The direction cosines of the line `x-y+2z=5, 3x+y+z=6` are

Options

A. `(-3)/(5sqrt(2)),5/(5sqrt(2)),4/(5sqrt(2))`
B. `3/(5sqrt(2)),(-5)/(5sqrt(2)),4/(5sqrt(2))`
C. `3/(5sqrt(2)),5/(5sqrt(2)),4/(5sqrt(2))`
D. none of these

Explanation

To find the direction cosines of the line formed by the intersection of the two planes given by the equations \(x - y + 2z = 5\) and \(3x + y + z = 6\), we can follow these steps:

### Step 1: Identify the normal vectors of the planes

The normal vector of a plane given by the equation \(Ax + By + Cz = D\) can be determined from the coefficients \(A\), \(B\), and \(C\).

- For the first plane \(x - y + 2z = 5\), the normal vector \(n_1\) is:
\[ n_1 = (1, -1, 2) \]

- For the second plane \(3x + y + z = 6\), the normal vector \(n_2\) is:
\[ n_2 = (3, 1, 1) \]

### Step 2: Compute the cross product of the normal vectors

The direction of the line of intersection of the two planes can be found by taking the cross product of the two normal vectors \(n_1\) and \(n_2\).

\[ b = n_1 \times n_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 3 & 1 & 1 \end{vmatrix} \]

Calculating the determinant:
\[ b = \hat{i} \begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} \]

Calculating each of the 2x2 determinants:
1. \(\begin{vmatrix} -1 & 2 \\ 1 & 1 \end{vmatrix} = (-1)(1) - (2)(1) = -1 - 2 = -3\)

2. \(\begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (2)(3) = 1 - 6 = -5\)

3. \(\begin{vmatrix} 1 & -1 \\ 3 & 1 \end{vmatrix} = (1)(1) - (-1)(3) = 1 + 3 = 4\)

Putting it all together:
\[ b = -3\hat{i} + 5\hat{j} + 4\hat{k} \]

### Step 3: Find the direction ratios

The direction ratios of the line are given by the vector:
\[ b = (-3, 5, 4) \]

### Step 4: Calculate the magnitude of the direction ratios

The magnitude of the vector \(b\) is:
\[ |b| = \sqrt{(-3)^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2} \]

### Step 5: Calculate the direction cosines

The direction cosines \(l, m, n\) can be calculated by dividing the direction ratios by the magnitude:
\[ l = \frac{-3}{5\sqrt{2}}, \quad m = \frac{5}{5\sqrt{2}}, \quad n = \frac{4}{5\sqrt{2}} \]

### Final Result

Thus, the direction cosines of the line are:
\[ \left( \frac{-3}{5\sqrt{2}}, \frac{5}{5\sqrt{2}}, \frac{4}{5\sqrt{2}} \right) \]