Finding the solution of three planes

  • Thread starter Thread starter gracelette
  • Start date Start date
  • Tags Tags
    Planes
AI Thread Summary
The discussion revolves around solving a system of equations representing three planes and determining the conditions under which they have no solutions, meet in a line, or intersect at a point. A key insight is that for no solutions, the parameter 'a' must not equal 3, as this leads to an inconsistency in the equations. For the planes to meet in a line, the last equation must simplify to 0z = 0, while a unique solution occurs when the last equation takes the form kz = constant, with k not equal to 0. Participants also discuss the importance of identifying pivots in the augmented matrix to ascertain the nature of the solutions. The conversation concludes with a resolution of the problem, indicating that the initial confusion has been clarified.
gracelette
Messages
2
Reaction score
0

Homework Statement


given the planes with equations:
x + y + 7z = -7
2x + 3y + 17z = -16
x + 2y + (a^2 + 1) z = 3a

find values for the constant a for which:
-there are no solutions
-the planes meet in a line. in this case find the parametric equation of the line
-meet at a point. then find the parametric equation of the point.

Homework Equations



gaussian elimination? I don't think there are any. =S

The Attempt at a Solution



I put the planes into an augmented matrix, and then into echelon form, and got (I think its correct) the general solution, which is attached. From the I can read off the equation of the line, right? The parts I'm really stuck on are parts one and three. I realize that for the planes to meet in a point, there has to be 3 pivots but beyond that I'm a bit stumped.

Well actually, I have a bit of a clue for part one, that a =/= 3 because then 0z = 18 which is inconsistent. is that right? But for part 3, I'm not sure how to do it, really.
 

Attachments

  • attempt at a solution.png
    attempt at a solution.png
    23.5 KB · Views: 543
Last edited:
Physics news on Phys.org
gracelette said:

Homework Statement


given the planes with equations:
x + y + 7z = -7
2x + 3y + 17z = -16
x + 2y + (a^2 + 1) z = 3a

find values for the constant a for which:
-there are no solutions
-the planes meet in a line. in this case find the parametric equation of the line
-meet at a point. then find the parametric equation of the point.

Homework Equations



gaussian elimination? I don't think there are any. =S

The Attempt at a Solution



I put the planes into an augmented matrix, and then into echelon form, and got (I think its correct) the general solution, which is attached. From the I can read off the equation of the line, right? The parts I'm really stuck on are parts one and three. I realize that for the planes to meet in a point, there has to be 3 pivots but beyond that I'm a bit stumped.

Well actually, I have a bit of a clue for part one, that a =/= 3 because then 0z = 18 which is inconsistent. is that right? But for part 3, I'm not sure how to do it, really.
You have a mistake. In your first augmented matrix, the bottom row is 0 0 a2 - 9 | 3a + 9

Right after that, you say let a = -3i. There are two real values that make a2 - 9 equal to 0.

For your other questions, there will be no solutions if the last equation is 0z = k, with k not equal to 0.
There will be multiple solutions (points along a line) if the last equation is 0z = 0.
Finally, there will be a unique solution (a single point) if the last equation is kz = <whatever>, with k not equal to 0.
 
Ooops. I guess this is what happens when I try do maths at obscene times of day. Thankyou, i worked it out now!
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

Similar threads

Back
Top