# Thread: Line of Intersection between two planes

1. ## Line of Intersection between two planes

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.

2. ## Re: Line of Intersection between two planes

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.

3. ## Re: Line of Intersection between two planes

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?

4. ## Re: Line of Intersection between two planes

So far, so good.

5. ## Re: Line of Intersection between two planes

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 $z=-6t$?

6. ## Re: Line of Intersection between two planes

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

Not sure where I went wrong :s

7. ## Re: Line of Intersection between two planes

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.

8. ## Re: Line of Intersection between two planes

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!

9. ## Re: Line of Intersection between two planes

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.

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

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

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

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

. . . . . . . . . . . . $x - y + z \;=\;5$
. . $(4-\tfrac{4}{3}t) - (\text{-}1-\tfrac{1}{3}t) + t \;=\;5$
. . . . . $4 - \tfrac{4}{3}t + 1 + \tfrac{1}{3}t + t \;=\;5$
. . . . . . . . . . . . . . . . . $5 \;=\;5 \;\;\;check!$