Question

Could some explain this too me:
Find two different planes whose intersection is the line x=1+t, y=2-t, z=3+2t. Write equations for each plane in the form Ax+By+Cz=D

Answer 1

I found this answer on yahoo, but i dont undersand why it is this way:

Eliminate t in a susbset of 2 equations. Do it twice and that's it;

x = - 2 + 3t, y = 3 - 2t, give 2x + 3y = 5

y = 3 - 2t, z = 5 + 4t. give 2y + z = 11.

Answer 2

each plane will contain at least the point give for the line; (1,2,3)

of course any other point on the line works, its just that this ones is easy to pull off

Answer 3

now we need to find 2 vectors that are parallel to the given vector of the line <1,-1,2>

hmmm, or we could establish the plane that this line is a normal to and pull out 2 vectors from it

Answer 4

these vectors in turn would be normals for 2 planes that are commoned at the given point

Answer 5

1(x-1)-1(y-2)+2(z-3)=0

x-1 -y+2 +2z -6 = 0

x -y +2z -5 = 0

now plug in some numbers to determine 2 point in this plane that are not colinear and you can create your 2 vectors

Answer 6

(5,0,0) (0,-5,0) seem to fit just fine

( 1,2,3) (5,0,0) (0,-5,0)
-1-2-3 -1-2-3 -1-2-3
------------------------------
<0,0,0> <4,-2,-3> <-1,-7,-3>

sooo

2 planes the connect on the given line are:

4(x-1)-2(y-2)-3(y-3) = 0

and

(x-1)+7(y-2)+3(z-3)=0