write equations for the planes that are parallel to x+3y-5z=2, using x, y, and z to express the answer?

here is what I got so far: because the plane are parallel to x+3y-5z=2, which can be expressed as normal vector=i+3j-5k, and point one=(0,0,2). Plane two would share the same normal vector plane one has, so n=i+3j-5k and Point two=(0,0,p)for a point on plane two. After that, I used projection vector(P1P2) onto vector n to get the distance for the planes that are parallel.

But I am never able to solve for p, did I do something wrong. Is there another way to do this problem?

There are many planes that are parallel to the given plane.

Two planes are parallel if, say, we have:

ax+by+cz+d_{1}=0 and

ax+by+cz+d_{2}=0

The distance between two parallel planes is given by:

d=\frac{|d_{1}-d_{2}|}{\sqrt{a^{2}+b^{2}+c^{2}}}

A plane that would be parallel to yours may be

x+3y-5z=0 if it passes through the origin (0,0,0).

what about if I change the question to this way: write equations for the planes that are parallel to x+3y-5z=2, and lie three units from it.

write equations for the planes that are parallel to x+3y-5z=2, and lie three units from it.

I get: normal vector=i+3j-5k, but how to I lie three units from the plane?

Well, the normal vector is <1, 3, -5>

A point on this plane is (2, 0, 0)

The line going through the plane which is perpendicular to it is therefore:

\vec{l} = \(2\\0\\0\) + \lambda \(1\\3\\-5\)

The unit vector of the perpendicular vector is:

\frac{1}{\sqrt{1+3^2+5^2}}\(1\\3\\-5\)

Three units, becomes:

\frac{3}{\sqrt{1+3^2+5^2}}\(1\\3\\-5\)

And the points will be:

\(x_1\\y_1\\z_1\) = \(2\\0\\0\) +\frac{3}{\sqrt{1+3^2+5^2}} \(1\\3\\-5\)

and

\(x_2\\y_2\\z_2\) = \(2\\0\\0\) -\frac{3}{\sqrt{1+3^2+5^2}} \(1\\3\\-5\)

Now that you have the points, just substitute them in:

x+3y-5z=d

for (x1, y1, z1) and (x2, y2, z2)