# Intersection of Planes

• Oct 17th 2009, 05:55 PM
skeske1234
Intersection of Planes
State whether each of the following pairs of planes intersect. If the planes do intersect, determine the eqtn of their line of intersection.

x-y+z-2=0
2x+y+z-4=0

my work:
eqtn 1 + eqtn 2
3x+2z-6=0
x=-2/3t +2

let z=t

-2/3t + 2 -y + t -2 =0
y=1/3 t
z=t

therefore parametric equation of line is:
x=2- 2/3 t
y=1/3 t
z=t

However, the answer in the back of the book has:

x=2-2t
y=t
z=3t

How do I get this answer? i think there's just one step I am missing to convert it to that that I am not getting.
• Oct 17th 2009, 06:10 PM
redsoxfan325
Quote:

Originally Posted by skeske1234
State whether each of the following pairs of planes intersect. If the planes do intersect, determine the eqtn of their line of intersection.

x-y+z-2=0
2x+y+z-4=0

my work:
eqtn 1 + eqtn 2
3x+2z-6=0
x=-2/3t +2

let z=t

-2/3t + 2 -y + t -2 =0
y=1/3 t
z=t

therefore parametric equation of line is:
x=2- 2/3 t
y=1/3 t
z=t

However, the answer in the back of the book has:

x=2-2t
y=t
z=3t

How do I get this answer? i think there's just one step I am missing to convert it to that that I am not getting.

Replace $t$ with $3t$. (Your value for $t$ can arbitrary, so long as it's linear.)

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A way to get directly to the answer in the book is as follows:

Set the equations equal to each other and simplify:

$x-y+z-2=2x+y+z-4 \implies x+2y-2=0 \implies x=2-2y$

Let $y=t$. Thus,

$x=2-2t$ and $y=t$. To find z we need to sub back into one of the equations; we'll use the first one:

$2-2t-t+z-2=0 \implies z=3t$

So the (parametric) equation of the line is:

$\langle 2-2t,t,3t\rangle=\langle 2,0,0\rangle+t\langle -2,1,3\rangle$

(which, in my opinion, is a better way to write it, because it gives you the direction vector of the line).