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.

Answer 2

Step 1: Understanding the Problem
We are given two planes with the equations:
Plane 1: \( x - 2y - 2z + 1 = 0 \)
Plane 2: \( 2x - 3y - 6z + 1 = 0 \)
The acute angle bisector of these planes is the plane \( P \), and we are asked to find which of the given points lies on this plane.
Step 2: General Approach
The acute angle bisector of two planes is given by the equation:
\[ \frac{x - 2y - 2z + 1}{\sqrt{1^2 + (-2)^2 + (-2)^2}} = \pm \frac{2x - 3y - 6z + 1}{\sqrt{2^2 + (-3)^2 + (-6)^2}} \] This equation represents the bisector, where we will take the positive sign for the acute angle. We will now calculate both sides.
Step 3: Simplify the Expressions
For the first plane, the normal vector is \( (1, -2, -2) \), so the magnitude of the normal vector is:
\[ \sqrt{1^2 + (-2)^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \] For the second plane, the normal vector is \( (2, -3, -6) \), and the magnitude of the normal vector is:
\[ \sqrt{2^2 + (-3)^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] Now, the equation of the acute angle bisector plane \( P \) becomes:
\[ \frac{x - 2y - 2z + 1}{3} = \frac{2x - 3y - 6z + 1}{7} \] Step 4: Solve the Equation
Cross-multiply to solve for the equation of the angle bisector plane:
\[ 7(x - 2y - 2z + 1) = 3(2x - 3y - 6z + 1) \] Expanding both sides:
\[ 7x - 14y - 14z + 7 = 6x - 9y - 18z + 3 \] Now, simplify and collect like terms:
\[ 7x - 6x - 14y + 9y - 14z + 18z + 7 - 3 = 0 \] \[ x - 5y + 4z + 4 = 0 \] This is the equation of the acute angle bisector plane \( P \).
Step 5: Check Which Point Lies on Plane \( P \)
We are given four points to check. Let’s substitute each point into the equation of the plane and see which satisfies it.
Let’s test the point \( (-2, 0, -\frac{1}{2}) \):
Substitute \( x = -2 \), \( y = 0 \), and \( z = -\frac{1}{2} \) into the equation \( x - 5y + 4z + 4 = 0 \):
\[ -2 - 5(0) + 4\left(-\frac{1}{2}\right) + 4 = -2 + 0 - 2 + 4 = 0 \] Thus, the point \( (-2, 0, -\frac{1}{2}) \) satisfies the equation of the acute angle bisector plane \( P \).
Step 6: Conclusion
The point \( (-2, 0, -\frac{1}{2}) \) lies on the plane \( P \).