Key Concepts and Formulas

• Perpendicular Bisector Plane: A plane that bisects a line segment $AB$ at right angles has two key properties:
  1. Its normal vector is parallel to the direction vector of the line segment $AB$.
  2. It passes through the midpoint of the line segment $AB$.
• Equation of a Plane: The general equation of a plane passing through a point $(x_0, y_0, z_0)$ and having a normal vector $\vec{n} = \langle a, b, c \rangle$ is given by: $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$
• Midpoint Formula: The midpoint $M$ of a line segment joining two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is calculated as: $M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)$
• Direction Vector: The direction vector of a line segment joining point $A(x_1, y_1, z_1)$ to point $B(x_2, y_2, z_2)$ is $\vec{AB} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle$.

Step-by-Step Solution

Let the given points be $A = (4, -2, 3)$ and $B = (2, 4, -1)$.

Step 1: Determine the Normal Vector of the Plane

• What and Why: Since the plane bisects the line segment $AB$ at right angles, it means the plane is perpendicular to the line segment $AB$. Therefore, the direction vector of the line segment $\vec{AB}$ (or $\vec{BA}$) will serve as the normal vector to the plane.
• Calculation: We calculate the direction vector $\vec{BA}$. Let $\vec{n}$ be the normal vector. $\vec{n} = \vec{BA} = \langle x_A - x_B, y_A - y_B, z_A - z_B \rangle$ $\vec{n} = \langle 4 - 2, -2 - 4, 3 - (-1) \rangle$ $\vec{n} = \langle 2, -6, 4 \rangle$ Thus, the direction ratios of the normal to the plane are $a = 2$, $b = -6$, $c = 4$.

Step 2: Find a Point on the Plane

• What and Why: The problem states that the plane bisects the line segment $AB$. This implies that the plane must pass through the exact midpoint of the segment $AB$. We use the midpoint formula to find this point.
• Calculation: Using the midpoint formula for points $A(4, -2, 3)$ and $B(2, 4, -1)$: $M = \left(\frac{4 + 2}{2}, \frac{-2 + 4}{2}, \frac{3 + (-1)}{2}\right)$ $M = \left(\frac{6}{2}, \frac{2}{2}, \frac{2}{2}\right)$ $M = (3, 1, 1)$ So, the plane passes through the point $(x_0, y_0, z_0) = (3, 1, 1)$.

Step 3: Formulate the Equation of the Plane

• What and Why: Now that we have the normal vector $\vec{n} = \langle 2, -6, 4 \rangle$ and a point on the plane $(3, 1, 1)$, we can substitute these values into the general equation of a plane to find its specific equation.
• Calculation: Substitute $a=2$, $b=-6$, $c=4$ and $x_0=3$, $y_0=1$, $z_0=1$ into the plane equation $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$: $2(x - 3) - 6(y - 1) + 4(z - 1) = 0$ Expand the equation: $2x - 6 - 6y + 6 + 4z - 4 = 0$ Combine the constant terms: $2x - 6y + 4z - 4 = 0$ To simplify the equation, divide the entire equation by 2: $x - 3y + 2z - 2 = 0$ Rearranging it, the equation of the plane is: $x - 3y + 2z = 2$

Step 4: Check the Given Options

• What and Why: The question asks which of the given points the plane passes through. We have the equation of the plane, so we will substitute the coordinates of each option into this equation to see which one satisfies it (i.e., makes the left-hand side equal to the right-hand side, which is 2).
• Checking Options:

  • (A) $(4, 0, 1)$: Substitute into $x - 3y + 2z$: $4 - 3(0) + 2(1) = 4 - 0 + 2 = 6$. Since $6 \neq 2$, option (A) is incorrect.

  • (B) $(0, -1, 1)$: Substitute into $x - 3y + 2z$: $0 - 3(-1) + 2(1) = 0 + 3 + 2 = 5$. Since $5 \neq 2$, option (B) is incorrect.

  • (C) $(0, 1, -1)$: Substitute into $x - 3y + 2z$: $0 - 3(1) + 2(-1) = 0 - 3 - 2 = -5$. Since $-5 \neq 2$, option (C) is incorrect.

  • (D) $(4, 0, -1)$: Substitute into $x - 3y + 2z$: $4 - 3(0) + 2(-1) = 4 - 0 - 2 = 2$. Since $2 = 2$, option (D) satisfies the equation.

Therefore, the plane passes through the point $(4, 0, -1)$.

Common Mistakes & Tips

• Confusing Normal Vector with Direction Vector: Remember that for a plane perpendicular to a line segment, the direction vector of the segment is the normal vector of the plane.
• Forgetting the Midpoint: A common error is to use one of the endpoints of the segment instead of the midpoint. The "bisects" part of the problem statement explicitly requires the midpoint.
• Algebraic Precision: Be meticulous with calculations, especially when dealing with negative signs in coordinate geometry. Double-check your arithmetic when expanding and simplifying the plane equation.
• Alternative Method (Equidistance): A point $P(x,y,z)$ lies on the perpendicular bisector plane of segment $AB$ if and only if its distance from $A$ is equal to its distance from $B$, i.e., $PA = PB$. This implies $PA^2 = PB^2$. Setting up the equation $(x-4)^2 + (y+2)^2 + (z-3)^2 = (x-2)^2 + (y-4)^2 + (z+1)^2$ and simplifying will also yield the plane equation $x - 3y + 2z = 2$. This method can serve as a robust verification.

Summary

To find the equation of a plane that perpendicularly bisects a line segment, we first determine the plane's normal vector by using the direction vector of the line segment. Then, we find a point that lies on the plane, which is the midpoint of the line segment. Using the normal vector and the midpoint, we construct the equation of the plane. Finally, we verify which of the given options satisfies this plane equation. Following these steps, the equation of the plane was found to be $x - 3y + 2z = 2$, and the point $(4, 0, -1)$ was the only option that satisfied this equation.

The final answer is $\boxed{\text{(D)}}$.