Find the distance between the given parallel planes.
5x−4y+z=10, 10x−8y+2z=3
here is an example of how to do this:
http://www.math.ucla.edu/~ronmiech/Calculus_Problems/32A/chap11/section5/718d65/718_65.html
To find the distance between two parallel planes, we need to find the perpendicular distance from one plane to the other.
Parallel planes have the same normal vector, so we can find the normal vector by comparing the coefficients of x, y, and z in the equation of the planes.
For the first plane 5x − 4y + z = 10, the normal vector is (5, -4, 1).
For the second plane 10x − 8y + 2z = 3, the normal vector is (10, -8, 2).
Now, we can find the distance between the planes using the formula:
distance = |(A1x1 + B1y1 + C1z1 - D1)| / sqrt(A1^2 + B1^2 + C1^2)
where (A1, B1, C1) is the normal vector of the first plane and (x1, y1, z1) is any point on the second plane.
Let's calculate the distance:
distance = |(5*0 + (-4)*0 + 1*0 - 10)| / sqrt(5^2 + (-4)^2 + 1^2)
= 10 / sqrt(25 + 16 + 1)
= 10 / sqrt(42)
= 10 / 6.480740698
≈ 1.541
Therefore, the distance between the given parallel planes is approximately 1.541 units.
To find the distance between two parallel planes, we need to find a perpendicular distance between the planes.
First, let's find the normal vectors of the planes. The coefficient of x, y, and z in the equations of the planes gives us the normal vector.
For the first plane, with the equation 5x−4y+z=10, the normal vector is (5, -4, 1).
For the second plane, with the equation 10x−8y+2z=3, the normal vector is (10, -8, 2).
Next, we can calculate the dot product of the normal vectors. The dot product of two perpendicular vectors is zero.
Dot product = (5 * 10) + (-4 * -8) + (1 * 2) = 50 + 32 + 2 = 84
The magnitude of the dot product gives us the product of the magnitudes of the normal vectors multiplied by the cosine of the angle between them.
|Dot product| = |(5, -4, 1)| * |(10, -8, 2)| * cosine(angle)
We can find the magnitude of a vector using the formula:
|v| = sqrt(v1^2 + v2^2 + v3^2)
| (5, -4, 1) | = sqrt(5^2 + (-4)^2 + 1^2) = sqrt(25 + 16 + 1) = sqrt(42)
|v1| = sqrt(42)
|v2| = sqrt(168)
|v1| * |v2| = sqrt(42) * sqrt(168) = sqrt(42 * 168) = sqrt(7056) = 84
Therefore,
|Dot product| = |(5, -4, 1)| * |(10, -8, 2)| * cosine(angle)
84 = 84 * cosine(angle)
Since cosine(angle) can only be between -1 and 1,
|Dot product| = |(5, -4, 1)| * |(10, -8, 2)| * cosine(angle) ≤ |(5, -4, 1)| * |(10, -8, 2)|
|Dot product| ≤ sqrt(42) * sqrt(168) = 84
Finally, we can calculate the distance between the two parallel planes by taking the absolute value of the distance between the two planes:
Distance = |Distance| = |constant term of plane 1 - constant term of plane 2| / |Dot product|
Distance = |10 - 3| / 84 = 7 / 84 = 1 / 12
Therefore, the distance between the given parallel planes is 1/12.
point in first plane
0,0 something
(0 , 0 , 10)
d = | 10*0 -8*0 +2*10 -3 | /sqrt(10^2+8^2+4^2)
= 17/13.4