# Equation of Parallel Plane Equidistant to a Point

## 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. $\stackrel{\rightarrow}{n}$ = <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 $\stackrel{\rightarrow}{PQ}$ = <1,2,-1>. I used this vector and the normal vector to find the scalar projection of $\stackrel{\rightarrow}{PQ}$ along $\stackrel{\rightarrow}{n}$. This came out to be $\frac{-2}{3}$.

So, $\frac{-2}{3}$ 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 $\frac{-2}{3}$, that means that $\stackrel{\rightarrow}{n}$$\bullet$$\stackrel{\rightarrow}{PQ}$ = -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 $\stackrel{\rightarrow}{PQ}$ 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!

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vela
Staff Emeritus
Homework Helper
Your professor should have used the unit-length normal vector when calculating where the point on the other plane is.

Oh yes, that makes sense. I was thinking that distance * normal vector would not give the proper distance, but I couldn't figure out why.

So is my method correct or incorrect?

And one last general question...is it correct to have a negative scalar projection, or should it be given as positive? Or does it not matter (so -2/3 and 2/3 is the same thing)?

vela
Staff Emeritus
Homework Helper
With the vectors you used, the scalar projection is -2/3. I wouldn't say that -2/3 is the distance of the point from the plane, however, since distances are positive quantities.

I don't think your answer is correct. I got 2x-y+2z=0 for the other plane. (But I may have messed up as well.)

LCKurtz
Here's another approach you might try. You have a unit normal $\hat n = \frac 1 3\langle 2,-1,2\rangle$, and using that in your formula gave $-\frac 2 3$. This just means your normal points to the other side of the plane than where the given point is.
You know the plane you seek must have the form $2x-y+2z=k$. It will be on the other side of the given point so the same normal will give a positive scalar projection. An easy point on that plane is $(0,-k,0)$. So do the same work and choose $k$ to get $\frac 2 3$. I think you will discover Vela's answer is correct.