Hello,
Q: Consider the surface:
x^2 + 3y^2 -z^2 + 3x = 1
With which of the following surfaces does the given surface have a curve of intersection that lies in a plane?
I know that two of the correct answers to this are:
2x^2 + 6y^2 - 2z^2 - 9y = 0
and 3x^2 + 9y^2 - 3z^2 - 8y = 0
I'm not at all sure of what procedure to use in this problem. I guess i'm not exactly sure what the question is asking by a 'curve of intersection'. So I tried working this problem out in reverse. Just eyeballing it, it looks like I can take the normal vector for the original surface (plane) to be <1,3,-1>. The 'correct' answers happen to be planes that all appear to have normal vectors that are scalar multiples of the original plane's normal vector. ie: <2,6,-2> and <3,9,-3>
If I remember right this means that all three planes are parallel. But I still don't understand why that makes these answers correct. Can you help me understand this?
Q: Consider the surface:
x^2 + 3y^2 -z^2 + 3x = 1
With which of the following surfaces does the given surface have a curve of intersection that lies in a plane?
I know that two of the correct answers to this are:
2x^2 + 6y^2 - 2z^2 - 9y = 0
and 3x^2 + 9y^2 - 3z^2 - 8y = 0
I'm not at all sure of what procedure to use in this problem. I guess i'm not exactly sure what the question is asking by a 'curve of intersection'. So I tried working this problem out in reverse. Just eyeballing it, it looks like I can take the normal vector for the original surface (plane) to be <1,3,-1>. The 'correct' answers happen to be planes that all appear to have normal vectors that are scalar multiples of the original plane's normal vector. ie: <2,6,-2> and <3,9,-3>
If I remember right this means that all three planes are parallel. But I still don't understand why that makes these answers correct. Can you help me understand this?