1 more calc question: How to find eq. of Plane given 2 parallel vectors?

[phatj]

So I have two parallel vectors, I'll just call them a<1,2,3> and b<2,4,6>.

How do I find the equation of the plane that contains these 2 parallel lines?

You cant axb because the cross product is 0 (|| lines), so I can't find an "n" normal vector.

Maybe Im looking too deeply into the problem?

 

[phatj]

.

 

[DeMeo]

Originally posted by: phatj
So I have two parallel vectors, I'll just call them a<1,2,3> and b<2,4,6>.

How do I find the equation of the plane that contains these 2 parallel lines?

You cant axb because the cross product is 0 (|| lines), so I can't find an "n" normal vector.

Maybe Im looking too deeply into the problem?

I beleive you will find the answer in episode 12 of Star Trek Voyager. Near the end of the show, the alien guy is talking to the dude with the tatooed face and he explains this whole theory in detail.

 

[phatj]

ok..

 

[Muzzan]

You can't determine "the" plane that contains two parallel vectors, since there is an infinite number of such planes. Imagine two parallel vectors in R^3. Consider a plane that contains both of them. Rotate that plane slightly around axis "defined" by those vectors. You now have another plane which also contains them. You need another vector and a point on the plane to be able to uniquely determine a plane.

 

[Howard]

Originally posted by: Muzzan
You can't determine "the" plane that contains two parallel vectors, since there is an infinite number of such planes. Imagine two parallel vectors in R^3. Consider a plane that contains both of them. Rotate that plane slightly around axis "defined" by those vectors. You now have another plane which also contains them. You need another vector and a point on the plane to be able to uniquely determine a plane.

He means parallel vectors that are NOT coincident, if you extend both (stretching the definition of a vector, but oh well).

 

[Howard]

Originally posted by: Howard

Originally posted by: Muzzan
You can't determine "the" plane that contains two parallel vectors, since there is an infinite number of such planes. Imagine two parallel vectors in R^3. Consider a plane that contains both of them. Rotate that plane slightly around axis "defined" by those vectors. You now have another plane which also contains them. You need another vector and a point on the plane to be able to uniquely determine a plane.

He means parallel vectors that are NOT coincident, if you extend both (stretching the definition of a vector, but oh well).

You know those bars the gymnasts spin around on? Those bars are the vectors.

Or, that's the meaning I'm getting.

 

[otispunkmeyer]

a.b = |a|.|b| cos(angle)

if a and b are perpendicular then their scalar product will be = 0 (zero)

nvm i didnt read properly

*hopes this is useful anyway*

 

[otispunkmeyer]

i know what u are wanting to do, and i have done it before, but i jus cant think right now, which means i better get revising!!!

 

[Muzzan]

Howard, my point still stands. He needs more information to be able to uniequly determine a plane. And I don't think he meant non-coincident vectors, if that were so, he wouldn't have specified them using coordinates (which assumes some sort of origin, and that the vectors are "lines" starting at the origin, etc).

To expand on what you said about gymnast bars (probably not the correct term )... If you know the gymnast bars' position in space, i.e. their start and ending points, then yes, you could determine a plane. Suppose one of the bars start at A and end at B, and that the other bar ends at C. Then the vectors B - A and B - C will lie in the plane, and will not be parallel (assuming nothing is degenerate). So you could determine a normal vector to the plane. (But you see, the above problem adds information about the "starting point" of both (quasi-)vectors, which when you specify them using coordinates is taken to be (0, 0, 0)...).

i know what u are wanting to do, and i have done it before, but i jus cant think right now, which means i better get revising!!!

If you have in fact done it before, you either did it wrongly, or you had two non-parallel vectors and a point (not just two parallel vectors).

 

[jamesave]

never recall calc consist of vector problem.

 

[Armitage]

Create a third vector C between any point on vector A and any point on vector B.
This vecotr also lies in the plane.
Now you can define the plane by A x C or B x C

 

[Muzzan]

But Armitage, if A and B are parallel then your vector C will also be parallel to both. (Talking about "points on vectors" makes my skin crawl btw, surely "scalar multiple" is what you meant though).

 

[Armitage]

Originally posted by: Muzzan
But Armitage, if A and B are parallel then your vector C will also be parallel to both. (Talking about "points on vectors" makes my skin crawl btw, surely "scalar multiple" is what you meant though).

No, if C passes through a point on A and a point on B, then it can't possibly be parallel to either unless the lines are coincident, in which case you can't define a single plane.

Why does "points on a vector" make your skin crawl? Seems pretty straight forward to me? But I'm an engineer not a mathemetician.

 

[chuckywang]

What Muzzan is saying is that vectors only have magnitude and direction, not position. Therefore, it makes no sense to say that "vectors are parallel." However, it makes perfect sense for "lines to be parallel" which is what I think the OP is trying to say. Now, two parallel lines will define a plane. If the OP can provide me with two LINES, not two VECTORS, then I can solve this problem.

 

[chuckywang]

Originally posted by: phatj
So I have two parallel vectors, I'll just call them a<1,2,3> and b<2,4,6>.

How do I find the equation of the plane that contains these 2 parallel lines?

You cant axb because the cross product is 0 (|| lines), so I can't find an "n" normal vector.

Maybe Im looking too deeply into the problem?

Right now, I'm not sure what "a<1,2,3> and b<2,4,6>" means. They look like vectors to me, not lines. Once this is cleared up, I can solve this problem.

 

[chuckywang]

Originally posted by: Muzzan
But Armitage, if A and B are parallel then your vector C will also be parallel to both. (Talking about "points on vectors" makes my skin crawl btw, surely "scalar multiple" is what you meant though).

Change the word "vector" to the word "line" and Armitage is correct in his method to solve this problem.

 

[UncleWai]

+1 vote for Armitage

 

[Muzzan]

What Muzzan is saying is that vectors only have magnitude and direction, not position. Therefore, it makes no sense to say that "vectors are parallel."

No, that's not really what I'm saying, because it does make perfect sense to talk about parallel vectors. Why, both Planetmath and Mathworld manage to define it...

(I curse this forum's markup language!)

 

[phatj]

Here is the context of the actual problem. I worded my OP incorrectly:

Prove that the line of intersection of the planes x+2y-z=2 and 3x+2y+2z=7 is parallel to the line x = 1+6t, y=3-5t, and z = 2-4t. Find an equation of the plane determined by these two lines.

 

[phatj]

.

 

[Muzzan]

Solving this system:

{ x + 2y - z = 2
{ 3x + 2y + 2z = 7

yields

x = 5/2 - 3/2t
y = -1/4 + 5/4t
z = t

where t is any real. Thus the line of intersection (call it L_1) has the equation (5/2, -1/4, 0) + t(-3/2, 5/4, 1), with a direction vector of A = (-3/2, 5/4, 1). The line given (call it L_2),

x = 1 + 6t
y = 3 - 5t
z = 2 - 4t

has the direction vector B = (6, -5, -4). These vectors are parallel, since A = -1/4 * B, so L_1 and L_2 are indeed parallel.

Determining the plane which contains both is now just a matter of picking two arbitrary points a, b on L_1, and an arbitrary point c on L_2, and forming the differences a - b and a - c. Then (a - b) x (a - c) will yield a normal to the plane, etc. (There are many variations of the last step, of course).