Line of Intersection between two planes

**IanCarney** · June 23rd 2011, 01:44 PM

How would I determine the line of intersection between plane 1: $4x + 2y + 6z -14 = 0$ and plane 2: $x - y + z - 5 = 0?$ ?

I don't remember exactly how to do it, and trying to find an explanation online just confused me even more

I was thinking somewhere along the lines of using the cross products of the normals, but I'm not sure what to do after that.

**Plato** · June 23rd 2011, 01:48 PM

Originally Posted by IanCarney

How would I determine the line of intersection between plane 1: $4x + 2y + 6z -14 = 0$ and plane 2: $x - y + z - 5 = 0?$ ?
I was thinking somewhere along the lines of using the cross products of the normals

That cross product is now the direction vector for the line of intersection.

Now we need to find some point that is on both planes.

**IanCarney** · June 23rd 2011, 01:50 PM

Originally Posted by Plato

That cross product is now the direction vector for the line of intersection.

Now we need to find some point that is on both planes.

Cross product would be (8, 2, -6), right?

For the point, are you supposed to let a variable equal zero and then find the other two variables?

**Plato** · June 23rd 2011, 01:56 PM

So far, so good.

**IanCarney** · June 23rd 2011, 01:58 PM

Okay so assuming I let $z=0$

$4x+2y=14$ and $x-y=5$

$y=-1$ and $x=-4$

Would the equation of the line be $x=8t-4$ , $y=2t-1$ , and $z=-6t$ ?

**IanCarney** · June 23rd 2011, 02:35 PM

Ugh, according to the book the answer is $x=8+4t$ , $y=t$ , $z=-3-3t$

Not sure where I went wrong :s

**Plato** · June 23rd 2011, 02:39 PM

Originally Posted by IanCarney

Ugh, according to the book the answer is $x=8+4t$ , $y=t$ , $z=-3-3t$

Those are different ways to write the same line.
Can you prove that?

EDIT: there may be a sign error in your point. Check the x value.

**IanCarney** · June 23rd 2011, 02:48 PM

Originally Posted by Plato

Those are different ways to write the same line.
Can you prove that?

EDIT: there may be a sign error in your point. Check the x value.

Which x value are you talking about?

Edit: Oh right, it's supposed to be $+4$ . Thanks for your help!

**Soroban** · June 23rd 2011, 06:15 PM

Hello, IanCarney!

I divided the first equation by 2 . . . Don't know why "they" didn't.

Here is a rather primitive method.

Determine the line of intersection of planes: . $\begin{array}{cccc} 2x + y + 3z &=& 7 & [1] \\ x - y + z &=& 5 & [2]\end{array}$

Add [1] and [2]: . $3x + 4z \:=\:12 \quad\Rightarrow\quad x \:=\:4 - \tfrac{4}{3}z$

Substitute into [1]: . $2(4 - \tfrac{4}{3}z) + y + 3z \:=\:7 \quad\Rightarrow\quad y \:=\:\text{-}1 - \tfrac{1}{3}x$

We have: . $\begin{Bmatrix}x &=& 4 - \frac{4}{3}z \\ \\[-4mm]y &=& \text{-}1 - \frac{1}{3}z \\ \\[-4mm] z &=& z \end{Bmatrix}$

On the right, replace $z$ with the parameter $t.$

. . . . . . . . . $\begin{Bmatrix}x &=& 4 - \frac{4}{3}t \\ \\[-4mm] y &=& \text{-}1 - \frac{1}{3}t \\ \\[-4mm] z &=& t \end{Bmatrix}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We can check our results.

Substitute our answers into [1]:

. . . . . . . . . . . $2x + y + 3z \;=\;7$

. . $2(4-\tfrac{4}{3}) + (\text{-}1 - \tfrac{1}{3}z) + 3t \:=\:7$

. . . . . $8 - \tfrac{8}{3}y -1 - \tfrac{1}{3}t + 3t \;=\;7$

. - - - - . . . . . . . . . . . . . $7 \;=\;7 \;\;\;check!$

Substitute our answers into [2]:

. . . . . . . . . . . . $x - y + z \;=\;5$

. . $(4-\tfrac{4}{3}t) - (\text{-}1-\tfrac{1}{3}t) + t \;=\;5$

. . . . . $4 - \tfrac{4}{3}t + 1 + \tfrac{1}{3}t + t \;=\;5$

. . . . . . . . . . . . . . . . . $5 \;=\;5 \;\;\;check!$

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