What is the distance between the parallel planes: ax+by+cz=d1 and ax+by+cz=d2?
Well, I did it but there is probably an easier way
A point in plane 1 is (0,0,d1/c)
A line through that point perpendicular to the plane is:
(x-0)/a = (y-0)/b = (z-d1/c)/c = t to make it parametric in t
then
x = at
y = bt
z = ct + d1/c
Where does that line hit the second plane?
a(at) + b(bt) +c(ct +d1/c) = d2
so
t = (d2-d1)/(a^2+b^2+c^2)
and at that point in that second plane
x2 = a t
y2 = b t
z2 = ct + d1/c
get (x2-0), (y2-0) , (z2-d1/c)
for distance formula
then
D = sqrt[(at)^2 + (bt)^2 + (ct)^2]
D = t sqrt (a^2+b^2+c^2)
D = [(d2-d1)/(a^2+b^2+c^2)]sqrt (a^2+b^2+c^2)
D = (d2-d1)/sqrt (a^2+b^2+c^2)
To find the distance between the parallel planes given by the equations ax + by + cz = d1 and ax + by + cz = d2, we can use the formula for the distance between a point and a plane.
The formula for the distance between a point (x0, y0, z0) and a plane ax + by + cz + d = 0 is given by:
Distance = |(ax0 + by0 + cz0 + d)| / sqrt(a^2 + b^2 + c^2)
In this case, both planes have the same values for a, b, and c. Therefore, we can take any point from one plane and use the formula to find the distance to the other plane.
Let's assume a point (x0, y0, z0) lies on the plane ax + by + cz = d1. Substituting these values into the formula, we get:
Distance = |(ax0 + by0 + cz0 + d2)| / sqrt(a^2 + b^2 + c^2)
So, the distance between the parallel planes is |(ax0 + by0 + cz0 + d2)| / sqrt(a^2 + b^2 + c^2), where (x0, y0, z0) is any point on one of the planes.
To find the distance between two parallel planes, you can use the formula for the distance from a point to a plane.
First, let's consider the equation of a plane in 3D space: ax + by + cz = d, where (a, b, c) is the normal vector to the plane.
Since both planes are parallel and have the same normal vector (a, b, c), we can say that the distance between the planes will be the same as the distance from any point on one plane to the other plane.
Now, to find the distance, we need to choose a point on one of the planes.
Let's assume a point P0 on the plane ax + by + cz = d1, such that (x0, y0, z0) are the coordinates of this point.
The distance from the point P0 to the plane ax + by + cz = d2 can be calculated using the following formula:
distance = |(ax0 + by0 + cz0 - d2) / sqrt(a^2 + b^2 + c^2)|
In this formula, ax0 + by0 + cz0 represents the signed distance from the point P0 to the plane ax + by + cz = d2. Taking the absolute value ensures that the distance is positive.
The denominator sqrt(a^2 + b^2 + c^2) is the magnitude of the normal vector (a, b, c), which represents the length of a perpendicular line from the origin to the plane.
By substituting appropriate values for (a, b, c, d2) and (x0, y0, z0), you can calculate the distance between the two parallel planes.
It's important to note that if the normal vector (a, b, c) is a unit vector, the formula simplifies to:
distance = |ax0 + by0 + cz0 - d2|