Equation for Line of Plane?

COME NOAH, HADRIEL, I guess not 21 anymore....XTOTHEL and uhhh. Maybe Devil's Sunrise idk.

Okay so my teacher gave us an extra credit question.

Find the equation of a plane perpendicular (orthogonal) to another plane that also contains two other points in space.

We know the equation of the plane that we are making to be perpendicular, Ax + By + Cz = D, and we know the position/coordinations of the two points we're trying to make our new plane contain: (x[SUB]1[/SUB], y[SUB]1[/SUB], z[SUB]1[/SUB]) and (x[SUB]2[/SUB], y[SUB]2[/SUB], z[SUB]2[/SUB]).

This is about all the information I can explain to you, it's hard to explain really because you need to know how an XYZ plane coordinate system works to really even grasp this question. If needed, I can try to use paint to draw a picture of what I mean but I know it's gonna come out crappy.

To give a seperate example, if we had wanted to make a plane, let's call it Q just to name it, perpendicular to some other plane W, and we wanted Q to contain one point in space, (x[SUB]1[/SUB], y[SUB]1[/SUB], z[SUB]1[/SUB]), then the equation of that plane would be:

 Spoiler

There would be no equation because you could rotate the plane Q around the point while still being perpendicular to W and have infinite planes like that.

But our question is specific, with two points, so I need the equation of that plane, Q.

Good luck me.

without actually doing the problem, this is a way to do it...

find the equation of the line orthogonal to Ax + By + Cz = D

pick a point on that line

then you have 3 points, from which you can define a plane

(make sure the point you pick is not also on the line defined by (x1,y1,z1) and (x2,y2,z2) or youll run into problems, not that you can really stumble on that doing it generally like this)

There are various ways you can go about this. Modular's is probably my favorite way to define a plane.

Another way is to construct the plane with 2 non-parallel lines, one of which is defined by the 2 points given, the other would be a line orthogonal to the given surface.
To find the 2nd line, you need the normal vector to Ax + By + Cz = D, and a point on that plane.

I actually forgot everything about this, maybe it's time I did some refreshing.

Essentially modular got it all. Get the normal of the 1st plant, and then use that with any other point-in-space to define the next plane.

If they give 2 points, then it seems like they're intentionally asking you do go by the R = point-vector + directional-vector1 + directional-vector2 method. Which is easier if you already have the normal as a line vector. Of course, see which form of a plane expression they want you to give... dot product? R? Cartesian?

Should also say that I'm getting old... and am not going to be able to answer very many questions in the near future. For various reasons.

Hadriel

! Yeah the plan was to use vector n, the normal to the plane we're making perpendicular .

That's very very interesting. I never really thought about doing it those ways, both Modular and Kalovale's ideas are really good.

That's enough for now, don't bother doing the work if you don't want to, I'll do it and give you the answer I got later just in case you're interested.

No Hadriel you must continue to post math for me.

Okay so. I'm pretty sure I have it done.

I have the two points already, (x[SUB]1[/SUB], y[SUB]1[/SUB], z[SUB]1[/SUB]) and (x[SUB]2[/SUB], y[SUB]2[/SUB], z[SUB]2[/SUB]).

I know from plane Ax + By + Cz = D that there is a normal vector n = Ai + Bj + Ck, which gives the position (A, B, C).

Now what I have three points I set up a matrix

[x y z 1
x1 y1 z1 1
x2 y2 z2 1
A B C 1] equal to =0

From there I'll make matrix:

[x-x1 y-y1 z-z1
x2-x1 y2-y1 z2-z1
A-x1 B-y1 C-z1] = 0

And solve the determinate of that. Hehehe I'll run it by my teacher, class is at 5 PM today. Thanks for the help you math geniuses!

Hehe got green light from professor today saying I did it right. I may try to solve the question using two vectors like Kalo said an hand it in for extra extra credit. Nice job guys.

i thought its nifty that theres a degree of freedom in the problem statement. you can rotate the ABCD plane in say... the theta direction of a spherical coordinate system and still end up with the same answer.