# Thread: 2 planes in 3 dimensions

1. ## 2 planes in 3 dimensions

Can someone please explain to me what two lines in 3-D would like?

(such as having 2 equations and 3 unknowns)

(also, my teacher said something about how it can be written as a vector with one independent variable. this doesn't make too much sense to me)

2. Originally Posted by TheBerkeleyBoss
Can someone please explain to me what two lines in 3-D would like?

(such as having 2 equations and 3 unknowns)

(also, my teacher said something about how it can be written as a vector with one independent variable. this doesn't make too much sense to me)
$x=x_0+ta_x$
$y=y_0+ta_y$
$z=z_0+ta_z$
So $\left(\begin{array}{cc}x\\y\\z\end{array}\right)=\ left(\begin{array}{cc}x_0\\y_0\\z_0\end{array}\rig ht)+t\left(\begin{array}{cc}a_x\\a_y\\a_z\end{arra y}\right)$,which is a mapping with one independent variable t.

3. Thank you for the prompt reply. I reviewed the section of Calc.III involving the parameterization of a line through 3-D space.

However, I still have a sticking point:

In class, we reduced a matrix created from three variables and two equations. We were left with:

$A + 2C = 0$

and

$B - 3C = 0$

Rearranged....

$A = -2C$ and $B = 3C$. C cannot be equal to zero.

My teacher than states that this has infinitely solutions. I have a hard time understanding this...

4. Originally Posted by TheBerkeleyBoss
Thank you for the prompt reply. I reviewed the section of Calc.III involving the parameterization of a line through 3-D space.

However, I still have a sticking point:

In class, we reduced a matrix created from three variables and two equations. We were left with:

$A + 2C = 0$

and

$B - 3C = 0$

Rearranged....

$A = -2C$ and $B = 3C$. C cannot be equal to zero.

My teacher than states that this has infinitely solutions. I have a hard time understanding this...
This is because every C determine an A and a B. But C have infinite choice and different C gernerate different (A,B). So it has infinite solutions.

5. Thank you again for your explanation. I'm just about fully understanding what he was trying to teach us.

He first wrote the general form of a line on the board, Ax + By - C = 0

Then he wrote "two points determine a line".

Then.... (x1, y1) = (7,5) , (x2, y2) = (4,3) .....find A,B,C

he set it up two equations:

$7A + 5B - C = 0$
$4A + 3B - C = 0$

then he put these equations in matrix form.

After putting it in RREF form, we were left with:

A = -2C , B = 3C, and C cannot equal zero ...you showed me earlier that I can write this as a vector with one independent variable.

However, what does this vector represent? Is it supposed to represent a place where two lines intersect? Since I was given two points, it only seems as if I have one line, so how could it represent an intersection?

Furthermore, seeing an equation set-up as 7A + 5B - C = 0 seems so strange to me. Most of the time, the "A" and "B" are given, but in this case the x and y are given. Very, very hard for me to understand conceptually.

Ahhhh.

6. Originally Posted by TheBerkeleyBoss
Thank you again for your explanation. I'm just about fully understanding what he was trying to teach us.

He first wrote the general form of a line on the board, Ax + By - C = 0

Then he wrote "two points determine a line".

Then.... (x1, y1) = (7,5) , (x2, y2) = (4,3) .....find A,B,C

he set it up two equations:

$7A + 5B - C = 0$
$4A + 3B - C = 0$

then he put these equations in matrix form.

After putting it in RREF form, we were left with:

A = -2C , B = 3C, and C cannot equal zero ...you showed me earlier that I can write this as a vector with one independent variable.

However, what does this vector represent? Is it supposed to represent a place where two lines intersect? Since I was given two points, it only seems as if I have one line, so how could it represent an intersection?

Furthermore, seeing an equation set-up as 7A + 5B - C = 0 seems so strange to me. Most of the time, the "A" and "B" are given, but in this case the x and y are given. Very, very hard for me to understand conceptually.

Ahhhh.
emm..what your teacher try to told you is to do the inverse procedure. Usually, the A,B,C are given. But this time, he ask you to calculate A,B,C given several pairs of (x,y),right?
What I want to tell to is that this "inverse procedure" ,mostly, is more important. Sometimes you just know the coordinates of two points, you have to describe the whole line..So you must master this kind of method.
Your example is on the 2D plane.
the form $Ax+By+C=0$is the general form for lines in a plane, that is, given any (A,B,C), $Ax+By+C=0$represents a line and any line in a plane have the equation $Ax+By+C=0$.
This equation means that every point on the line satisfies the equation. And every pair of (x,y)satisfying it is on the line.
Now you are give two points (7,5),(4,3),how to calculate A,B,C?
The only way is to use the previous property of the equation $Ax+By+C=0$.
So you substitude (7,5),(4,3) for (x,y), then you will get two linear equations..
Then solve it, you get A=2C,B=-3C.
So $2Cx-3Cy+C=0$.
if C=0, then the equation is meaningless..
So C!=0.
Note that $C(2x-3y+1)=0$
So $2x-3y+1=0$, this will be the eqation of the line...

For the meaning of the vectors, you may understand the first one as the "origin",the second one as the "direction".

7. well put! I understand now!

8. Originally Posted by ynj
$x=x_0+ta_x$
$y=y_0+ta_y$
$z=z_0+ta_z$
So $\left(\begin{array}{cc}x\\y\\z\end{array}\right)=\ left(\begin{array}{cc}x_0\\y_0\\z_0\end{array}\rig ht)+t\left(\begin{array}{cc}a_x\\a_y\\a_z\end{arra y}\right)$,which is a mapping with one independent variable t.
ynj
All of your posts above make a very good explanation.
I'm borrowing your thoughts to assist another.
Thanks.