The equation of the line passing through `(1, 1, 1)` and perpendicular to the line of intersection of the planes `x+2y-4z=0 and 2x-y+2z=0` is

Options

A. `(x-1)/(5)=(1-y)/(1)=(z-1)/(2)`
B. `(x-1)/(-5)=(1-y)/(1)=(z-1)/(2)`
C. `(x-1)/(0)=(1-y)/(-10)=(z-1)/(-5)`
D. `(x-1)/(-10)=(y+2)/(0)=(z-2)/(-5)`

Answer

To find the equation of the line passing through the point \( (1, 1, 1) \) and perpendicular to the line of intersection of the planes given by the equations \( x + 2y - 4z = 0 \) and \( 2x - y + 2z = 0 \), we can follow these steps:

### Step 1: Find the normal vectors of the planes
The normal vector of the first plane \( x + 2y - 4z = 0 \) is given by the coefficients of \( x, y, z \):
\[ \mathbf{n_1} = (1, 2, -4) \]
The normal vector of the second plane \( 2x - y + 2z = 0 \) is:
\[ \mathbf{n_2} = (2, -1, 2) \]

### Step 2: Find the direction vector of the line of intersection
To find the direction vector of the line of intersection of the two planes, we take the cross product of the normal vectors \( \mathbf{n_1} \) and \( \mathbf{n_2} \):
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -4 \\ 2 & -1 & 2 \end{vmatrix} \]

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

Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & -4 \\ -1 & 2 \end{vmatrix} = (2)(2) - (-4)(-1) = 4 - 4 = 0 \)
2. \( \begin{vmatrix} 1 & -4 \\ 2 & 2 \end{vmatrix} = (1)(2) - (-4)(2) = 2 + 8 = 10 \)
3. \( \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (1)(-1) - (2)(2) = -1 - 4 = -5 \)

Thus, we have:
\[ \mathbf{d} = 0 \mathbf{i} - 10 \mathbf{j} - 5 \mathbf{k} = (0, -10, -5) \]

### Step 3: Normalize the direction vector
To simplify, we can divide by -5:
\[ \mathbf{d} = (0, 2, 1) \]

### Step 4: Find the direction ratios of the required line
The line we are looking for is perpendicular to the line of intersection, so its direction ratios \( (L, M, N) \) must satisfy:
\[ 0L + 2M + 1N = 0 \]
This simplifies to:
\[ 2M + N = 0 \implies N = -2M \]

### Step 5: Choose a value for \( M \)
Let’s choose \( M = 1 \):
\[ N = -2(1) = -2 \]
Thus, the direction ratios of the required line are \( (L, 1, -2) \).

### Step 6: Write the equation of the line
The line passes through the point \( (1, 1, 1) \) and has direction ratios \( (L, 1, -2) \). The parametric equations of the line can be written as:
\[ \frac{x - 1}{L} = \frac{y - 1}{1} = \frac{z - 1}{-2} \]

### Step 7: Finalize the equation
Since \( L \) can be any non-zero value, we can set \( L = 1 \) for simplicity. Thus, the equation of the line becomes:
\[ \frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{-2} \]

This can be expressed as:
\[ x - 1 = y - 1 = \frac{1 - z}{2} \]

### Summary
The equation of the line passing through \( (1, 1, 1) \) and perpendicular to the line of intersection of the given planes is:
\[ \frac{x - 1}{1} = \frac{y - 1}{1} = \frac{z - 1}{-2} \]