Question

Let the acute angle bisector of the two planes \( x - 2y - 2z + 1 = 0 \) and \( 2x - 3y - 6z + 1 = 0 \) be the plane \( P \). Then which of the following points lies on \( P \)?

• A. \( (3, 1, -\frac{1}{2}) \)
• B. \( (-2, 0, -\frac{1}{2}) \)
• C. \( (0, 2, -4) \)
• D. \( (4, 0, -2) \)

Hint:

When solving for points on the angle bisector of two planes, ensure you correctly apply the formula for the angle bisector and test the points by substituting their coordinates.

Answer 1

The Correct Option is B

The equation of the acute angle bisector of two planes is given by the following formula: \[ \frac{x - 2y - 2z + 1}{\sqrt{1^2 + (-2)^2 + (-2)^2}} = \pm \frac{2x - 3y - 6z + 1}{\sqrt{2^2 + (-3)^2 + (-6)^2}} \] Substituting the coordinates of each point in the options, we find that the point \( (-2, 0, -\frac{1}{2}) \) satisfies the equation of the plane \( P \). Thus, the point lies on the acute angle bisector. Given the equations of planes: \[ P_1: x - 2y - 2z + 1 = 0 \] \[ P_2: 2x - 3y - 6z + 1 = 0 \] The equation of the plane bisectors is given by: \[ \frac{x - 2y - 2z + 1}{\sqrt{1^2 + (-2)^2 + (-2)^2}} = \pm \frac{2x - 3y - 6z + 1}{\sqrt{2^2 + (-3)^2 + (-6)^2}} \] \[ \frac{x - 2y - 2z + 1}{3} = \pm \frac{2x - 3y - 6z + 1}{7} \] Since \( a_1a_2 + b_1b_2 + c_1c_2 = 20>0 \), we choose the negative sign for the acute bisector: \[ \frac{x - 2y - 2z + 1}{3} = -\frac{2x - 3y - 6z + 1}{7} \] \[ 7(x - 2y - 2z + 1) = -3(2x - 3y - 6z + 1) \] \[ 7x - 14y - 14z + 7 = -6x + 9y + 18z - 3 \] \[ 13x - 23y - 32z + 10 = 0 \] The point \((-2, 0, -\frac{1}{2})\) satisfies this equation.