find the equation of the plane passes through the middle point of the join of the points (2,-3,1) and (4,5,-3) and is perpendicular to the line joning the points?
To find the equation of the plane, we first need to determine the normal vector of the plane. The normal vector is perpendicular to the plane and can be found by taking the cross product of the direction vector of the line joining the two points.
Step 1: Find the direction vector of the line joining the points (2,-3,1) and (4,5,-3).
Let's call these two points A (2,-3,1) and B (4,5,-3). The direction vector is given by subtracting the coordinates of point A from point B:
AB = (4, 5, -3) - (2, -3, 1)
= (2, 8, -4)
Step 2: Calculate the cross product of the direction vector AB with any vector perpendicular to the line AB. This will give us the normal vector of the plane.
Since we want the plane to be perpendicular to the line AB, we can choose any vector perpendicular to AB. One such vector is (8, -2, 0).
Taking the cross product of AB and (8, -2, 0), we get:
n = AB × (8, -2, 0)
= [(8)(-4) - (-2)(0), (-4)(0) - (8)(0), (2)(0) - (8)(-2)]
= (-32, 0, 16)
Step 3: Find the middle point of the join of the points A and B.
The middle point is given by taking the average of the coordinates of points A and B:
Middle point = [(2+4)/2, (-3+5)/2, (1-3)/2]
= (3, 1, -1)
Step 4: Write the equation of the plane using the normal vector and the coordinates of the middle point.
The equation of a plane can be written in the form:
Ax + By + Cz + D = 0, where A, B, C are the components of the normal vector, and (x, y, z) are the coordinates of a point on the plane.
Using the normal vector (-32, 0, 16) and the middle point (3, 1, -1), the equation of the plane is:
-32x + 0y + 16z + D = 0
Step 5: Find the value of D.
To find the value of D, substitute the coordinates of the middle point (3, 1, -1) into the equation:
-32(3) + 0(1) + 16(-1) + D = 0
-96 -16 + D = 0
D = 112
The equation of the plane that passes through the middle point of the join of the points (2,-3,1) and (4,5,-3) and is perpendicular to the line joining the points is:
-32x + 16z + 112 = 0
To find the equation of a plane passing through the midpoint of two given points (A and B) and perpendicular to the line joining those points, follow these steps:
Step 1: Find the midpoint of the line segment AB.
- The midpoint of two points (x₁, y₁, z₁) and (x₂, y₂, z₂) can be found using the formula:
Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2)
For the points A(2, -3, 1) and B(4, 5, -3), the midpoint is:
Midpoint = ((2 + 4)/2, (-3 + 5)/2, (1 - 3)/2) = (3, 1, -1)
Step 2: Find the direction vector of the line AB.
- The direction vector of the line AB can be obtained by subtracting the coordinates of point A from point B.
Direction vector = (x₂ - x₁, y₂ - y₁, z₂ - z₁)
For points A(2, -3, 1) and B(4, 5, -3):
Direction vector = (4 - 2, 5 - (-3), -3 - 1) = (2, 8, -4)
Step 3: Obtain the normal vector of the plane.
- Since the plane is perpendicular to the line AB, the direction vector of the line AB will also be the normal vector of the plane.
Normal vector of the plane = (2, 8, -4)
Step 4: Use the point-normal form to write the equation of the plane.
- The point-normal form of the equation of a plane is:
Ax + By + Cz = D,
where (A, B, C) are the components of the normal vector and (x, y, z) are the coordinates of any point on the plane.
Using the point (3, 1, -1) and the normal vector (2, 8, -4) in the equation, we get:
2x + 8y - 4z = D
To find D, substitute the coordinates (3, 1, -1) into the equation:
2(3) + 8(1) - 4(-1) = D
6 + 8 + 4 = D
D = 18
Therefore, the equation of the plane passing through the midpoint of the line AB and perpendicular to the line is:
2x + 8y - 4z = 18