To find the equation of the line of projection of the given line onto the specified plane, we will follow these steps:
### Step 1: Identify the equations of the given line and plane
The equations of the line are given as:
1. \( 3xy + 2z - 1 = 0 \)
2. \( x + 2y - z = 2 \)
The equation of the plane is:
\( 3x + 2y + z = 0 \)
### Step 2: Express the line in parametric form
To express the line in parametric form, we need to find a point on the line and the direction vector of the line.
From the second equation \( x + 2y - z = 2 \), we can express \( z \) in terms of \( x \) and \( y \):
\[ z = x + 2y - 2 \]
Substituting this into the first equation:
\[ 3xy + 2(x + 2y - 2) - 1 = 0 \]
\[ 3xy + 2x + 4y - 4 - 1 = 0 \]
\[ 3xy + 2x + 4y - 5 = 0 \]
This equation can be solved for \( y \) in terms of \( x \) or vice versa, but for simplicity, we can choose specific values for \( x \) and \( y \) to find a point on the line.
Let’s choose \( x = 1 \):
\[ 3(1)y + 2(1) + 4y - 5 = 0 \]
\[ 3y + 2 + 4y - 5 = 0 \]
\[ 7y - 3 = 0 \]
\[ y = \frac{3}{7} \]
Now substituting \( y = \frac{3}{7} \) back to find \( z \):
\[ z = 1 + 2\left(\frac{3}{7}\right) - 2 = 1 + \frac{6}{7} - 2 = \frac{1}{7} \]
Thus, a point on the line is \( (1, \frac{3}{7}, \frac{1}{7}) \).
### Step 3: Find the direction vector of the line
To find the direction vector of the line, we can differentiate the parametric equations or use the coefficients from the equations. The direction vector can be derived from the coefficients of \( x \) and \( y \) in the line equations.
From the equation \( x + 2y - z = 2 \):
- The direction vector can be taken as \( (1, 2, 1) \).
### Step 4: Find the normal vector of the plane
The normal vector \( \vec{n} \) of the plane \( 3x + 2y + z = 0 \) is given by the coefficients of \( x, y, z \):
\[ \vec{n} = (3, 2, 1) \]
### Step 5: Project the direction vector onto the plane
To find the projection of the direction vector \( \vec{d} = (1, 2, 1) \) onto the plane, we can use the formula:
\[ \vec{d}_{\text{proj}} = \vec{d} - \frac{\vec{d} \cdot \vec{n}}{\vec{n} \cdot \vec{n}} \vec{n} \]
Calculating \( \vec{d} \cdot \vec{n} \):
\[ \vec{d} \cdot \vec{n} = 1 \cdot 3 + 2 \cdot 2 + 1 \cdot 1 = 3 + 4 + 1 = 8 \]
Calculating \( \vec{n} \cdot \vec{n} \):
\[ \vec{n} \cdot \vec{n} = 3^2 + 2^2 + 1^2 = 9 + 4 + 1 = 14 \]
Now substituting back:
\[ \vec{d}_{\text{proj}} = (1, 2, 1) - \frac{8}{14}(3, 2, 1) \]
\[ = (1, 2, 1) - \left(\frac{24}{14}, \frac{16}{14}, \frac{8}{14}\right) \]
\[ = (1 - \frac{12}{7}, 2 - \frac{8}{7}, 1 - \frac{4}{7}) \]
\[ = \left(\frac{7 - 12}{7}, \frac{14 - 8}{7}, \frac{7 - 4}{7}\right) \]
\[ = \left(-\frac{5}{7}, \frac{6}{7}, \frac{3}{7}\right) \]
### Step 6: Write the equation of the line of projection
The line of projection can be expressed in parametric form using the point \( (1, \frac{3}{7}, \frac{1}{7}) \) and the projected direction vector \( \left(-\frac{5}{7}, \frac{6}{7}, \frac{3}{7}\right) \).
Let \( t \) be the parameter:
\[ x = 1 - \frac{5}{7}t \]
\[ y = \frac{3}{7} + \frac{6}{7}t \]
\[ z = \frac{1}{7} + \frac{3}{7}t \]
### Final Answer
The equation of the line of projection is:
\[
\begin{align*}
x &= 1 - \frac{5}{7}t \\
y &= \frac{3}{7} + \frac{6}{7}t \\
z &= \frac{1}{7} + \frac{3}{7}t
\end{align*}
\]
