finding the equation of the intersection
1. The problem statement, all variables and given/known data
2. Relevant equations
3. 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
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
then substitute y into plane 2 to get
x-8(-6/13z - 21/13) + z=14 and simplifies
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.
Re: finding the equation of the intersection
5×14 - 28 = 42, not 32.
|All times are GMT -5. The time now is 08:08 AM.|
Powered by vBulletin Copyright ©2000 - 2013, Jelsoft Enterprises Ltd.
© 2012 Physics Forums