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
my answer was 0.111 but it was wrong
To verify that the two planes are parallel, we can compare their normal vectors.
The normal vector of the first plane is <2,0,-5>, and the normal vector of the second plane is also <2,0,-5>. Since the normal vectors are parallel, the planes are parallel.
To find the distance between the parallel planes, we can choose a point on one of the planes and find its perpendicular distance to the other plane.
Let's choose a point on the first plane, for example, (0,1,1). The distance from this point to the second plane can be found by using the formula for the distance between a point and a plane:
distance = |(ax + by + cz + d)/sqrt(a^2 + b^2 + c^2)|
where the coefficients (a, b, c) are the components of the normal vector of the second plane, and (x, y, z) are the coordinates of the chosen point. The constant term d is not needed for this calculation since both planes have the same d value (which is 0 in this case).
Plugging in the values, we get:
distance = |(2(0) + 0(1) - 5(1) + 6)/sqrt(2^2 + 0^2 + (-5)^2)|
distance = |-1/5|
distance = 0.2
Therefore, the distance between the two planes is 0.2 units.