Question

The equation of the plane passing through the point \( (1, 2, 2) \) and perpendicular to the planes \[ x - y + 2z = 3 \quad \text{and} \quad 2x - 2y + z + 12 = 0 \] is:

• A. \( x - 2y + 2z - 1 = 0 \)
• B. \( 2x - 3y + 4z - 4 = 0 \)
• C. \( x + y + z - 5 = 0 \)
• D. \( x + y - 3 = 0 \)

Hint:

If a plane is perpendicular to two other planes, its normal is the cross product of the two normals.

Answer 1

The Correct Option is D

Let normals to the given planes be: \[ \vec{n}_1 = \langle 1, -1, 2 \rangle, \quad \vec{n}_2 = \langle 2, -2, 1 \rangle \] Since required plane is perpendicular to both planes, its normal vector is the cross product of \( \vec{n}_1 \times \vec{n}_2 \): \[ \vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 2
2 & -2 & 1 \end{vmatrix} = (-1)(1) - (2)(-2) \hat{i} - (1)(1) - (2)(2) \hat{j} + (1)(-2) - (-1)(2) \hat{k} = \langle 0, -5, 0 \rangle \Rightarrow \text{Normal: } \hat{i} + \hat{j} \] Equation of plane through \( (1, 2, 2) \) and normal \( (1, 1, 0) \): \[ (x - 1) + (y - 2) = 0 \Rightarrow x + y - 3 = 0 \]