quick basic question! no solving required!

[Gavriell]

Calculus 3 - can the following question have different solutions, cuz I have tried different methods to solve, but always get different answers?? The question is ->(Find the line of the intersection of two planes?)
Also how can I check my answer if there are different solutions?

 

[tkhunny]

Equations of lines can be expressed in different ways, but that doesn't make them different lines.

Please demonstrate.

 

[Dr.Peterson]

If you are given the equations of two planes, then there is one solution, which may be a line, or may be empty or may be a plane (if the two planes are actually the same plane). Different methods should not yield different answers, if done correctly. If you get different equations, you should be able to show that they are the same line.

Did you check all your answers (by checking whether every point in your solution set is in both planes)? You should be able to find which, if any, are correct.

You may need to show us the problem and your work, so we can tell you where you are going wrong.

 

[Gavriell]

here is the problem with my solutions

 

[Dr.Peterson]

If you plug your expressions in solution (a) into the equations, you find that they are true. This answer is correct.

Solution (b) fails the test; you didn't show where you got the point (-7, 4, 1). That point is not on either plane. Check whatever work you did there to find an explanation for the error.

Your vector k on the second page is correct; that appears to be the source of k in solution (b).

In your work on the second page, you made a silly error in the last step. The equations you really obtained are x = 1/3 + 2t, y = -1 + 6t, and z = 6t. These pass the substitution test, so this is another correct solution, based on a different parameterization. You should be able to derive (a) from (c) or vice versa.

But I hope you don't seriously believe that 1/3 + 2t = 7/3 t, and -1 + 6t = 5t!

 

[pka]

Here are the given planes:
\(\displaystyle \Pi_1: 3x-5y+4z=6\\\Pi_2: 3x-3y+2z=4\) the normals are \(\displaystyle n_1: <3,-5,4>\\n_2: <3,-3,2>\)
As the normals are not parallel there are two planes with a line of intersection.
Find a point that is on both planes and use \(\displaystyle n_1\times n_2=<2,6,6>\) as the direction vector of the line.