- #1

pry_or

- 9

- 0

## Homework Statement

Find the equation of the plane parallel to 2x - y + 2z = 4 if the point P = (3,2,-1) is equidistant from both planes.

## Homework Equations

a(x-x0) + b(y-y0) + c(z-z0) = 0

## The Attempt at a Solution

I found the normal vector to the given plane first, as this is quite easy. [itex]\stackrel{\rightarrow}{n}[/itex] = <2,-1,2>, which is also the normal vector for the plane I am finding since they are parallel.

After that, I found a point that is on the given plane. For simplicity, I picked Q = (2,0,0). I then found the equation of the vector from the given point and the point I found, which is [itex]\stackrel{\rightarrow}{PQ}[/itex] = <1,2,-1>. I used this vector and the normal vector to find the scalar projection of [itex]\stackrel{\rightarrow}{PQ}[/itex] along [itex]\stackrel{\rightarrow}{n}[/itex]. This came out to be [itex]\frac{-2}{3}[/itex].

So, [itex]\frac{-2}{3}[/itex] is the distance from the point to both planes. This is where I got stuck.

On my first attempt, I tried to find a point that, when I used it to find the scalar projection again, gave me the same distance.

If the scalar projection is [itex]\frac{-2}{3}[/itex], that means that [itex]\stackrel{\rightarrow}{n}[/itex][itex]\bullet[/itex][itex]\stackrel{\rightarrow}{PQ}[/itex] = -2. So:

2a - b + 2c = -2

Now, I'm not sure if I can pick any point that makes that true and it gives me a valid equation. I assume that as long as the scalar projection is the same, and the normal vector is the same, the equation would be a valid, parallel plane?

I used the point (-1,0,0), which gives me the equation 2(x+1)- y + 2z = 0

I would like to know if this is correct. Since afterward, I looked at the work of my professor and he used a different method that I. Here's what he did.

He started out the same, by picking a point on the given plane and finding [itex]\stackrel{\rightarrow}{PQ}[/itex] and finding the scalar projection. After that, he used this equation:

point on parallel plane = (starting point) + distance<normal vector>

So he got...

(3,2,-1) - 2/3<2,-1,2> = (5/3, 8/3, -7/3)

He then used that point to write his equation. His method makes sense to me intuitively, and seems more "concrete" than my method. The problem is, I decided to calculate the scalar projection with his calculated point and I something 2 instead of -2/3. This would imply to me that the second plane is not equidistant to the point.

So what I would like to know here, which method is correct, if either? Was I wrong to assume that I should just guess a point and that point would work? Was his equation wrong? Are we both wrong?

Thanks!