verify that the two planes are parallel, and find the distance between the planes. (round your answer to three decimal places)
2x-5z=5
2x-5z=6
To verify that the two planes are parallel, we need to check if their normal vectors are parallel. The normal vector of a plane with the equation of the form ax + by + cz = d is (a, b, c).
Both planes have the same normal vector (2, 0, -5), so they are parallel.
To find the distance between the planes, we can choose a point on one of the planes and find the distance between this point and the other plane.
Let's choose the point (0, 0, 1) on the plane 2x - 5z = 5. The distance between this point and the plane 2x - 5z = 6 is:
|2(0) - 5(1) - 6| / sqrt(2^2 + 0^2 + (-5)^2) = |-11| / sqrt(29) = 11/sqrt(29)
So, the distance between the two planes is approximately 1.999 units when rounded to three decimal places.
To verify if the two planes are parallel, we need to compare their normal vectors. The normal vectors of a plane are the coefficients of x, y, and z in the equation of the plane.
Let's compare the normal vectors of the two planes:
Plane 1: 2x - 5z = 5
Normal vector 1: (2, 0, -5)
Plane 2: 2x - 5z = 6
Normal vector 2: (2, 0, -5)
Since the two normal vectors are the same (parallel vectors), we can conclude that the planes are parallel.
To find the distance between the two parallel planes, we can use the formula:
Distance = |c1 - c2| / √(a^2 + b^2 + c^2)
where a, b, and c are the coefficients of x, y, and z respectively in the equation of any one of the planes, and c1 and c2 are the constant terms in the equations of the two planes.
Using Plane 1, a = 2, b = 0, c = -5, c1 = 5, and c2 = 6, we can plug these values into the formula:
Distance = |5 - 6| / √(2^2 + 0^2 + (-5)^2)
= |-1| / √(4 + 25)
= 1 / √29
≈ 0.183 (rounded to three decimal places)
Therefore, the distance between the two parallel planes is approximately 0.183 units.
To verify if two planes are parallel, we need to compare their normal vectors.
Convert the given equations to the standard form of a plane: Ax + By + Cz = D.
Equation 1: 2x - 5z = 5
Rearrange: 2x - 5z - 5 = 0
Equation 2: 2x - 5z = 6
Rearrange: 2x - 5z - 6 = 0
Now, let's compare the coefficients of x, y, and z for both equations:
For x: The coefficient is 2 in both equations.
For y: There is no term with y in either equation.
For z: The coefficient is -5 in both equations.
The normal vector for the first plane is <2, 0, -5>.
The normal vector for the second plane is also <2, 0, -5>.
Since the normal vectors are the same, the two planes are parallel.
To find the distance between the two parallel planes, we can use the formula:
Distance = |D2 - D1| / √(A^2 + B^2 + C^2)
Where D1 and D2 are the constants on the right side of the equations (in standard form) for the two planes.
For the first equation: D1 = 5
For the second equation: D2 = 6
Plugging in the values, we get:
Distance = |6 - 5| / √(2^2 + 0^2 + (-5)^2)
= |1| / √(4 + 25)
= 1 / √29
≈ 0.183 (rounded to three decimal places)
Therefore, the distance between the two planes is approximately 0.183.