac{1}{2} \).

3. **Substitute Values:**
Substitute the values into the formula:
\[ \frac{1 \cdot 1 + 1 \cdot (-1) + 1 \cdot n}{\sqrt{1^2 + 1^2 + 1^2} \sqrt{1^2 + (-1)^2 + n^2}} = \frac{1}{2} \]
This simplifies to:
\[ \frac{1 - 1 + n}{\sqrt{3} \sqrt{2 + n^2}} = \frac{1}{2} \]
Which further simplifies to:
\[ \frac{n}{\sqrt{3} \sqrt{2 + n^2}} = \frac{1}{2} \]

4. **Cross Multiply:**
Cross-multiplying gives:
\[ 2n = \sqrt{3} \sqrt{2 + n^2} \]

5. **Square Both Sides:**
Squaring both sides results in:
\[ 4n^2 = 3(2 + n^2) \]
Expanding this gives:
\[ 4n^2 = 6 + 3n^2 \]

6. **Rearrange the Equation:**
Rearranging the equation leads to:
\[ 4n^2 - 3n^2 - 6 = 0 \]
Simplifying gives:
\[ n^2 - 6 = 0 \]

7. **Solve for \( n \):**
Thus, we find:
\[ n^2 = 6 \implies n = \pm \sqrt{6} \]

### Final Answer:
The values of \( n \) are \( n = \sqrt{6} \) and \( n = -\sqrt{6} \).