The equation of the plane which passes through
the points (2, 1, -1) and (-1, 3, 4) and
perpendicular to the plane `x- 2y + 4z = 0` is

Options

A. `18x + 17y + 4z = 49`
B. `18x - 17y + 4z = 49`
C. `18x + 17y - 4z + 49 = 0`
D. `18x +17y + 4z -49 = 0`

Answer

To find the equation of the plane that passes through the points \( A(2, 1, -1) \) and \( B(-1, 3, 4) \) and is perpendicular to the plane given by the equation \( x - 2y + 4z = 0 \), we can follow these steps:

### Step 1: Find the normal vector of the given plane
The equation of the plane \( x - 2y + 4z = 0 \) can be written in the form \( Ax + By + Cz + D = 0 \). The coefficients \( A, B, C \) give us the normal vector \( \mathbf{n} \) of the plane.

From the equation, we have:
\[ \mathbf{n} = \langle 1, -2, 4 \rangle \]

### Step 2: Find the direction vector of line segment AB
Next, we find the direction vector \( \overrightarrow{AB} \) from point \( A \) to point \( B \). This is calculated as:
\[ \overrightarrow{AB} = B - A = (-1 - 2, 3 - 1, 4 - (-1)) = (-3, 2, 5) \]

### Step 3: Find the cross product of the two vectors
The plane we are looking for must contain the vector \( \overrightarrow{AB} \) and be perpendicular to the normal vector \( \mathbf{n} \). To find a normal vector \( \mathbf{N} \) for the new plane, we take the cross product of \( \overrightarrow{AB} \) and \( \mathbf{n} \):
\[ \mathbf{N} = \overrightarrow{AB} \times \mathbf{n} \]
Calculating the cross product:
\[ \mathbf{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 2 & 5 \\ 1 & -2 & 4 \end{vmatrix} \]

Calculating this determinant:
\[ \mathbf{N} = \mathbf{i} \begin{vmatrix} 2 & 5 \\ -2 & 4 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -3 & 5 \\ 1 & 4 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -3 & 2 \\ 1 & -2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 2 & 5 \\ -2 & 4 \end{vmatrix} = (2)(4) - (5)(-2) = 8 + 10 = 18 \)
2. \( \begin{vmatrix} -3 & 5 \\ 1 & 4 \end{vmatrix} = (-3)(4) - (5)(1) = -12 - 5 = -17 \)
3. \( \begin{vmatrix} -3 & 2 \\ 1 & -2 \end{vmatrix} = (-3)(-2) - (2)(1) = 6 - 2 = 4 \)

Thus, we have:
\[ \mathbf{N} = 18\mathbf{i} + 17\mathbf{j} + 4\mathbf{k} \]

### Step 4: Use the point-normal form of the plane equation
The equation of a plane in point-normal form is given by:
\[ N_x(x - x_0) + N_y(y - y_0) + N_z(z - z_0) = 0 \]
where \( (x_0, y_0, z_0) \) is a point on the plane (we can use point \( A(2, 1, -1) \)) and \( (N_x, N_y, N_z) \) are the components of the normal vector \( \mathbf{N} \).

Substituting the values:
\[ 18(x - 2) + 17(y - 1) + 4(z + 1) = 0 \]

Expanding this:
\[ 18x - 36 + 17y - 17 + 4z + 4 = 0 \]
\[ 18x + 17y + 4z - 49 = 0 \]

### Final Answer
The equation of the plane is:
\[ 18x + 17y + 4z - 49 = 0 \]