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 represented by the equations \(x - y + z - 5 = 0\) and \(x - 3y - 6 = 0\), we can follow these steps:

### Step 1: Identify the equations of the lines
We have two equations:
1. \(x - y + z - 5 = 0\)
2. \(x - 3y - 6 = 0\)

### Step 2: Rewrite the equations in parametric form
To find the direction ratios, we can express these equations in a parametric form.

From the first equation \(x - y + z - 5 = 0\), we can express \(z\) in terms of \(x\) and \(y\):
\[ z = 5 - x + y \]

From the second equation \(x - 3y - 6 = 0\), we can express \(x\) in terms of \(y\):
\[ x = 3y + 6 \]

### Step 3: Substitute to find a relationship
Substituting \(x = 3y + 6\) into the first equation:
\[ z = 5 - (3y + 6) + y \]
\[ z = 5 - 3y - 6 + y = -2y - 1 \]

### Step 4: Parametric equations
Now we can express \(x\), \(y\), and \(z\) in terms of a parameter \(t\):
Let \(y = t\). Then:
\[ x = 3t + 6 \]
\[ z = -2t - 1 \]

So, the parametric equations of the line are:
\[ x = 3t + 6, \quad y = t, \quad z = -2t - 1 \]

### Step 5: Find direction ratios
The direction ratios can be derived from the coefficients of \(t\) in the parametric equations:
- The coefficient of \(t\) in \(x\) is \(3\),
- The coefficient of \(t\) in \(y\) is \(1\),
- The coefficient of \(t\) in \(z\) is \(-2\).

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

### Step 6: Conclusion
Therefore, the direction ratios of the line represented by the given equations are \(3, 1, -2\).

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