Perpendicular Bisector Plane Equation And Intersection Point Of Two Lines

In the realm of three-dimensional geometry, a fundamental problem involves determining the equation of a plane that perpendicularly bisects the line segment joining two given points. This plane, known as the perpendicular bisector plane, holds significant geometric properties and finds applications in various fields, including computer graphics, engineering, and physics. To embark on this mathematical journey, let's consider two points in space, A(2, 3, 5) and B(5, -2, 7). Our objective is to find the equation of the plane that not only bisects the line segment AB but also intersects it at a right angle.

The equation of a plane is generally represented in the form Ax + By + Cz + D = 0, where A, B, C, and D are constants that define the plane's orientation and position in space. To determine these constants for our specific perpendicular bisector plane, we need to leverage the geometric properties of the bisector and the given points. The first crucial observation is that the midpoint of the line segment AB must lie on the bisector plane. This midpoint, denoted as M, can be calculated by averaging the coordinates of A and B. Mathematically, the coordinates of M are given by M = ((2+5)/2, (3-2)/2, (5+7)/2) = (7/2, 1/2, 6). Since M lies on the plane, its coordinates must satisfy the plane's equation. This gives us our first equation involving A, B, C, and D.

Furthermore, the direction vector of the line segment AB is perpendicular to the normal vector of the bisector plane. The direction vector, denoted as ${\vec{AB}}$, can be found by subtracting the coordinates of A from the coordinates of B, yielding ${\vec{AB}}$ = (5-2, -2-3, 7-5) = (3, -5, 2). The normal vector of the plane, denoted as ${\vec{n}}$, is given by (A, B, C). The condition of perpendicularity implies that the dot product of ${\vec{AB}}$ and ${\vec{n}}$ must be zero. This gives us another equation involving A, B, and C. With two equations in hand, we need one more to uniquely determine the constants A, B, C, and D. The final equation comes from substituting the coordinates of the midpoint M into the plane's equation. This ensures that the midpoint lies on the plane, fulfilling the bisection requirement. By solving these three equations simultaneously, we can obtain the values of A, B, C, and D, thus defining the equation of the perpendicular bisector plane.

Let's delve into the step-by-step calculations to solidify our understanding. First, we calculate the midpoint M as described earlier, obtaining M = (7/2, 1/2, 6). Next, we determine the direction vector ${\vec{AB}}$ as (3, -5, 2). Now, we express the condition of perpendicularity as the dot product of ${\vec{AB}}$ and ${\vec{n}}$ being zero: 3A - 5B + 2C = 0. This is our first equation. Substituting the coordinates of M into the plane equation, we get A(7/2) + B(1/2) + C(6) + D = 0, which simplifies to 7A + B + 12C + 2D = 0. This is our second equation. Now we have a system of equations:

1. 3A - 5B + 2C = 0
2. 7A + B + 12C + 2D = 0

We need to find another equation to solve for A, B, C, and D. This comes from the condition that the midpoint lies on the plane. Substituting M into the plane equation Ax + By + Cz + D = 0 yields the equation 7A/2 + B/2 + 6C + D = 0, or 7A + B + 12C + 2D = 0. Notice this is the same as equation (2). We need a third independent equation. Since the normal vector (A, B, C) is parallel to the direction vector (3, -5, 2), we can write A = 3k, B = -5k, and C = 2k for some constant k. Substituting these into the plane equation, we get 3kx - 5ky + 2kz + D = 0. The normal vector of the plane is perpendicular to the vector connecting the two points. This gives us the equation 3A - 5B + 2C = 0. Substituting the midpoint (7/2, 1/2, 6) into the plane equation Ax + By + Cz + D = 0, we get 7A/2 + B/2 + 6C + D = 0. We now have two equations: 3A - 5B + 2C = 0 and 7A/2 + B/2 + 6C + D = 0. Solving these equations will give us the coefficients A, B, C, and D, which define the perpendicular bisector plane. Simplifying the equations and solving for the coefficients, we find that the equation of the plane is 3x - 5y + 2z = 19.

Therefore, the equation of the plane that bisects perpendicularly the join of (2, 3, 5) and (5, -2, 7) is 3x - 5y + 2z = 19. This plane is unique and satisfies the given conditions. The steps involved in finding this equation highlight the interplay between geometric concepts and algebraic techniques. Understanding the properties of planes, lines, and vectors is crucial in solving such problems. This methodology can be extended to more complex scenarios involving different geometric objects and constraints, making it a fundamental tool in various mathematical and engineering applications.

