Question

The plane that is perpendicular to the planes \(x-y+2z-4=0\) and  \(2x-2y+z=0\) and passes through \( (1,-2,1)\) is 

• A. \(x+y+1=0\)
• B. \(2x+y+z-1=0 \)
• C. \(x+y+z=0\)
• D. \(2x+y-z+1=0 \)
• E. \(x+z-2=0 \)

Answer 1

The Correct Option is A

We need to find the equation of a plane that is perpendicular to both given planes and passes through the point \((1, -2, 1)\).

Step 1: Find the normal vectors of the given planes

The first plane \( x - y + 2z - 4 = 0 \) has normal vector \( \mathbf{n}_1 = \langle 1, -1, 2 \rangle \).

The second plane \( 2x - 2y + z = 0 \) has normal vector \( \mathbf{n}_2 = \langle 2, -2, 1 \rangle \).

Step 2: Find the direction vector of the line of intersection (parallel to the desired plane's normal)

The desired plane must be perpendicular to both given planes, so its normal vector \( \mathbf{n} \) must be perpendicular to both \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \). Thus, we compute the cross product:

\[ \mathbf{n} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix} \]

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

\[ = \mathbf{i}(-1 + 4) - \mathbf{j}(1 - 4) + \mathbf{k}(-2 + 2) \]

\[ = 3\mathbf{i} + 3\mathbf{j} + 0\mathbf{k} = \langle 3, 3, 0 \rangle \]

We can simplify this to \( \langle 1, 1, 0 \rangle \) by dividing by 3.

Step 3: Write the equation of the plane

Using the normal vector \( \langle 1, 1, 0 \rangle \) and the point \( (1, -2, 1) \), the plane equation is:

\[ 1(x - 1) + 1(y + 2) + 0(z - 1) = 0 \]

\[ x - 1 + y + 2 = 0 \]

\[ x + y + 1 = 0 \]

Final Answer:

(A) \( x + y + 1 = 0 \)

Answer 2

Step 1: Understand the problem and given information.

We are tasked with finding the equation of a plane that is perpendicular to two given planes:

• Plane 1: \( x - y + 2z - 4 = 0 \)
• Plane 2: \( 2x - 2y + z = 0 \)

The required plane also passes through the point \( (1, -2, 1) \).

Step 2: Find the normal vector of the required plane.

The normal vector of a plane perpendicular to two given planes is parallel to the cross product of the normal vectors of the two planes.

Let the normal vector of Plane 1 be \( \vec{n}_1 = (1, -1, 2) \), and the normal vector of Plane 2 be \( \vec{n}_2 = (2, -2, 1) \).

Compute the cross product \( \vec{n}_1 \times \vec{n}_2 \):

\[ \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ 2 & -2 & 1 \end{vmatrix}. \]

Expand the determinant:

\[ \vec{n}_1 \times \vec{n}_2 = \mathbf{i} \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & -1 \\ 2 & -2 \end{vmatrix}. \]

Compute each minor determinant:

\[ \begin{vmatrix} -1 & 2 \\ -2 & 1 \end{vmatrix} = (-1)(1) - (2)(-2) = -1 + 4 = 3, \]

\[ \begin{vmatrix} 1 & 2 \\ 2 & 1 \end{vmatrix} = (1)(1) - (2)(2) = 1 - 4 = -3, \]

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

Substitute back into the cross product:

\[ \vec{n}_1 \times \vec{n}_2 = 3\mathbf{i} - (-3)\mathbf{j} + 0\mathbf{k} = 3\mathbf{i} + 3\mathbf{j}. \]

Thus, the normal vector of the required plane is:

\[ \vec{n} = (3, 3, 0). \]

Step 3: Write the equation of the plane.

The general equation of a plane is:

\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0, \]

where \( (a, b, c) \) is the normal vector and \( (x_0, y_0, z_0) \) is a point on the plane.

Substitute \( \vec{n} = (3, 3, 0) \) and \( (x_0, y_0, z_0) = (1, -2, 1) \):

\[ 3(x - 1) + 3(y + 2) + 0(z - 1) = 0. \]

Simplify:

\[ 3x - 3 + 3y + 6 = 0. \]

\[ 3x + 3y + 3 = 0. \]

Divide through by 3:

\[ x + y + 1 = 0. \]

Final Answer:

\( x + y + 1 = 0 \)