equations of two planes parallel to a plane?

Question

Find the equation of the two planes parallel to and sqrt5 units away from -2y+z-1.

please help... Can you tell me your solutions too so I can learn to do it by myself?

sorry. the equation of the plane is
-2y+z-1=0

Answer 1

can you please correct the equation of your plane? You did not put down an equation.

If the planes are parallel they will have the same normal so put down your proper equation so we can help u further.

EDIT: okay, so your normal then is (0, -2, 1) and your equations have form -2y + z = c.
Since the planes are sqrt 5 units away from the given plane, one will contain the point (0, -2, 2) and the other will contain the point (0, 2, 0)
thus the required planes are
-2y + z = 6 and -2y + z = -4

Answer 2

Find the equation of the two planes parallel to and √5 units away from -2y + z - 1 = 0.

The normal vector n, can be taken from the coefficients.
n = <0, -2, 1>

|| n || = √[0² + (-2)² + 1²] = √5

First pick a point on the plane and then move a distance √5 units in each direction from it along the normal vector. That will be a point in the parallel plane. The parallel planes will have the same normal vector and therefore the same coefficients.

Pick the point on the plane P(0, 0, 1).
Now move a distance of ±√5 in the direction of n to find points in the other planes.

Q = P + n = <0, 0, 1> + <0, -2, 1> = Q(0, -2, 2)
R = P - n = <0, 0, 1> - <0, -2, 1> = R(0, 2, 0)

The equations of the parallel planes are of the form:
-2y + z + k = 0

Solve for k. Plug in the point Q(0, -2, 2).
-2(-2) + 2 + k = 0
k = -6

The equation of the parallel plane is:
-2y + z - 6 = 0

Solve for k. Plug in the point R(0, 2, 0).
-2*2 + 0 + k = 0
k = 4

The equation of the parallel plane is:
-2y + z + 4 = 0
___________

The equation of the two parallel planes is:

-2y + z - 6 = 0
-2y + z + 4 = 0

Answer 3

What Is A Parallel Plane