The plane `x-y-z=4` is rotated through an angle `90^(@)` about its line of intersection with the plane `x+y+2z=4`. Then the equation of the plane in its new position is

Options

A. `x+y+4z=20`
B. `x+5y+4z=20`
C. `x+y-4z=20`
D. `5x+y+4z=20`

Answer

To find the equation of the plane after it has been rotated through an angle of \(90^\circ\) about its line of intersection with another plane, we can follow these steps:

### Step 1: Identify the equations of the planes
The equations of the given planes are:
1. Plane 1: \(x - y - z = 4\)
2. Plane 2: \(x + y + 2z = 4\)

### Step 2: Find the normal vectors of the planes
The normal vector of Plane 1, \( \vec{n_1} \), can be derived from its equation:
\[ \vec{n_1} = (1, -1, -1) \]
The normal vector of Plane 2, \( \vec{n_2} \), is:
\[ \vec{n_2} = (1, 1, 2) \]

### Step 3: Find the line of intersection of the two planes
To find the line of intersection, we can solve the two equations simultaneously. We can express one variable in terms of the others. For instance, from Plane 1:
\[ z = x - y - 4 \]
Substituting \(z\) into Plane 2:
\[ x + y + 2(x - y - 4) = 4 \]
Simplifying this gives:
\[ x + y + 2x - 2y - 8 = 4 \implies 3x - y = 12 \implies y = 3x - 12 \]
Now substituting \(y\) back into the expression for \(z\):
\[ z = x - (3x - 12) - 4 = -2x + 8 \]
Thus, the parametric equations of the line of intersection can be expressed as:
\[ x = t, \quad y = 3t - 12, \quad z = -2t + 8 \]

### Step 4: Find the direction vector of the line of intersection
The direction vector \( \vec{d} \) of the line of intersection can be obtained from the coefficients of \(t\):
\[ \vec{d} = (1, 3, -2) \]

### Step 5: Find the new normal vector after rotation
When Plane 1 is rotated \(90^\circ\) about the line of intersection, the new normal vector \( \vec{n'} \) can be found using the cross product of \( \vec{n_1} \) and \( \vec{d} \):
\[ \vec{n'} = \vec{n_1} \times \vec{d} \]
Calculating the cross product:
\[ \vec{n'} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 1 & 3 & -2 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{n'} = \hat{i}((-1)(-2) - (-1)(3)) - \hat{j}(1(-2) - (-1)(1)) + \hat{k}(1(3) - (-1)(1)) \]
\[ = \hat{i}(2 + 3) - \hat{j}(-2 + 1) + \hat{k}(3 + 1) \]
\[ = 5\hat{i} + 1\hat{j} + 4\hat{k} = (5, 1, 4) \]

### Step 6: Write the equation of the new plane
The equation of a plane can be expressed as:
\[ 5(x - x_0) + 1(y - y_0) + 4(z - z_0) = 0 \]
Using a point on the line of intersection, for example when \(t = 0\):
\[ x_0 = 0, \quad y_0 = -12, \quad z_0 = 8 \]
Substituting these values into the plane equation gives:
\[ 5(x - 0) + 1(y + 12) + 4(z - 8) = 0 \]
Simplifying this:
\[ 5x + y + 12 + 4z - 32 = 0 \implies 5x + y + 4z - 20 = 0 \]
Thus, the equation of the new plane is:
\[ 5x + y + 4z = 20 \]

### Final Answer
The equation of the plane in its new position is:
\[ 5x + y + 4z = 20 \]