Equation of the line passing through the point `(1,2,3)` and parellel to the plane `2x+3y+z+5=0` is

Options

A. `(x-1)/-1=(y-2)/1=(z-3)/-1`
B. `(x-1)/2=(y-2)/3=(z-3)/1`
C. `(x-1)/1=(y-2)/4=(z-3)/7`
D. `(x-1)/3=(y-2)/4=(z-3)/2`

Answer

To find the equation of the line passing through the point \( (1, 2, 3) \) and parallel to the plane given by the equation \( 2x + 3y + z + 5 = 0 \), we can follow these steps:

### Step 1: Identify the normal vector of the plane
The normal vector \( \mathbf{n} \) of the plane can be derived from the coefficients of \( x, y, \) and \( z \) in the plane equation. For the plane \( 2x + 3y + z + 5 = 0 \), the normal vector is:
\[ \mathbf{n} = (2, 3, 1) \]

### Step 2: Determine the direction vector of the line
Since the line is parallel to the plane, its direction vector \( \mathbf{d} \) must be perpendicular to the normal vector of the plane. To find a suitable direction vector, we can choose any vector that is not a scalar multiple of \( \mathbf{n} \). For simplicity, we can choose:
\[ \mathbf{d} = (-3, 2, 0) \]
This choice is arbitrary, as there are infinitely many vectors that can be chosen as long as they are not parallel to \( \mathbf{n} \).

### Step 3: Write the parametric equations of the line
Using the point \( (1, 2, 3) \) and the direction vector \( (-3, 2, 0) \), we can write the parametric equations of the line:
\[ x = 1 - 3t \]
\[ y = 2 + 2t \]
\[ z = 3 + 0t \]
where \( t \) is a parameter.

### Step 4: Convert to symmetric form
The symmetric form of the line can be expressed as:
\[ \frac{x - 1}{-3} = \frac{y - 2}{2} = \frac{z - 3}{0} \]
Since the \( z \) component does not change with \( t \), we can represent it as:
\[ \frac{x - 1}{-3} = \frac{y - 2}{2} = k \]
where \( k \) is a parameter representing the ratio.

### Final Equation of the Line
Thus, the equation of the line passing through the point \( (1, 2, 3) \) and parallel to the plane \( 2x + 3y + z + 5 = 0 \) can be written as:
\[ \frac{x - 1}{-3} = \frac{y - 2}{2} = z - 3 \]