Finding the equation of a plane given two lines

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SUMMARY

The discussion focuses on finding the equation of a plane that contains two given lines represented in vector form. The user initially calculated the cross product of the direction vectors of the lines, resulting in the normal vector <-2,-2,0>, and derived the equation of the plane as -2(x-1)-2(y-1)=0. However, this equation simplifies to the equivalent form x+y=2, demonstrating that multiple representations exist for the same plane. The conversation emphasizes that while there are various forms to express a plane's equation, they can be equivalent.

PREREQUISITES
  • Understanding of vector equations and their representations in three-dimensional space.
  • Knowledge of cross product calculations and their geometric interpretations.
  • Familiarity with the general equation of a plane in both point-normal and standard forms.
  • Ability to simplify algebraic expressions and recognize equivalent equations.
NEXT STEPS
  • Study the properties of vector equations of lines and planes in three-dimensional geometry.
  • Learn about the geometric interpretation of the cross product in relation to planes.
  • Explore different forms of the equation of a plane, including point-normal and standard forms.
  • Practice simplifying and transforming equations to identify equivalent representations.
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Students studying geometry, particularly those focusing on vector calculus and three-dimensional space, as well as educators seeking to clarify the concept of plane equations.

navalava
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Homework Statement


Find an equation of the plane that contains these lines:
r=<1,1,0>+t<1,-1,2>
r=<2,0,2>+s<-1,1,0>

Homework Equations





The Attempt at a Solution


I took the cross product of <1,-1,2> and <-1,1,0> to get <-2,-2,0>.
I used the point (1,1,0) to get the equation of the plane:
-2(x-1)-2(y-1)=0

But the correct answer is supposed to be x+y=2.

I'd really appreciate it if someone could show me what I'm doing wrong. Thanks!
 
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-2(x-1)-2(y-1)=0 and x+y=2 are the same plane. Can you figure out a way to show that?
 
Your answer simplifies to x + y = 2.
 
As with equations of lines in three dimensions, it should be noted that there is not a unique equation for a given plane.
 
glebovg said:
As with equations of lines in three dimensions, it should be noted that there is not a unique equation for a given plane.

I thought that is only true for planes in the form of

a(x - x_0) + b(y - y_0) + c(z - z0) = 0

but not for a'x + b'y + c'z = d
 
There are various ways of describing a plane. The two equations you provided are essentially equivalent.
 
glebovg said:
There are various ways of describing a plane. The two equations you provided are essentially equivalent.

No, but the first one, you can choose any intercept and it can still be reduced to my second one. By just observation the first one "isn't unique", even though the answer will be the same.
 
Since there are infinitely many points on the plane you may choose any point as long as it is on the plane.
 
flyingpig said:
I thought that is only true for planes in the form of

a(x - x_0) + b(y - y_0) + c(z - z0) = 0

but not for a'x + b'y + c'z = d

For the second form you can still multiply by any constant and it's the same plane. That's not quite unique.
 
  • #10
If you multiply it by a constant what you obtain is a different equation of the plane.
 

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