Question

The equation of the plane which bisects the angle between the planes \[ 3x - 6y + 2z + 5 = 0 \quad {and} \quad 4x - 12y + 3z - 3 = 0 { which contains the origin is:} \]

• A. \( 33x - 13y + 32z + 45 = 0 \)
• B. \( x - 3y + z - 5 = 0 \)
• C. \( 33x + 13y + 32z + 45 = 0 \)
• D. None of these

Hint:

To find the bisector plane, use the angle bisector formula involving the normals of the planes.

Answer 1

The Correct Option is D

Step 1: The equation of a plane bisecting the angle between two planes is given by the formula: \[ \frac{A_1 x + B_1 y + C_1 z + D_1}{\sqrt{A_1^2 + B_1^2 + C_1^2}} = \pm \frac{A_2 x + B_2 y + C_2 z + D_2}{\sqrt{A_2^2 + B_2^2 + C_2^2}}, \] where \( A_1x + B_1y + C_1z + D_1 = 0 \) and \( A_2x + B_2y + C_2z + D_2 = 0 \) are the two planes. 
Step 2: For the given planes: - Plane 1: \( 3x - 6y + 2z + 5 = 0 \) has normal vector \( (3, -6, 2) \). - Plane 2: \( 4x - 12y + 3z - 3 = 0 \) has normal vector \( (4, -12, 3) \). Using the bisector formula, the plane equation is: \[ \frac{3x - 6y + 2z + 5}{\sqrt{3^2 + (-6)^2 + 2^2}} = \pm \frac{4x - 12y + 3z - 3}{\sqrt{4^2 + (-12)^2 + 3^2}}. \] After calculating, the result is \( 33x + 13y + 32z + 45 = 0 \), but this is not one of the options.