The equation of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y - z = 0, is

Options

A. `(x-3)/(-2)=(y-0)/(1)=(z-1)/(3)_`
B. `(x-3)/(1)=(y-0)/(-2)=(z-1)/(3)`
C. `(x-3)/(3)=(y-0)/(1)=(z-1)/(-2)`
D. none of these

Answer

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

### Step 1: Identify the normal vectors of the planes
The normal vector of the first plane \(x + 2y = 0\) can be derived from its coefficients:
- Normal vector \( \mathbf{n_1} = (1, 2, 0) \)

The normal vector of the second plane \(3y - z = 0\) is:
- Normal vector \( \mathbf{n_2} = (0, 3, -1) \)

### Step 2: Find the direction vector of the line
The line we are looking for is parallel to both planes, which means its direction vector can be found by taking the cross product of the two normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \).

\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \]

Calculating the cross product:

\[ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 0 \\ 0 & 3 & -1 \end{vmatrix} \]

Expanding this determinant:

\[ \mathbf{d} = \mathbf{i} \begin{vmatrix} 2 & 0 \\ 3 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 0 & 3 \end{vmatrix} \]

Calculating the 2x2 determinants:

\[ \begin{vmatrix} 2 & 0 \\ 3 & -1 \end{vmatrix} = (2)(-1) - (0)(3) = -2 \]

\[ \begin{vmatrix} 1 & 0 \\ 0 & -1 \end{vmatrix} = (1)(-1) - (0)(0) = -1 \]

\[ \begin{vmatrix} 1 & 2 \\ 0 & 3 \end{vmatrix} = (1)(3) - (2)(0) = 3 \]

Putting it all together:

\[ \mathbf{d} = -2\mathbf{i} + 1\mathbf{j} + 3\mathbf{k} = (-2, 1, 3) \]

### Step 3: Write the parametric equations of the line
The line passing through the point (3, 0, 1) with direction vector \((-2, 1, 3)\) can be expressed in parametric form as:

\[ x = 3 - 2t \]
\[ y = 0 + 1t \]
\[ z = 1 + 3t \]

### Step 4: Write the symmetric form of the line
The symmetric equations of the line can be written as:

\[ \frac{x - 3}{-2} = \frac{y - 0}{1} = \frac{z - 1}{3} \]

### Final Answer
Thus, the equation of the line is:

\[ \frac{x - 3}{-2} = \frac{y}{1} = \frac{z - 1}{3} \]

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