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Equation for the plane

  • #1

Homework Statement


Problem: Please find an equation for the plane that contains the point <3, -2, 4> and that includes the line given by (x-3)/2 = (y+1)/-1, z=5 (in symmetric form). Simplify

Homework Equations


I'm really not sure where to start and what process to take to arrive to my answer.

The Attempt at a Solution


My first attempt at this problem was to turn the symmetric equation into a vector equation <2t-3, -t-1, and 5> then take the normal vector <2, -1, 0> but then I'm not sure if I should be doing that in the first place.
 

Answers and Replies

  • #2
Orodruin
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What you have done is one of the steps in a possible way of finding a solution. However, it is not clear to me that you have a plan of attack. You need to ask yourself what information you need to define a plane and then try to extract that information from the problem. For example, what is your idea behind finding the tangent vector of the line?
 
  • #3
What you have done is one of the steps in a possible way of finding a solution. However, it is not clear to me that you have a plan of attack. You need to ask yourself what information you need to define a plane and then try to extract that information from the problem. For example, what is your idea behind finding the tangent vector of the line?
So I know that I need 2 vectors for a plane and currently I'm given a point and a equation of the line. This is the point where I'm unsure of what to do since I don't know how to utilize the tangent vector of the line.
 
  • #4
Orodruin
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So you have one vector in the plane. Can you think of a way to get a second (linearly independent) based on the information you have available? What are you planning to do once you have two vectors in the plane?
 
  • #5
So you have one vector in the plane. Can you think of a way to get a second (linearly independent) based on the information you have available? What are you planning to do once you have two vectors in the plane?
Could I possibly set t = 1 and get a point for the line and then use the point and the result I got from t=1 to get a vector? I know I want to get the normal vector from two vectors to get the equation of the plane. Not sure how to get the second vector that's linearly independent. (Just to clarify: the first vector on the plane is the line right?)
 
  • #6
Orodruin
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Could I possibly set t = 1 and get a point for the line and then use the point and the result I got from t=1 to get a vector?
Does that vector necessarily lead to a tangent vector to the plane?

Just to clarify: the first vector on the plane is the line right?
More accurately, its tangent vector. If the line is to be in the plane, its tangent vector must be parallel to the plane.
 
  • #7
Does that vector necessarily lead to a tangent vector to the plane?


More accurately, its tangent vector. If the line is to be in the plane, its tangent vector must be parallel to the plane.
So I've thought about this a lot more. This is my thought process. P1: (3,-2,4), P2: (3, -1, 5), P3: t=1: (5, -2, 5). I can then find the vector P1 to P2 and P1 to P3. Use the two vectors I found to do a cross product which will give me the normal vector, which I can then use to come up with the equation of the plane.
 
  • #8
Orodruin
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So I've thought about this a lot more. This is my thought process. P1: (3,-2,4), P2: (3, -1, 5), P3: t=1: (5, -2, 5). I can then find the vector P1 to P2 and P1 to P3. Use the two vectors I found to do a cross product which will give me the normal vector, which I can then use to come up with the equation of the plane.
This is a valid procedure (I assume you got P2 and P3 by picking two points on the line). Note that you could just as well replace either of the two vectors you propose by the vector from P2 to P3, which will just be (proportional to) the tangent vector of the line. I would probably just have taken an arbitrary point on the line and used the tangent vector of the line and the difference vector from the arbitrary point to P1.
 
  • #9
This is a valid procedure (I assume you got P2 and P3 by picking two points on the line). Note that you could just as well replace either of the two vectors you propose by the vector from P2 to P3, which will just be (proportional to) the tangent vector of the line. I would probably just have taken an arbitrary point on the line and used the tangent vector of the line and the difference vector from the arbitrary point to P1.
Thank you so much for your patience and time Orodruin. Been stuck on this problem for a long time.
 
  • #10
Ray Vickson
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So I've thought about this a lot more. This is my thought process. P1: (3,-2,4), P2: (3, -1, 5), P3: t=1: (5, -2, 5). I can then find the vector P1 to P2 and P1 to P3. Use the two vectors I found to do a cross product which will give me the normal vector, which I can then use to come up with the equation of the plane.
What is stopping you from doing that?
 
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