Find the equation of line of intersection of the planes `3x-y+ z=1 and x + 4 y -2 z =2.`

Answer

To find the equation of the line of intersection of the planes given by the equations \(3x - y + z = 1\) and \(x + 4y - 2z = 2\), we can follow these steps:

### Step 1: Write the equations of the planes in standard form
The equations of the planes are already in standard form:
1. Plane 1: \(3x - y + z - 1 = 0\) (let's denote this as \(P_1\))
2. Plane 2: \(x + 4y - 2z - 2 = 0\) (let's denote this as \(P_2\))

### Step 2: Find the normal vectors of the planes
The normal vector of plane \(P_1\) is given by the coefficients of \(x\), \(y\), and \(z\) in its equation:
- Normal vector of \(P_1\): \(\vec{n_1} = (3, -1, 1)\)

The normal vector of plane \(P_2\) is:
- Normal vector of \(P_2\): \(\vec{n_2} = (1, 4, -2)\)

### Step 3: Find the direction vector of the line of intersection
The direction vector \(\vec{d}\) of the line of intersection can be found by taking the cross product of the normal vectors \(\vec{n_1}\) and \(\vec{n_2}\):
\[ \vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 4 & -2 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{d} = \hat{i} \begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} -1 & 1 \\ 4 & -2 \end{vmatrix} = (-1)(-2) - (1)(4) = 2 - 4 = -2\)

2. \(\begin{vmatrix} 3 & 1 \\ 1 & -2 \end{vmatrix} = (3)(-2) - (1)(1) = -6 - 1 = -7\)

3. \(\begin{vmatrix} 3 & -1 \\ 1 & 4 \end{vmatrix} = (3)(4) - (-1)(1) = 12 + 1 = 13\)

Thus, we have:
\[ \vec{d} = -2\hat{i} + 7\hat{j} + 13\hat{k} = (-2, 7, 13) \]

### Step 4: Find a point on the line of intersection
To find a point on the line, we can solve the system of equations formed by the two planes. We can express \(z\) in terms of \(x\) and \(y\) using one of the plane equations. Let's use the first plane:
\[ z = 1 - 3x + y \]

Substituting \(z\) into the second plane's equation:
\[ x + 4y - 2(1 - 3x + y) = 2 \]
Simplifying this:
\[ x + 4y - 2 + 6x - 2y = 2 \]
\[ 7x + 2y - 4 = 2 \]
\[ 7x + 2y = 6 \quad \text{(Equation 1)} \]

Now, we can choose a value for \(x\) to find corresponding \(y\) and \(z\). Let’s take \(x = 0\):
\[ 7(0) + 2y = 6 \implies 2y = 6 \implies y = 3 \]
Substituting \(x = 0\) and \(y = 3\) into the equation for \(z\):
\[ z = 1 - 3(0) + 3 = 4 \]
Thus, a point on the line is \((0, 3, 4)\).

### Step 5: Write the parametric equations of the line
The parametric equations of the line can be expressed as:
\[ x = 0 - 2t = -2t \]
\[ y = 3 + 7t \]
\[ z = 4 + 13t \]

### Final Answer
The equation of the line of intersection of the two planes is given by:
\[ \begin{cases} x = -2t \\ y = 3 + 7t \\ z = 4 + 13t \end{cases} \]