The direction cosines of a line which is perpendicular to both the lines whose direction ratios are `1,-1,0 and 2,-1,1` are

Options

A. `-1,-1,1`
B. `(-1)/(sqrt(3)),(-1)/(sqrt(3)),(1)/(sqrt(3))`
C. `1,-1,-1`
D. `(1)/(sqrt(3)),(-1)/(sqrt(3)),(-1)/(sqrt(3))`

Answer

To find the direction cosines of a line that is perpendicular to both given lines with direction ratios \(1, -1, 0\) and \(2, -1, 1\), we can use the cross product of the two vectors represented by these direction ratios. Here’s a step-by-step solution:

### Step 1: Define the vectors
Let the first vector \( \mathbf{A} \) be represented by the direction ratios \( (1, -1, 0) \) and the second vector \( \mathbf{B} \) be represented by the direction ratios \( (2, -1, 1) \).

\[ \mathbf{A} = \begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}, \quad \mathbf{B} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \]

### Step 2: Calculate the cross product \( \mathbf{A} \times \mathbf{B} \)
The cross product of two vectors can be calculated using the determinant of a matrix formed by the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) and the components of the vectors \( \mathbf{A} \) and \( \mathbf{B} \).

\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 0 \\ 2 & -1 & 1 \end{vmatrix} \]

### Step 3: Expand the determinant
Using the determinant expansion, we have:

\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \mathbf{i} \):
\[ \begin{vmatrix} -1 & 0 \\ -1 & 1 \end{vmatrix} = (-1)(1) - (0)(-1) = -1 \]

2. For \( \mathbf{j} \):
\[ \begin{vmatrix} 1 & 0 \\ 2 & 1 \end{vmatrix} = (1)(1) - (0)(2) = 1 \]

3. For \( \mathbf{k} \):
\[ \begin{vmatrix} 1 & -1 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (-1)(2) = -1 + 2 = 1 \]

Putting it all together:

\[ \mathbf{A} \times \mathbf{B} = -1 \mathbf{i} - 1 \mathbf{j} + 1 \mathbf{k} = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} \]

### Step 4: Normalize the resulting vector
To find the direction cosines, we need to normalize the vector \( \mathbf{C} = \begin{pmatrix} -1 \\ -1 \\ 1 \end{pmatrix} \).

The magnitude of \( \mathbf{C} \) is calculated as follows:

\[ |\mathbf{C}| = \sqrt{(-1)^2 + (-1)^2 + (1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

### Step 5: Compute the direction cosines
The direction cosines are obtained by dividing each component of \( \mathbf{C} \) by its magnitude:

\[ \text{Direction cosines} = \left( \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \]

### Final Answer
The direction cosines of the line that is perpendicular to both given lines are:

\[ \left( \frac{-1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \]