Proving the line lies on the plane

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SUMMARY

The discussion focuses on determining whether the line defined by the equation (x, y, z) = (5, -4, 6) + u(1, 4, -1) lies within the plane defined by (x, y, z) = (3, 0, 2) + s(1, 1, -1) + t(2, -1, 1). The solution involves substituting u=0 into the line's equation to find the point (5, -4, 6) and then establishing that this point can be expressed as a linear combination of the vectors defining the plane. The key conclusion is that proving the vector difference (5, -4, 6) - (3, 0, 2) is a linear combination of the plane's direction vectors suffices to show the line lies in the plane.

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  • Basic concepts of planes and lines in vector geometry
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Students and educators in mathematics, particularly those studying vector geometry, linear algebra, and related fields. This discussion is beneficial for anyone looking to deepen their understanding of the relationship between lines and planes in three-dimensional space.

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Homework Statement


Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Justify your answer algebraically.


Homework Equations




The Attempt at a Solution


I started by getting the parametric equation of (x, y, z) = (5, -4, 6) + u(1,4,-1)
x=5+u
y=-4+4u
z=6-u
I then subbed in u=0 to get a set of points
x=5
y=-4
z=6

I then got the parametric equation for (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)
x=3+s+2t
y=0+s-t
z=2-s+t

I decided to use the points (5,-4,6)
 
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Hi soulja101! :smile:
soulja101 said:
Does the line with equation (x, y, z) = (5, -4, 6) + u(1,4,-1) lie in the plane with equation (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)? Justify your answer algebraically.

I started by getting the parametric equation of (x, y, z) = (5, -4, 6) + u(1,4,-1)
x=5+u
y=-4+4u
z=6-u
I then subbed in u=0 to get a set of points
x=5
y=-4
z=6

This is very long-winded :rolleyes:

you could just put u = 0 in (x, y, z) = (5, -4, 6) + u(1,4,-1), giving (5, -4, 6) immediately :wink:
I then got the parametric equation for (x, y, z) = (3, 0, 2) + s(1,1,-1) + t(2, -1, 1)
x=3+s+2t
y=0+s-t
z=2-s+t

why make it so complicated?

all you have to prove is that (5, -4, 6) minus (3, 0, 2) is a linear combination of (1,1,-1) and (2, -1, 1), and then the same for (1,4,-1) :smile:
 

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