Help needed with 3d geometry problem: important

[elphin]

help needed with 3d geometry problem: important!

Homework Statement

find the equations of the two planes through the origin which are parallel to the line
(x - 1)/2 = (-y-3) = (-z-1)/2
and at a distance of 5/3 from it. also, show that the planes are perpendicular

Homework Equations

The Attempt at a Solution

my attempt at the solution

plane equations passing through (0,0,0) are ax+by+cz=0 (&) a'x+b'y+c'z=0
now, 2.a - b - 2.c = 0 (since direction ratios of given line is perpendicular to the normal of the plane)
also by distance formula (a.(1) + b.(-3) +c.(-1))/(a^2 + b^2 + c^2)^(1/2) = 5/3
the a - 3.b - c = 5/3 (since a^2 + b^2 + c^2 = 1)
and now i am stuck...

FYI : [the answer is 2x+2y+z=0; x-2y+2z=0]

 

[Dick]

You have got the equations you need to find the normals. 2*a-b-2*c=0, a-3*b-c=5/3 and a^2+b^2+c^2=1. If you solve those will get normals for the two planes. The books answer doesn't assume a^2+b^2+c^2=1 so be prepared to multiply the normal by a constant to get the books answers.

 

[elphin]

the problem is the equations are unsolvable

2*a - b - 2*c = 0
the we get (a - c) = b/2
substitute this in a - 3*b - c =5/3
we get b = -2/3
now substitute b = -2/3 in a^2 + b^2 + c^2 = 1
we get a^2 + c^2 = -1/3 !

 

[elphin]

this means that my basic equations are wrong.. i am missing a conceptual link here.. please help...!

 

[Ray Vickson]

the problem is the equations are unsolvable

2*a - b - 2*c = 0
the we get (a - c) = b/2
substitute this in a - 3*b - c =5/3
we get b = -2/3
now substitute b = -2/3 in a^2 + b^2 + c^2 = 1
we get a^2 + c^2 = -1/3 !

No, you don't get this. You should get a^2 + c^2 = 1 - (2/3)^2 = 5/9.

RGV

 

[elphin]

ooops .. sorry .. you are right ..

let me sum up this situation in an equation...

sleepless night + math => (2/3)^2 = (4/3)

thanks dude...