Given planes -9x + 4y + 8z = 14 and 4x -5y + 6z = 8. Let P(x0, y0, z0) be a point on the line of intersection of the planes and suppose that x0 = 12. Find y0 and z0
we are given 2 vectors to work with, and thats about it right
go ahead and let x0 be 12 in the plane equations to get these into line formats
-9(12) + 4y + 8z = 14 -108 +4y +8z = 14 4y = -8z - 94 y = -2z - 94/4 4(12) -5y + 6z = 8 48 -5y +6z = 8 -5y = -6z -40 y = 6/5 z +8 right?
you gotta let me know when i forget how to add :) lets try that again
-9(12) + 4y + 8z = 14 -108 +4y +8z = 14 4y = -8z +122 y = -2z + 122/4
i see that can be reduced more y = -z + 61/2
the rest looks good; y = 6/5 z +8 y = -z + 61/2 now the we have reduced them to lines; we can find where they cross by completing the system of equations we have developed
do you agree? or is there another method you had in mind that you needed help on?
ops sorry, was working another problem. let me see 1 sec
hrm kinda following this, it looks like the same thing just worked in a different way did you come up with 223/11 for y0? maybe my math is wrong but it didn't accept 223/11 for the answer
dunno yet :) lets check
6z -5y = -40 -2z -2y = -61 y = (-366-80)/(-12-10) = 223/11 .... yep, so either the format is off, or the method is twisted in my head
hrm so frustrating.. there aren't any problems like this in our book
z = (-61(-5)-2(40))/(10+12) = 225/22 if so, but then the next step would be to plug it all into the plane equations to test
in essense, we have a picture of the zy plane taken at x=12
so the intersection of the zy instance at x=12 should result in the point they are looking for
what do you mean
i mean, you are prolly used to looking at the xy plane from the point of z=0 when you graph
there are a whole lot of other planes in front of and behind that xy plane at other z spots that may or maynot be the same depending on the equation of each plane in particular
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