# Math Help - urgent question!!! intersection of planes

1. ## urgent question!!! intersection of planes

Okay I'm freaking out. I'm reviewing for my final tomorrow, and I realized I don't get the plane portion of the final. I can't find my notes on it.

So for intersection of a plane. I'm gonna make up a problem and show how far I can get into it.

Intersection between 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7.

So you take the cross product of [2, 3, 4] and [5, 2, 3], right? How do you find this cross product? is it [10, 6, 12] or something else?

Then after you find that...say it is [1,2,3] even though it isn't.

So set x to 0 for the two planes. Then solve for y and z. Say it turns out to be [0, 6, 7].

then it is [0 6, 7] + t(1,2,3). is that right?

Okay urgent how do you find a cross product?? I have no idea how.

2. Originally Posted by view360
So for intersection of a plane. I'm gonna make up a problem and show how far I can get into it.

Intersection between 2x + 3y - 4z = 12 and 5x + 2y + 3z = 7.

So you take the cross product of [2, 3, 4] and [5, 2, 3], right? How do you find this cross product? is it [10, 6, 12] or something else?

Then after you find that...say it is [1,2,3] even though it isn't.

So set x to 0 for the two planes. Then solve for y and z. Say it turns out to be [0, 6, 7].

then it is [0 6, 7] + t(1,2,3). is that right?

Okay urgent how do you find a cross product?? I have no idea how.
I hope that you are more careful on your exam than you are here.
The normal is $\left\langle {2,3, - 4} \right\rangle$ not [MATh]\left\langle {2,3, 4} \right\rangle [/tex]

3. Sorry, was trying to figure out the problem on my paper, and I was trying it both ways. I must have mixed them up.

Can anyone help me with this? I can't find my notes on how to cross product, and I tried googling it but I did not learn anything even close to what it is pulling up?

4. Hello, view360!

I have an entire different approach.
I'll use my own example . . .

Intersection between: . $2x - y +z \:=\: 3\:\text{ and }\:3x + 2y - 3z \:=\: 9$

We know that the intersection of two planes is a line.

We try to solve the system of equations: . $\begin{array}{cccc}2x - y + z &=& 3 & [1] \\ 3x + 2y - 3z &=& 9 & [2] \end{array}$

Eliminate one of the variables.

. . $\begin{array}{cccc}\text{Multiply [1] by 2:} & 4x - 2y + 2z &=& 6 \\
\text{Add [2]:} & 3x + 2y - 3z &=& 9 \end{array}$

And we have: . $7x - z \:=\:15 \quad\Rightarrow\quad{\color{blue} z \:=\:7x - 15}$

Substitute into [1]: . $2x - y + (7x - 15) \:=\:3 \quad\Rightarrow\quad 9x - y \:=\:18 \quad\Rightarrow\quad{\color{blue} y \:=\:9x - 18}$

We have these equations: . $\begin{array}{ccc}x &=&x \\ y &=& 9x - 18 \\ z &=& 7x-15\end{array}$

On the right side, replace $x$ with a parameter $t\!:\;\;\begin{Bmatrix}x &=& t \\ y &=& 9t - 18 \\ z&=& 7t-15 \end{Bmatrix}$

There are the parametric equations of the line of intersection.

5. Originally Posted by Soroban
Hello, view360!

I have an entire different approach.
I'll use my own example . . .

We know that the intersection of two planes is a line.

We try to solve the system of equations: . $\begin{array}{cccc}2x - y + z &=& 3 & [1] \\ 3x + 2y - 3z &=& 9 & [2] \end{array}$

Eliminate one of the variables.

. . $\begin{array}{cccc}\text{Multiply [1] by 2:} & 4x - 2y + 2z &=& 6 \\
\text{Add [2]:} & 3x + 2y - 3z &=& 9 \end{array}$

And we have: . $7x - z \:=\:15 \quad\Rightarrow\quad{\color{blue} z \:=\:7x - 15}$

Substitute into [1]: . $2x - y + (7x - 15) \:=\:3 \quad\Rightarrow\quad 9x - y \:=\:18 \quad\Rightarrow\quad{\color{blue} y \:=\:9x - 18}$

We have these equations: . $\begin{array}{ccc}x &=&x \\ y &=& 9x - 18 \\ z &=& 7x-15\end{array}$

On the right side, replace $x$ with a parameter $t\!:\;\;\begin{Bmatrix}x &=& t \\ y &=& 9t - 18 \\ z&=& 7t-15 \end{Bmatrix}$

There are the parametric equations of the line of intersection.

That makes so much more sense! I like your method a lot, finding x, y, and z just is so much more logical, lol.

one more question -- is this the type of thing you can put into [x,y,z] + t(a,b,c) form? Cuz for most of our plane things I remember we used that.

6. Hello again, view360!

Is this the type of thing you can put into $(x_1,y_1,z_1) + t(a,b,c)$ form? . Yes!

We have: . $\begin{Bmatrix}x &=& t \\ y &=& 9t-18 \\ z &=&7t - 15 \end{Bmatrix}$

I like to re-write them: . $\begin{Bmatrix} x &=& 0 + 1t \\ y &=& \text{-}18 + 9t \\ z &=& \text{-}15 + 7y \end{Bmatrix}$

Therefore: . $(x,y,z) \;=\;(0,\text{-}18,\text{-}15) + t(1,9,7)$