Question

A plane \( (\pi) \) passing through the point \( (1,2,-3) \) is perpendicular to the planes \( x + y - z + 4 = 0 \) and \( 2x - y + z + 1 = 0 \). If the equation of the plane \( (\pi) \) is \( ax + by + cz + 1 = 0 \), then \( a^2 + b^2 + c^2 \) is equal to:

• A. \( 4 \)
• B. \( 3 \)
• C. \( 2 \)
• D. \( 1 \)

Hint:

The normal to a plane can be found by taking the cross product of the normal vectors of two given perpendicular planes.
- Always check for scaling factors when finding normal vectors.

Answer 1

The Correct Option is C

A plane perpendicular to two given planes will have its normal vector parallel to the cross product of the normal vectors of the given planes.

Step 1: Find the normal vectors

The normal to the plane \( x + y - z + 4 = 0 \) is:

\[ \mathbf{n_1} = (1,1,-1) \]

The normal to the plane \( 2x - y + z + 1 = 0 \) is:

\[ \mathbf{n_2} = (2,-1,1) \]

Step 2: Compute the cross product \( \mathbf{n_1} \times \mathbf{n_2} \)

\[ \mathbf{n} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -1 \\ 2 & -1 & 1 \end{vmatrix} \]

Expanding the determinant:

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

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

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

\[ = 0\mathbf{i} -3\mathbf{j} -3\mathbf{k} \]

\[ = (0,-3,-3) \]

Step 3: Compute \( a^2 + b^2 + c^2 \)

Since the normal to the required plane is parallel to \( (0,-3,-3) \), we take:

\[ a = 0, \quad b = -3, \quad c = -3 \]

Thus,

\[ a^2 + b^2 + c^2 = 0^2 + (-3)^2 + (-3)^2 = 0 + 9 + 9 = 18 \]

Since normal vectors can be scaled, dividing by 9, we get:

\[ \frac{18}{9} = 2 \]

Thus, the correct answer is \( \boxed{2} \).