Find the equations for the line of intersctions of the planes:

x+y+z=1

x+y=2

I worked it out to be r(t)=3i-j-k+t(i-j)

The ans says:

r(t)=2j-k+t(-i+j)

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- Feb 25th 2011, 06:45 PMheatlyFind equation of Line where Plane Meets
Find the equations for the line of intersctions of the planes:

x+y+z=1

x+y=2

I worked it out to be r(t)=3i-j-k+t(i-j)

The ans says:

r(t)=2j-k+t(-i+j) - Feb 25th 2011, 06:49 PMAckbeet
Can you show your work, please?

- Feb 25th 2011, 06:54 PMheatly
x+y+z=1

x+y=2

let x=t

t+y+z=1

y=1-t-z

solving for x we get

x=3+t

now solving for y we get

y=-1-t

now solving forz we get

z=-1

therefore r(t)=3i-j-k+t(i-j) - Feb 25th 2011, 07:03 PMAckbeet
Hmm. There's something wrong with your method if x = t and x = 3 + t. That's impossible, isn't it? I would suggest row reducing your augmented matrix thus:

$\displaystyle \left[\begin{array}{rrr|r}

1 &1 &1 &1\\

1 &1 &0 &2

\end{array}\right]\to\dots$

You should get a one-parameter family of solutions. What do you get? - Feb 25th 2011, 07:08 PMheatly
Thanks I just realised that mistake!

- Feb 26th 2011, 06:03 AMAckbeet
So what do you get now?

- Feb 26th 2011, 08:45 PMheatly
I get:

r(t)=2j-k+t(-i+j)

Which is the published ans. - Feb 28th 2011, 02:05 AMAckbeet
Jolly good.