The vector equation of the line passing through the point (-1, -1 ,2) and parallel to the line 2x - 2= 3y +1 = 6z -2 is

Options

A. `(11)/(70)`
B. `(70)/(11)`
C. ` - (10)/(11)`
D. `(10)/(11)`.

Answer

To find the vector equation of the line passing through the point (-1, -1, 2) and parallel to the given line, we can follow these steps:

### Step 1: Identify the given line's equation
The line is given in the form:
\[ 2x - 2 = 3y + 1 = 6z - 2 \]

### Step 2: Rewrite the equation in parametric form
We can express the equation in parametric form. Let's set:
\[ 2x - 2 = t \]
From this, we can express \( x \):
\[ x = \frac{t}{2} + 1 \]

Next, for \( y \):
\[ 3y + 1 = t \]
\[ y = \frac{t - 1}{3} \]

And for \( z \):
\[ 6z - 2 = t \]
\[ z = \frac{t + 2}{6} \]

### Step 3: Determine the direction ratios
From the parametric equations, we can find the direction ratios of the line:
- For \( x \): \( \frac{1}{2} \)
- For \( y \): \( \frac{1}{3} \)
- For \( z \): \( \frac{1}{6} \)

Thus, the direction ratios can be represented as the vector:
\[ \mathbf{b} = \left( \frac{1}{2}, \frac{1}{3}, \frac{1}{6} \right) \]

### Step 4: Scale the direction ratios
To simplify calculations, we can multiply the direction ratios by 6:
\[ \mathbf{b} = (3, 2, 1) \]

### Step 5: Write the vector equation of the line
The vector equation of a line can be expressed as:
\[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} \]
Where:
- \( \mathbf{a} \) is the position vector of the point (-1, -1, 2), which is:
\[ \mathbf{a} = -1 \mathbf{i} - 1 \mathbf{j} + 2 \mathbf{k} \]
- \( \lambda \) is a parameter.

Substituting \( \mathbf{a} \) and \( \mathbf{b} \):
\[ \mathbf{r} = (-1 \mathbf{i} - 1 \mathbf{j} + 2 \mathbf{k}) + \lambda (3 \mathbf{i} + 2 \mathbf{j} + 1 \mathbf{k}) \]

### Final vector equation
Thus, the vector equation of the line is:
\[ \mathbf{r} = (-1 + 3\lambda) \mathbf{i} + (-1 + 2\lambda) \mathbf{j} + (2 + \lambda) \mathbf{k} \]