Find the equation of the plane equidistant from the points (2,8,3) and (4,12,9)
The plane must pass through the midpoint of the given points which is
(3,10,6)
A direction vector through the 2 points is
[2, 4, 6] or reduced to [1,2,3]
This must be a normal to the plane, thus the
equation of the plane would be
x + 2y + 3z = c
but (3,10,6) must lie on it, so
3 + 20 + 18 = c
c = 41
x + 2y + 3z = 41
To find the equation of the plane equidistant from two points, we can follow the following steps:
Step 1: Find the midpoint between the two given points.
Step 2: Find the normal vector of the plane.
Step 3: Use the midpoint and the normal vector to write the equation of the plane.
Let's perform these steps one by one:
Step 1: Find the midpoint between the two given points.
The midpoint between two points (x1, y1, z1) and (x2, y2, z2) is given by the formula:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2)
Given points: (2, 8, 3) and (4, 12, 9)
Midpoint = ((2 + 4)/2, (8 + 12)/2, (3 + 9)/2)
Midpoint = (3, 10, 6)
Step 2: Find the normal vector of the plane.
To find the normal vector, we can consider the vector joining the two given points (2, 8, 3) and (4, 12, 9). This vector will be orthogonal (perpendicular) to the plane we are trying to find.
Vector joining the given points = <4 - 2, 12 - 8, 9 - 3> = <2, 4, 6>
The normal vector of the plane is the same as the vector joining the given points, i.e., <2, 4, 6>.
Step 3: Use the midpoint and the normal vector to write the equation of the plane.
The equation of a plane in 3D can be written as:
Ax + By + Cz + D = 0
where (A, B, C) represents the normal vector of the plane, and (x, y, z) represents any point lying on the plane.
Substituting the values from the midpoint and the normal vector, we get:
2(x - 3) + 4(y - 10) + 6(z - 6) + D = 0
Simplifying, we have:
2x - 6 + 4y - 40 + 6z - 36 + D = 0
2x + 4y + 6z - 82 + D = 0
We can let D = 82 to eliminate the constant term, resulting in the equation:
2x + 4y + 6z = 82
Therefore, the equation of the plane equidistant from the points (2,8,3) and (4,12,9) is 2x + 4y + 6z = 82.
To find the equation of the plane equidistant from two points, we need to find the midpoint of the line segment connecting the two points and use it as the point on the plane. Let's find the midpoint first.
Midpoint formula:
The midpoint between two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by:
(mid_x, mid_y, mid_z) = ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2).
Let's calculate the midpoint using the given points:
Midpoint = ((2 + 4)/2, (8 + 12)/2, (3 + 9)/2)
= (3, 10, 6).
Now that we have the midpoint (3, 10, 6), we can use it as a point on the plane.
To find the equation of the plane, we also need a normal vector perpendicular to the plane. One way to find the normal vector is by taking the cross product of the vectors formed by the two points.
Vector formula:
Given two points (x₁, y₁, z₁) and (x₂, y₂, z₂), the vector formed by these points can be obtained by subtracting the coordinates of the first point from the coordinates of the second point:
vector = (x₂ - x₁, y₂ - y₁, z₂ - z₁).
Using the given points:
vector₁ = (4 - 2, 12 - 8, 9 - 3)
= (2, 4, 6).
Now we can take the cross product of vector₁ and the vector from the midpoint to find the normal vector:
Cross product formula:
Given two vectors (a, b, c) and (d, e, f), the cross product is given by:
normal_vector = (bf - ce, cd - af, ae - bd).
Using vector₁ = (2, 4, 6) and vector₂ = (3 - 2, 10 - 8, 6 - 3) = (1, 2, 3):
normal_vector = (4*3 - 2*6, 6*1 - 2*3, 2*6 - 4*1)
= (12 - 12, 6 - 6, 12 - 4)
= (0, 0, 8).
Now we have the midpoint (3, 10, 6) and the normal vector (0, 0, 8) for the plane.
The equation of a plane in the form Ax + By + Cz + D = 0 can be found using the point-normal form, where (A, B, C) is the normal vector and (x, y, z) are the coordinates of a point on the plane.
Substituting the midpoint (3, 10, 6) and normal vector (0, 0, 8) into the equation, we get:
0*x + 0*y + 8*z + D = 0.
Simplifying the equation, we find:
8z + D = 0,
8z = -D,
z = -D/8.
Therefore, the equation of the plane is z = -D/8.