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

Options

A. `3,1,-2`
B. `2,-4,1`
C. `(3)/(sqrt14), (1)/(sqrt14), (-2)/(sqrt14)`
D. `(2)/(sqrt14), (-4)/(sqrt14), (1)/(sqrt14)`

Answer

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

### Step 1: Write the equations in a more manageable form
We have two equations:
1. \( x - y + z - 5 = 0 \)
2. \( x - 3y - 6 = 0 \)

### Step 2: Express one variable in terms of another
From the second equation, we can express \( x \) in terms of \( y \):
\[ x = 3y + 6 \]

### Step 3: Substitute \( x \) into the first equation
Now, substitute \( x = 3y + 6 \) into the first equation:
\[ (3y + 6) - y + z - 5 = 0 \]
Simplifying this gives:
\[ 3y + 6 - y + z - 5 = 0 \\ 2y + z + 1 = 0 \\ z = -2y - 1 \]

### Step 4: Express \( y \) in terms of \( z \)
From the equation \( z = -2y - 1 \), we can express \( y \) in terms of \( z \):
\[ 2y = -z - 1 \\ y = -\frac{z + 1}{2} \]

### Step 5: Write the parametric equations
Now we have:
- \( x = 3y + 6 \)
- \( y = y \)
- \( z = -2y - 1 \)

We can express \( y \) in terms of a parameter \( t \):
Let \( y = t \), then:
\[ x = 3t + 6 \\ z = -2t - 1 \]

### Step 6: Write the parametric equations in standard form
The parametric equations can be written as:
\[ \begin{align*} x &= 3t + 6 \\ y &= t \\ z &= -2t - 1 \end{align*} \]

### Step 7: Identify the direction ratios
From the parametric equations, the coefficients of \( t \) give us the direction ratios of the line:
- For \( x \): coefficient is 3
- For \( y \): coefficient is 1
- For \( z \): coefficient is -2

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

### Final Answer
The direction ratios of the line are \( (3, 1, -2) \).

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