Linear Algebra(parametric equations)

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Homework Help Overview

The discussion revolves around determining the parametric equations of a line that passes through a specific point and is perpendicular to a given plane defined by the equation 2x - 5y = 6. The subject area is linear algebra, particularly focusing on vector representation and geometric interpretations of lines and planes.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the use of the point-parallel form for the line, questioning how to derive the direction vector when only the plane equation is provided. There is discussion about finding the normal vector to the plane and its relationship to the line's direction.

Discussion Status

Participants are actively engaging with the problem, with some suggesting methods to find the normal vector and others expressing uncertainty about the next steps. There is a recognition that the normal vector can be derived directly from the plane's equation, and some guidance has been offered regarding the properties of normal vectors in relation to plane equations.

Contextual Notes

Some participants note a lack of familiarity with finding normals to planes and express difficulty in locating relevant information. The discussion includes references to the general form of plane equations and their implications for identifying normal vectors.

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The question states:
Determine the parametric equations of the line passing through the point P(2,1,0) and perpendicular to the plane 2x-5y=6.

The equation to obviously use is the "point-parallel" I'm guessing. X= P +tV where P is the point and V is a vector, and X=(x,y,z). But how would I use this since there isn't a vector given for V and instead is the plane 2x-5y=6? Maybe this isn't the one I need to use. Please help.
 
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The direction vector for the line is the normal to the plane. Do you know how to find the normal to a plane?
 
Dick said:
The direction vector for the line is the normal to the plane. Do you know how to find the normal to a plane?

n . (x - p) ?
 
You can also write that as n.(x,y,z)=constant. What's a normal for your plane?
 
Dick said:
You can also write that as n.(x,y,z)=constant. What's a normal for your plane?

hm I honestly have no idea how to do this problem. I could set it up like this but then what would I do next?

N . (( x,y,z) - (2,1,0)= 0

x-2
y-1
z-0 ?
 
You don't need the point P to find the normal to the plane. Please look up how to find the normal to a plane ax+by+cz=constant.
 
Dick said:
You don't need the point P to find the normal to the plane. Please look up how to find the normal to a plane ax+by+cz=constant.

Ok I am not really finding too much on the topic.. I did however find one thing that said you want to use the cross product to find the norm to the plane.
 
would the norm happen to be (0,0,-4)
 
No, one choice for the normal would be (2,-5,0). Now where did I get that?
 
  • #10
Dick said:
No, one choice for the normal would be (2,-5,0). Now where did I get that?

ahhh from the equation itself. so that would be the case everytime?
 
  • #11
ur5pointos2sl said:
ahhh from the equation itself. so that would be the case everytime?


Whenever looking at the equation of a plane in the form ax + by + cz + d = 0, the vector (a,b,c) is always the vector that is perpendicular to the plane (which is a normal vector). Since you are trying to find the direction vector for a line perpendicular to the given plane, finding the direction vector is easy. You also have the point P, so you should be able to answer it now.
 
  • #12
ur5pointos2sl said:
ahhh from the equation itself. so that would be the case everytime?

I'd hoped you could look this up but, yes, a normal to ax+by+cz=constant is (a,b,c). For your homework, explain to me why.
 

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