how do i find the distance from the point (1,0,2) to the line of intersection of two planes?

Question

the two planes are: x+y-2z +1 = 0 and 2x-y-2z-1=0. i know that you need to find the intersection and i can go on from there but how can you figure out the intersecting line if you have three unknowns and only two equations? yeah well please if u kow u can help please do. thanks in advance.

Answer 1

maybe:
x+y-2z+1=2x-y-2z-1
-x+2y+2=0
x=2y+2---------------remember this

than standard:
x+y-2z+1=0 add to equation under
2x-y-2z-1=0
------------------
3x-4z=0-----------------------remember this
x=4/3 z and z=3/4 x

than:
x+y-2z+1=0 / multiply by (-2) and add to equation under
2x-y-2z-1=0
----------------------
-3y - 6 z -3 = 0 or 3y+6z= - 3 -----------remember this


now we have
3y+6z= - 3 / multiply by (2/3) and add to equation under
3x - 4z= 0
------------------------
3x+2y= - 2

so than you use this final equation
3x+2y= - 2
and remember that x=2y+2

so you have:
6y+6+2y= -2
8y= -8
y= -1



and finaly from previous you have that
x=2y+2 and that is 0
z=3/4x and that is 0


so (x,y,z) = (0,-1,0)

Or I did somewhere wrong....at least I tried

Answer 2

ok in the event that they're parallel they do no longer intersect in any respect, in the event that they do no longer seem to be parallel they intersect as a line. what different selection is there? 3 planes can intersect as a factor or as a line. replace into the question approximately 3 planes? i would not confront your instructor yet might recheck the question and if it asks approximately 2 planes intersecting i might ask for a proof, on account which you don't get it. he could desire to have made a mistake whilst writing out the try or have been given it out of a defective textual content cloth or you have misinterpret.

Answer 3

distance formula you dont need to worry about the y cordinant because it is zero, so you use the square root of (the x value squared plus the z value squared). at least thats what i think

Answer 4

x + y - 2z = 0
2x - y - 2z = 1
x - 2y = 1
let t = y
x = 2t + 1
2t + 1 + t = 2z
z = (3t+1)/2
(x,y,z) = (2t+1,t,(3t+1)/2)
(-1,-1,-1)
(3,1,2)
(7,3,5)
(11,5,8)