The discussion revolves around finding the equations of a plane that passes through a specific point and contains a given line in three-dimensional space. The subject area includes vector geometry and the properties of planes and lines in Cartesian coordinates.
Some participants express confusion regarding the vectors derived from the line equations and their implications for defining the plane. There is a recognition that additional information may be necessary to uniquely determine the plane's equation, indicating an ongoing exploration of the problem's requirements.
Participants note that having a point and a line is insufficient to define a unique plane, suggesting that the problem statement may lack completeness.
dodgedanpei said:Homework Statement
The plane that passes through the point (1, 6, 4) and contains the line
x = 1 + 2t; y = 2 - 3t; z = 3 - t where t is an element of R
Homework Equations
x = 1 + 2t; y = 2 - 3t; z = 3 - t
The Attempt at a Solution
Let L be the solution.
L = (1,6,4) - ?
t = (x -1)/ 2 = (2-y)/3 = 3-z
This doesn't make any sense at all. First off, <1, 2, 3> is a vector from the origin to the point (1, 2, 3) on the line. Second, the vector t<2, -3, -1> = <2t, -3t, -t> is a vector that has the same direction as the line.dodgedanpei said:Well I tried making 2 vectors by using the 3 equations.
I got
vector x = t(2,-3,-1) = (1,2,3)
dodgedanpei said:But the two vectors are meant to be s(0,-4,7) + t(-8,0,25) , where s and t are real numbers.