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

Options

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

Answer

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 will use the formula for the cosine of the angle between two lines in 3D space.

### Step-by-step Solution:

1. **Identify Direction Ratios:**
- For the first line, the direction ratios are \( l_1 = 1, m_1 = 1, n_1 = 1 \).
- For the second line, the direction ratios are \( l_2 = 1, m_2 = -1, n_2 = n \).

2. **Use the Cosine Formula:**
The cosine of the angle \( \theta \) between two lines can be calculated using the formula:
\[ \cos \theta = \frac{l_1 l_2 + m_1 m_2 + n_1 n_2}{\sqrt{l_1^2 + m_1^2 + n_1^2} \sqrt{l_2^2 + m_2^2 + n_2^2}} \]

3. **Substitute Known Values:**
Given that \( \theta = 60^\circ \), we know that \( \cos 60^\circ = \frac{1}{2} \). Now substitute the values:
\[ \frac{1 \cdot 1 + 1 \cdot (-1) + 1 \cdot n}{\sqrt{1^2 + 1^2 + 1^2} \sqrt{1^2 + (-1)^2 + n^2}} = \frac{1}{2} \]

4. **Calculate the Denominator:**
- For the first line:
\[ \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]
- For the second line:
\[ \sqrt{1^2 + (-1)^2 + n^2} = \sqrt{1 + 1 + n^2} = \sqrt{2 + n^2} \]

5. **Set Up the Equation:**
Substitute these into the equation:
\[ \frac{1 - 1 + n}{\sqrt{3} \cdot \sqrt{2 + n^2}} = \frac{1}{2} \]
This simplifies to:
\[ \frac{n}{\sqrt{3} \cdot \sqrt{2 + n^2}} = \frac{1}{2} \]

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

7. **Square Both Sides:**
Squaring both sides to eliminate the square root:
\[ 4n^2 = 3(2 + n^2) \]

8. **Expand and Rearrange:**
Expanding the right side:
\[ 4n^2 = 6 + 3n^2 \]
Rearranging gives:
\[ 4n^2 - 3n^2 = 6 \implies n^2 = 6 \]

9. **Solve for \( n \):**
Taking the square root of both sides:
\[ n = \pm \sqrt{6} \]

### Final Answer:
The value of \( n \) is \( \sqrt{6} \) or \( -\sqrt{6} \).