5. Determine an equation for the plane that is exactly between the points A(-1, 2, 4) and B(3, 1, -4).
vector AB must be the normal to the plane
vector AB = (4,-1,-8)
so the plane must have equation:
4x - y - 8z = c
but the midpoint of AB must lie on it, and that midpoint is (1, 3/2, 0)
sub it in ...
4 - 3/2 - 0 = c
c = 5/2
plane equation:
4x - y - 8z = 5/2 or in only integers
8x - 2y - 16z = 5
To find the equation for the plane that is exactly between two points, you can begin by finding the midpoint between those points. Once you have the midpoint, you can use it to determine the normal vector of the plane, which you will need to write the equation.
Step 1: Find the midpoint between points A and B.
The midpoint formula is given by:
midpoint = ((x1 + x2) / 2, (y1 + y2) / 2, (z1 + z2) / 2)
Let's substitute the coordinates of points A and B into the midpoint formula:
midpoint = ((-1 + 3) / 2, (2 + 1) / 2, (4 + (-4)) / 2)
= (2 / 2, 3 / 2, 0 / 2)
= (1, 3/2, 0)
So, the midpoint between points A and B is (1, 3/2, 0).
Step 2: Determine the normal vector of the plane.
The normal vector of a plane is perpendicular to the plane. We can find it by taking the vector between A and B and then finding a vector perpendicular to it.
To find the vector between A and B, subtract the coordinates of point A from the coordinates of point B:
vector AB = (3 - (-1), 1 - 2, -4 - 4)
= (4, -1, -8)
Now, we need to find a vector that is perpendicular to AB. One way to do this is to take the cross product of AB with any other vector. Let's choose the vector (1, 0, 0) for simplicity:
cross product = (4, -1, -8) x (1, 0, 0)
To find the cross product, we calculate the determinant of the following matrix:
| i j k |
| 4 -1 -8 |
| 1 0 0 |
By expanding along the first column, we get:
i * (0 - 0) - j * (-8 - 0) + k * (-1 - 4)
= -8k + 5j
So, the cross product is (-8k + 5j). This is a normal vector perpendicular to the plane.
Step 3: Write the equation of the plane.
Now that we have the normal vector and the midpoint, we can write the equation for the plane. The equation of a plane is in the form Ax + By + Cz = D, where A, B, C are coefficients, and D is the distance from the origin.
Using the normal vector (-8k + 5j), we have A = 0, B = 5, C = -8.
Substituting the coordinates of the midpoint (1, 3/2, 0) into the equation, we can solve for D:
0 * 1 + 5 * (3/2) + -8 * 0 = D
(15/2) = D
So, the equation of the plane that is exactly between the two points A(-1, 2, 4) and B(3, 1, -4) is:
5y + (15/2) = 0.