2. Showing That the Lines rac{x-1}{2} = rac{y-2}{3} = rac{z-3}{4} and rac{x-4}{5} = rac{y-1}{2} = z Intersect and Finding the Point of Intersection

In the vast landscape of three-dimensional geometry, the intersection of lines is a fundamental concept that arises in numerous applications, from computer graphics and robotics to structural engineering and physics. Determining whether two lines intersect and, if so, finding their point of intersection, is a crucial skill in various fields. In this section, we embark on a journey to demonstrate that the lines defined by the equations ${\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}}$ and ${\frac{x-4}{5} = \frac{y-1}{2} = z}$ indeed intersect, and to pinpoint the precise coordinates of their meeting point.

The equations of lines in three-dimensional space can be expressed in several forms, including the symmetric form used here. The symmetric form provides a concise representation of a line's direction and a point it passes through. To ascertain whether two lines intersect, we must first understand the conditions that must be met. Geometrically, two lines intersect if they share a common point in space. Algebraically, this translates to the existence of a set of coordinates (x, y, z) that satisfies the equations of both lines simultaneously. To find such a point, we can parameterize each line and then equate the corresponding coordinates.

Let's begin by parameterizing the first line, ${\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}}$. We introduce a parameter, say t, and set each fraction equal to t. This yields the following parametric equations: x = 2t + 1, y = 3t + 2, and z = 4t + 3. These equations describe all the points on the first line as t varies. Similarly, we parameterize the second line, ${\frac{x-4}{5} = \frac{y-1}{2} = z}$. Here, let's use the parameter s. Setting each fraction equal to s, we obtain x = 5s + 4, y = 2s + 1, and z = s. These equations describe all the points on the second line as s varies. If the two lines intersect, there must exist values of t and s for which the coordinates (x, y, z) are the same for both lines. This gives us a system of equations:

1. 2t + 1 = 5s + 4
2. 3t + 2 = 2s + 1
3. 4t + 3 = s

This system consists of three equations with two unknowns, t and s. To determine if a solution exists, we can solve any two equations for t and s and then substitute these values into the third equation. If the third equation is satisfied, then the system is consistent, and the lines intersect. If the third equation is not satisfied, then the system is inconsistent, and the lines do not intersect. Let's solve the first two equations for t and s. From the first equation, we have 2t - 5s = 3. From the second equation, we have 3t - 2s = -1. Multiplying the first equation by 3 and the second equation by 2, we get 6t - 15s = 9 and 6t - 4s = -2. Subtracting the second equation from the first, we obtain -11s = 11, which gives s = -1. Substituting s = -1 into the second equation, we get 3t - 2(-1) = -1, which simplifies to 3t + 2 = -1. Solving for t, we find t = -1.

Now, we substitute t = -1 and s = -1 into the third equation, 4t + 3 = s. Plugging in the values, we get 4(-1) + 3 = -1, which simplifies to -1 = -1. Since the third equation is satisfied, the system is consistent, and the lines intersect. To find the point of intersection, we can substitute either t = -1 into the parametric equations of the first line or s = -1 into the parametric equations of the second line. Using the first line, we get x = 2(-1) + 1 = -1, y = 3(-1) + 2 = -1, and z = 4(-1) + 3 = -1. Therefore, the point of intersection is (-1, -1, -1). Similarly, using the second line, we get x = 5(-1) + 4 = -1, y = 2(-1) + 1 = -1, and z = -1. This confirms that the point of intersection is indeed (-1, -1, -1).

In conclusion, we have successfully demonstrated that the lines ${\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}}$ and ${\frac{x-4}{5} = \frac{y-1}{2} = z}$ intersect, and we have found their point of intersection to be (-1, -1, -1). The process involved parameterizing the lines, setting up a system of equations, solving for the parameters, and verifying the solution. This methodology provides a powerful tool for analyzing the intersection of lines in three-dimensional space and has broad applications in various scientific and engineering disciplines.

This analysis not only confirms the intersection but also reinforces the utility of parametric equations in handling spatial geometry problems. The ability to represent lines and curves parametrically is a cornerstone of advanced geometric computations and simulations, making this concept essential for students and professionals alike.

Conclusion

In this exploration, we have navigated through two essential problems in three-dimensional geometry: finding the equation of a plane that perpendicularly bisects the join of two points and determining the intersection point of two lines. These problems underscore the importance of understanding spatial relationships and the algebraic tools used to describe them. The methods discussed, from calculating midpoints and normal vectors to parameterizing lines and solving systems of equations, are fundamental in various fields, including engineering, computer graphics, and physics. Mastering these concepts provides a strong foundation for tackling more complex geometric challenges and applications.