Finding the equation of the intersection

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The discussion revolves around finding the intersection of three planes defined by their equations. The user initially derived two contradictory equations from the planes, leading to the conclusion that no common solution exists. However, the textbook provides a parametric solution, suggesting that the planes are consistent despite the contradictions found in the elimination process. The user questions why one equation was chosen over the other and why the process continued after reaching a contradiction. Ultimately, the parametric equations derived do satisfy all three planes, indicating a line of intersection rather than a single point.
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Homework Statement


Plane 1:2x-3y+8z=7
Plane 2:x-8y+z=14
Plane 3:5x-14y+17z=28

Homework Equations


N/A


The Attempt at a Solution


I took plane 1 and subtracted (plane 2)x2 to get
13y+6z=-21 (I will refer to this as equation 1)

Then I took (plane 2)x5 and subtracted plane 3 to get
-26y-12z=32
which simplifies to
13y+6z=-16 (I will refer to this as equation 2)

Then I was stumped because the two equations contradict each other and state 0=-5
which leads me to believe that there is no values of x,y,z that can satisfy all planes

although the answers in the book say
"consistent [14/13,-21/13,0] + t[-61/13,-6/13,1]"

I understand how they get the answer I just don't understand why, if equation 1 and 2 do not provide a valid statement why do they continue and find the parametric equations

like let z = t
13y=-6t -21
y=-6/13t -21/13
then substitute y into plane 2 to get
x-8(-6/13z - 21/13) + z=14 and simplifies
x=61/13z +14/13
then z=0+t
which gets the correct parametric equations which I tested by using different values of t and the points do satisfy all planes
I guess the real question I have is why do they chose "equation 1" instead of "equation 2" which would yield a different result if you were to use that to fidn the parametric equations, which I'm assuming the parametric equations don't work and why do you continue after the elimination of equation 1 and 2 which results in 0=-5.
 
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cdot9 said:

Homework Statement


Plane 1:2x-3y+8z=7
Plane 2:x-8y+z=14
Plane 3:5x-14y+17z=28

Homework Equations


N/A


The Attempt at a Solution


I took plane 1 and subtracted (plane 2)x2 to get
13y+6z=-21 (I will refer to this as equation 1)

Then I took (plane 2)x5 and subtracted plane 3 to get
-26y-12z=32
Hello cdot9. Welcome to PF !

5×14 - 28 = 42, not 32.
which simplifies to
13y+6z=-16 (I will refer to this as equation 2)

Then I was stumped because the two equations contradict each other and state 0=-5
which leads me to believe that there is no values of x,y,z that can satisfy all planes

although the answers in the book say
"consistent [14/13,-21/13,0] + t[-61/13,-6/13,1]"

I understand how they get the answer I just don't understand why, if equation 1 and 2 do not provide a valid statement why do they continue and find the parametric equations

like let z = t
13y=-6t -21
y=-6/13t -21/13
then substitute y into plane 2 to get
x-8(-6/13z - 21/13) + z=14 and simplifies
x=61/13z +14/13
then z=0+t
which gets the correct parametric equations which I tested by using different values of t and the points do satisfy all planes
I guess the real question I have is why do they chose "equation 1" instead of "equation 2" which would yield a different result if you were to use that to fidn the parametric equations, which I'm assuming the parametric equations don't work and why do you continue after the elimination of equation 1 and 2 which results in 0=-5.
 
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...
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