EPTMy problem is given a plane represented by a point, p, and a normal vector, n. How do I find the two planes that are perpendicular to my original plane?
Need Basic Math Help
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RekcaH
United States190 Posts
My problem is given a plane represented by a point, p, and a normal vector, n. How do I find the two planes that are perpendicular to my original plane? | ||
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thunk
United States6233 Posts
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Cloud
Sexico5880 Posts
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CDRdude
United States5625 Posts
On May 24 2008 11:51 RekcaH wrote: I'm starting to wish I paid attention during freshman math classes, I never thought how important it would actually be later on. My problem is given a plane represented by a point, p, and a normal vector, n. How do I find the two planes that are perpendicular to my original plane? Your question is kind of confusing. There are infinite planes perpendicular to any given plane. | ||
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RekcaH
United States190 Posts
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randombum
United States2378 Posts
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RekcaH
United States190 Posts
I'm think I may have found a solution. Just take the given normal vector and rotate it so it's along the x axis. Then I know the vectors <0,1,0> and <0,0,1> are perpendicular to it. Then just apply the opposite rotation to <0,1,0> and <0,0,1>. | ||
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wanderer
United States641 Posts
(actually, since we have a vector, the origin is our third point and conveniently enough it is also where our planes will intersect) A normal point is always perpendicular to something, that's why its called normal, so finding out what it is perpendicular to is the first part of the problem. edit another tip: Whenever you have an n-dimensional plane, and you find a normal vector to the plane, you've found the perpendicular plane! Hope that helps. | ||
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rgfdxm
United States239 Posts
The people upthread objecting that there are an infinite number of planes perpendicular to our plane are right, so I'm assuming you haven't written the problem correctly and when you say two orthogonal planes they mean two planes orthogonal to our original plane and to each other. This does narrow the form of our possible solutions to a set of two planes perpendicular to the original plane p, as your formulation says, although it does still leave infinitely many such solution pairs. I'll just solve for an easy one and show why that's valid. What we need, then, are three vectors that are all perpendicular to one another. Each of these vectors will be the normal vector to one of our three planes, and voila. So how to find an easy perpendicular vector to our first vector? A cross-product! That's guaranteed to give us a vector perpendicular to our original vector. But what to take a cross-product of? How about something easy like i = <1,0,0>? Take n x i = m, and m can be the normal of our second plane. Then take n x m = l to get a vector l perpendicular to both our original n and our new m. Now we have 3 perpendicular vectors and therefore three orthogonal planes. If you have some objection to the arbitrariness of choosing i as our vector to cross with n, consider that our solution is certainly not unique. Choosing a different vector than i gives us a different set of orthogonal planes, but that's ok as long as that's what the question was asking for. It makes sense that there are an infinite number of solutions, so as long as the problem gives no further restrictions on the answer we can pick any one of them we like. Cross products with i, j, or k are easy so we might as well use one of those. To see why it makes sense that there are an infinite number of pairs of planes orthogonal with each other and with p, consider the following. If you have any set of three mutually orthogonal vectors (say a, b, and c), consider what happens when you rotate that set of vectors about an axis parallel with one of them (say a). The other two spin around that axis and take on the coordinates of infinitely many pairs of vectors in the plane defined by a. In reverse, if all we have is vector a, we can choose any arbitrary vector d in the plane that it defines (since all vectors in that plane by definition are perpendicular to a) and then rotate about the a axis to put d where b used to be. Now it's clear that a x d = e takes the place of c, so if (a,b,c) are a solution then so must be (a,d,e). So if any vector perpendicular to our original one will work as our second vector, then all we need is to take the cross product of p with any vector we feel like. | ||
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evanthebouncy!
United States12816 Posts
Take the normal vector you have, a vector yes? Now, take that point, and that vector, you make a line. You know how to do that yes? pt-slope form or some shit u call it. Anyways make a line out of it, call it line L. Now, ANY PLANE that contains line L will be perpendicular to the original plane. You're probably wondering "Oh great how the fuck do I make a plane that contains line L?" easy, here's (one way, not the best but for me most intuitive) how: You take line L, it is made of a point, your point p, and a vector, your normal vector. Make a random point outside line L, any point would do, call it point q. Make a line from point p and point q, you know, 2 points make a line, so make it, call it line K. You take line K, it is made of a point, point q, and a vector, call it v. Take your normal vector, take your new vector v, and cross product them. You get another vector, call it r. Now take point p, a line, and r, a new NORMAL vector, and make plane out of that. To get a different plane, pick different point q from step above, should do. by all means send me pm if you are confused as fuck I'll go over it w/ u in detail and why and how and good shits | ||
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