Question

Find an equation of the plane that contains all the points that are equidistant from the given points. \((-3,1,2), \quad(6,-2,4)\)

Short answer

The equation of the plane containing all the points equidistant from (-3,1,2) and (6,-2,4) is: 9x -3y +2z -6 = 0.

Step-by-step solution

Calculate the Midpoint

First, find the midpoint of the segment joining the points (-3,1,2) and (6,-2,4). This can be done by averaging the x, y and z-coordinates of the two points separately. The formula for midpoint is: \((\frac{{x1+x2}}{2},\frac{{y1+y2}}{2},\frac{{z1+z2}}{2})\). So the midpoint M = \((\frac{{-3+6}}{2},\frac{{1+(-2)}}{2},\frac{{2+4}}{2})=(\frac{3}{2}, -\frac{1}{2}, 3)\)

Calculate the Normal Vector

Now calculate the vector between the two given points which will act as the normal to the plane. This vector is obtained as the difference between coordinates of the two points: \( \vec{n}=(x_2-x_1, y_2-y_1, z_2-z_1 )\). So, \( \vec{n}=(6-(-3), -2-1, 4-2)=(9, -3, 2)\)

Find the Equation of the Plane

To find the equation of a plane given that the plane contains a point \((x_0, y_0, z_0)\) and its normal vector is \(\vec{n}=(A,B,C)\), use the general form of a plane: \(Ax + By + Cz + D = 0\). Substitute \(\vec{n}, x_0, y_0, z_0\) and solve for D, then bring in the normal vector's components and solve for D based on \((x_0, y_0, z_0)\) and \(\vec{n}\): D = -A*x_0 - B*y_0 - C*z_0 = -9*\frac{3}{2} - (-3)*(-\frac{1}{2}) - 2*3= -6. Hence, the equation of the plane becomes: \(9x -3y +2z -6 = 0\)

Key concepts

Understanding the Midpoint Formula

In geometry, the midpoint of a line segment is the point that divides the segment into two equal parts. The midpoint formula is crucial when we need to find a point that is equidistant from the endpoints of the segment. For a given segment with endpoints \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) in three-dimensional space, the formula to calculate its midpoint \( M \) is given by the following equation:

\[ M = \left( \frac{{x_1+x2}}{2}, \frac{{y_1+y2}}{2}, \frac{{z_1+z2}}{2} \right) \]
By applying the midpoint formula, students can easily find the center of a region, which will be essential in more advanced geometry problems, such as identifying the plane that is equidistant from two points.

Calculating the Normal Vector

The concept of a vector in mathematics and physics refers to an object that has both a magnitude and a direction. When we talk about a plane in three-dimensional space, a normal vector to that plane is a vector that is perpendicular to every line lying on the plane. Therefore, having a normal vector is important because it helps characterize the plane fully.

To find the normal vector \(\vec{n}\) given two points in space, we subtract the corresponding components of the points. If \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) are our points, then the normal vector is calculated as:

\[ \vec{n} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \]
Understanding how to calculate the normal vector is a foundational skill for understanding other geometric constructs and analyses in higher dimensions.

General Form of a Plane

A plane in three-dimensional space can be described using its general form equation, which is expressed as:

\[ Ax + By + Cz + D = 0 \]
Here, \( A \) , \( B \) , and \( C \) are the components of the normal vector to the plane, and \( D \) is the scalar that helps anchor the plane in space. To determine the exact plane, we need a point \( (x_0, y_0, z_0) \) that the plane passes through and the normal vector \( \vec{n} \) as mentioned previously. To find \( D \), we simply plug our known point and the normal vector into the plane's equation:

\[ D = - Ax_0 - By_0 - Cz_0 \]
By solving for \( D \), we can formulate the complete equation of the plane that includes all the points equidistant from the given points. This method is incredibly useful in numerous applications, including computer graphics, physics, and many other fields.