# Mathematical Concept: Finding Multiple Potential Planes

Registered User regular
edited October 2008
There's a concept we have to learn for Calculus III that involves finding multiple potential planes that pass through certain points and that are a certain distance from another point.

In other words, only two points are given, whereas you need three for a plane. The last point is not given, but you are told that either of the possible planes (there can only be two in my understanding) is a certain distance from a single point.

So I suppose I have the following questions:

1) How do you find the distance from a single point to a plane?
2) How do you use that information to develop an equation that will give you both potential planes?
3) If you know of better places to be asking these kinds of conceptual questions, what are they?

My teacher's advice only goes so far as "there are two answers" and "draw a good picture to figure this out", but I'm still not able to come across what needs to be done. Thanks for any and all help you can provide!

P.S. If you need more specific examples or the concepts I have used so far (none of which I can put into a coherent answer, mind you) then I'll try to assist you. I'll also clarify anything that's too vague.

Mugenmidget on

## Posts

• Registered User regular
edited September 2008
(the following post is under the assumption that you are given 3 points- two that must be in the desired planes, and one that must be a certain distance away from said planes. If I misunderstood, disregard it, I guess.)

To find the distance from a single point to a plane, you need the normal of the plane- any line segment, usually a unit vector, perpendicular to the plane. Then you use this to find a line segment in this direction which passes through the point in question- the distance between it and where it passes through the plane is obviously what you're looking for.

I'll post more if I have time, but for now I should be doing my own calc 4 homework.

Tarantio on
• Registered User regular
edited September 2008
Well, you could check out this for a start. It uses the dot product to get the distance of a point to a line (which you could do since you seem to have that info), and a point to the plane. Problem with the latter though is you do not have the equation of the plane.

If this third point you are given is listed as the closest distance to the plane, well now you are given two vectors, one that's on the plane, and one that's perpendicular to the plane. Though it's late so I might be mistaken, but you should be able to get a third vector using the cross product of those two you found, and that will give two vectors on the same plane, which should allow you to get the equation of the plane.

Sorry I cannot give you a complete answer, but I think you should have something to start with. And usually when they say you could draw a picture to figure it out, when you actually draw this picture, it becomes obvious the answer (usually ;p).

Tzyr on
"Perfection is achieved, not when there is nothing more to add, but when there is nothing left to take away. "
• Registered User regular
edited September 2008
All you need for the equation of a plane is a normal vector and a point on the plane (or a line along the plane and a point on the plane but not on the line, or three points on the plane not all on one line, or two lines along the plane)

If you have the distance between your point and the plane, and then use that dot product method to find the distance to the line of the other two points, then those two lengths are two sides of a right triangle, and you can use trigonometry to find the angle between the plane of the three points you have and the plane you're trying to find. I just can't think of exactly how to apply that angle to find the equation of the plane you want.

Tarantio on
• Registered User regular
edited September 2008
Tarantio wrote: »
(the following post is under the assumption that you are given 3 points- two that must be in the desired planes, and one that must be a certain distance away from said planes. If I misunderstood, disregard it, I guess.)

To find the distance from a single point to a plane, you need the normal of the plane- any line segment, usually a unit vector, perpendicular to the plane. Then you use this to find a line segment in this direction which passes through the point in question- the distance between it and where it passes through the plane is obviously what you're looking for.

I'll post more if I have time, but for now I should be doing my own calc 4 homework.
Thanks for the help, I appreciate you taking time out of your own work.

I guess my problem is then how do I find a normal to a plane when I can't necessarily find two vectors in it. I'm sure the distance formula has to be thrown in somewhere and that's why I'll have two answers (since the formula would give the same distance even if the signs on the components were changed).

Basically, here's something very similar to the problem:

"Find the possible planes that pass through the points (1,1,2), (2,0,1) and are a distance of 1/2 from the point (1,2,3)."

So if I'm crossing vectors to find a normal, but only know two points for sure, how do I set it up with the information I'm given about the distance from that point? I assume once I figure that much out, the vector normal to the plane provides the coefficients for my plane equation, which I can then combine with whichever point I used in both vectors to get my standard equation.

This boils down to me having a lot of trouble visualizing the setup for this particular problem. I feel like I might have the general approach but I can't really execute it.

Mugenmidget on
• Registered User regular
edited September 2008
Alright, I spent too long on that post and missed the recent replies. And I find myself thoroughly confused with this new information. :-P

I feel like the trigonometric approach might be complicating things, but I really can't say for sure since I obviously don't know how to solve this in the first place. I'll see if I can plot/sketch some sort of 3D picture of this and post it here.

Mugenmidget on
• Registered User regular
edited September 2008
Well, here's how I visualize it:

The two points on the plane are essentially a line- treat them as one line that passes through both points, and you'll end up with the same answer.

So, now you've got a line and a plane. There are an infinite number of planes that pass through the line, you need the two that are a certain distance from your point.

I imagine a line perpendicular to the first line, which, if rotated with the first line as its axis, would intersect with your point. Now, the two places where that line gets x distance away from the point (say, intersects with a sphere of radius x centered around your point) are where the line we added is along the planes we're looking for.

For a more concrete example, imagine a marble held between the pages of a book. The pages of the book are the planes, the center of the marble is the point, the radius of the marble is the distance from the point to the planes, and the spine of the book is the line which holds the other two points.

Tarantio on
• Registered User regular
edited October 2008
Tarantio wrote: »
Well, here's how I visualize it:

The two points on the plane are essentially a line- treat them as one line that passes through both points, and you'll end up with the same answer.

So, now you've got a line and a plane. There are an infinite number of planes that pass through the line, you need the two that are a certain distance from your point.

I imagine a line perpendicular to the first line, which, if rotated with the first line as its axis, would intersect with your point. Now, the two places where that line gets x distance away from the point (say, intersects with a sphere of radius x centered around your point) are where the line we added is along the planes we're looking for.

For a more concrete example, imagine a marble held between the pages of a book. The pages of the book are the planes, the center of the marble is the point, the radius of the marble is the distance from the point to the planes, and the spine of the book is the line which holds the other two points.
I can envision it a lot better now thanks to your explanation, but I'm no closer to finding out how to calculate this. Any ideas?

Mugenmidget on