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ElijahRockers
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Homework Statement
Find two different planes whose intersection is the line [itex]x=-7+t, y=9-t, z=6+2t[/itex]. Write equations for each plane in the form [itex]Ax+By+Cz=D[/itex].
Homework Equations
The solutions are
x+y=2;
2y+z=24;
but I do not know how to obtain these answers.
(This is not for class work, I am just trying to teach myself.)
The Attempt at a Solution
When t = 0, the point on the line (and also a point on both planes) is P=(-7,9,6).
I understand that any point on the line must also be a point on both planes. What I really need is to find the 'tilt' or normal vector of each plane, though when I visualize it, I can see an infinite number of intersecting planes that could share this line. Am I visualizing it wrong?
Understanding that a vector parallel to the line is [itex]L=<i-j+2k>[/itex] and is also the result of the cross product between the normal vectors of each plane, I tried to work the cross product backwards to obtain the two normal vectors. This didn't seem to get me anywhere.
[itex]n_{1}=<A_{1}i+B_{1}j+C_{1}k>[/itex]
[itex]n_{2}=<A_{2}i+B_{2}j+C_{2}k>[/itex]
[itex]n_{1} \times n_{2} = <i-j+2k>[/itex]
I wound up with three equations and 6 variables.
[itex]B_{1}C_{2}-B_{2}C_{1}=1[/itex]
[itex]A_{2}C_{1}-A_{1}C_{2}=-1[/itex]
[itex]A_{1}B_{2}-A_{2}B_{1}=2[/itex]
Looking back on the process of finding the line when given the two planes, I notice that I set z to 0 so solve for x in terms of y, and did some substitution to eventually find a point on the line, but I am unsure of what the analogous 'backwards' process of this would be.
Any insight would be appreciated.
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