Determine the scalar equation of a plane where any point on that plane is equidistant to the points A(-1, -4, 0) and B(5, 2, 4)
So it must go through the midpoint of AB
and the direction of AB would become the normal of the plane.
Can you take it from there?
Sorry, but how does that work?
Do you know the meaning of normal?
Do you know how to find a midpoint given the two endpoints?
I don't think you will able to do any of these rather straightforward problems
if you don't know these type of concepts.
If you are looking for just answers, you have come to the wrong website.
Again … make good use of https://www.khanacademy.org/
To find the scalar equation of a plane, we can use the following steps:
Step 1: Find the midpoint of points A and B.
Step 2: Calculate the direction vector of the line AB.
Step 3: Determine a normal vector to the plane.
Step 4: Use the coordinates of the midpoint and the normal vector to write the scalar equation of the plane.
Step 1: Find the midpoint of points A and B.
The midpoint M can be calculated using the formula:
M = [(x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2]
For points A(-1, -4, 0) and B(5, 2, 4), the midpoint M is calculated as follows:
M = [((-1 + 5)/2), ((-4 + 2)/2), ((0 + 4)/2)]
= [4/2, -2/2, 4/2]
= [2, -1, 2]
So the midpoint M is (2, -1, 2).
Step 2: Calculate the direction vector of the line AB.
The direction vector d can be found by subtracting the coordinates of point A from point B:
d = B - A
For points A(-1, -4, 0) and B(5, 2, 4), the direction vector d is given by:
d = [(5 - (-1)), (2 - (-4)), (4 - 0)]
= [6, 6, 4]
So the direction vector d is (6, 6, 4).
Step 3: Determine a normal vector to the plane.
To find a normal vector to the plane, we can take the cross product of the direction vector d with any vector not parallel to it. A simple choice is the vector (1, 0, 0).
Taking the cross product of d = (6, 6, 4) and (1, 0, 0), we get:
n = d x (1, 0, 0)
Using the cross product formula, we have:
n = [(6 * 0) - (6 * 0), (6 * 0) - (4 * 0), (6 * 0) - (4 * 1)]
= [0, 0, -4]
So a normal vector to the plane is (0, 0, -4).
Step 4: Use the coordinates of the midpoint and the normal vector to write the scalar equation of the plane.
The scalar equation of a plane is given by the formula:
Ax + By + Cz = D
Using the midpoint M = (2, -1, 2) and the normal vector n = (0, 0, -4), we can substitute the values into the equation:
0(x - 2) + 0(y + 1) + (-4)(z - 2) = 0
Simplifying the equation gives us:
-4z + 8 = 0
-4z = -8
z = 2
So the scalar equation of the plane is:
0(x - 2) + 0(y + 1) + (-4)(z - 2) = 0