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