Question

For points \(\mathrm{A}(1,2,3), \mathrm{B}(5,4,1)\), the equation of plane which is perpendicular bisector of \(\underline{A B}\) is (A) \(x+2 y-7+z=0\) (B) \(2 x+y-z=7\) (C) \(x+2 y+z+7=0\) (D) \(2 x-2 y-z=7\)

Short answer

The equation of the plane which is the perpendicular bisector of the line segment AB is (B) \(2x + y - z = 7\).

Step-by-step solution

Find the midpoint of the line segment AB

To find the midpoint M of the line segment AB, use the midpoint formula: \(M(x,y,z) = \left(\frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2}\right)\) For points A(1,2,3) and B(5,4,1): \(M(x,y,z) = \left(\frac{1 + 5}{2}, \frac{2 + 4}{2}, \frac{3 + 1}{2}\right) = (3, 3, 2)\)

Find the direction vector of the line segment AB

To find the direction vector of the line segment AB, subtract the coordinates of point A from point B: \(\vec{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle\) For points A(1,2,3) and B(5,4,1): \(\vec{AB} = \langle 5 - 1, 4 - 2, 1 - 3 \rangle = \langle 4, 2, -2 \rangle\) This direction vector \(\langle 4, 2, -2 \rangle\) will be normal to the plane.

Find the equation of the plane

Using the point-normal form of the equation of a plane: \(A(x - x_M) + B(y - y_M) + C(z - z_M) = 0\) Where (A, B, C) are the components of the normal vector, and M is the midpoint of the line segment AB. Using the normal vector \(\langle 4, 2, -2 \rangle\) and midpoint M(3, 3, 2): \(4(x - 3) + 2(y - 3) - 2(z - 2) = 0\) Expanding and simplifying, we get: \(4x + 2y - 2z = 12\) So the equation of the plane is: \(2x + y - z = 6\) Comparing this equation to the given options, we can see that it is equivalent to option (B) multiplied by a constant (-2), which means option (B) is the correct answer: Answer: (B) \(2x + y - z = 7\)

Key concepts

Perpendicular Bisectors

A perpendicular bisector of a line segment is a geometric concept that refers to a line which both divides a line segment into two equal parts and is perpendicular (at a right angle) to the line segment. This is a significant concept in coordinate geometry as it helps in understanding symmetry and is often used in geometric constructions and proofs.

In three-dimensional geometry, a perpendicular bisector can be represented as a plane. The plane will pass through the midpoint of a line segment, and its normal (or perpendicular) direction will be aligned with the direction vector of the line segment.

For instance, if you have a line segment defined by points \(A(1, 2, 3)\) and \(B(5, 4, 1)\), the perpendicular bisector can be a plane that will pass through the midpoint of \(A\) and \(B\) and is perpendicular to line \(AB\). This concept helps in various applications like finding distances between points and constructing geometric shapes with precise measurements.

Midpoint Formula

The midpoint formula is used to find the central point between two points in a coordinate plane or space. This formula is fundamental in geometry and can be applied across different dimensions.

For a line segment with endpoints \(A(x_A, y_A, z_A)\) and \(B(x_B, y_B, z_B)\), the midpoint \(M\) is calculated by averaging the coordinates from both endpoints:

• \(M(x, y, z) = \left( \frac{x_A + x_B}{2}, \frac{y_A + y_B}{2}, \frac{z_A + z_B}{2} \right)\)

In our example, with points \(A(1, 2, 3)\) and \(B(5, 4, 1)\), plugging these into the formula gives the midpoint \(M(3, 3, 2)\). This midpoint is essential in finding the equation of the perpendicular bisector plane, as it is the point through which the plane must pass. Understanding how to derive the midpoint helps in solving a variety of problems related to line segments and planes.

Direction Vectors

Direction vectors are essential in defining and understanding geometric entities like lines and planes in spatial geometry. A direction vector provides the orientation of a line segment; it tells us the direction in which the segment runs by pointing from one end to another.

To find a direction vector for a line segment from point \(A\) to point \(B\), you subtract the coordinates of point \(A\) from those of point \(B\):

• \(\vec{AB} = \langle x_B - x_A, y_B - y_A, z_B - z_A \rangle\)

For example, for points \(A(1, 2, 3)\) and \(B(5, 4, 1)\), the direction vector is \(\langle 4, 2, -2 \rangle\). This vector not only tells us the direction of line \(AB\) in space but also serves as the normal vector for the perpendicular bisector plane.

Having a good grasp of direction vectors allows you to manipulate and create relations between different geometric constructs in three dimensions.

Equation of a Plane

The equation of a plane is a mathematical expression describing a flat, two-dimensional surface extending infinitely in three-dimensional space. In the context of coordinate geometry, planes are often represented in terms of their normal vectors and a point they pass through. This is usually given as

• \(A(x - x_0) + B(y - y_0) + C(z - z_0) = 0\)

where \((A, B, C)\) is the normal vector, providing the plane's orientation, and \((x_0, y_0, z_0)\) is a point on the plane.

In practical applications, such as our exercise example with line \(AB\) having a direction vector \(\langle 4, 2, -2 \rangle\), we used this vector as the plane's normal vector. Combined with the midpoint \(M(3, 3, 2)\), these give rise to the perpendicular bisector's equation:

• \(4(x - 3) + 2(y - 3) - 2(z - 2) = 0\)

Simplifying this equation helps identify the correct plane description, as seen in the provided multiple-choice options. Knowing how to form and interpret the equation of a plane is crucial in visualizing geometric problems and solutions in space.