The direction ratios of the line `x-y+z-5=0=x-3y-6` are

Options

A. `3, 1, -2`
B. `2, -4, 1`
C. `(3)/(sqrt(14)), (1)/(sqrt(14)), (-2)/(sqrt(14))`
D. `(2)/(sqrt(21)), (-4)/(sqrt(21)), (1)/(sqrt(21))`

Answer

To find the direction ratios of the line defined by the equations \( x - y + z - 5 = 0 \) and \( x - 3y - 6 = 0 \), we will follow these steps:

### Step 1: Identify the equations of the planes
The given equations represent two planes:
1. \( x - y + z - 5 = 0 \) (Plane 1)
2. \( x - 3y - 6 = 0 \) (Plane 2)

### Step 2: Determine the normal vectors of the planes
The direction ratios of the line of intersection of two planes can be found using the cross product of their normal vectors.

- For Plane 1, the normal vector \( \mathbf{n_1} \) can be derived from the coefficients of \( x, y, z \):
\[ \mathbf{n_1} = (1, -1, 1) \]

- For Plane 2, the normal vector \( \mathbf{n_2} \) is:
\[ \mathbf{n_2} = (1, -3, 0) \]

### Step 3: Calculate the cross product of the normal vectors
To find the direction ratios of the line, we need to calculate the cross product \( \mathbf{n_1} \times \mathbf{n_2} \).

\[ \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 1 & -3 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ = \mathbf{i} \left((-1) \cdot 0 - 1 \cdot (-3)\right) - \mathbf{j} \left(1 \cdot 0 - 1 \cdot 1\right) + \mathbf{k} \left(1 \cdot (-3) - (-1) \cdot 1\right) \]
\[ = \mathbf{i} (0 + 3) - \mathbf{j} (0 - 1) + \mathbf{k} (-3 + 1) \]
\[ = 3\mathbf{i} + 1\mathbf{j} - 2\mathbf{k} \]

Thus, the direction ratios of the line are:
\[ (3, 1, -2) \]

### Step 4: Conclusion
The direction ratios of the line are \( 3, 1, -2 \).

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