1. ## Confused about equation of a line

Equation says this:
$\frac{x-3}{-2}=\frac{y+1}{-1}=\frac{z}{3}$

But the book tells me it can also be written like this:
$d: \left\{\begin{matrix}
x-2y-5=0\\3y+z+3=0
\end{matrix}\right.$

I got confused on how that could be so I tried to get the second expression of a line from the first expression of the line by trying cross multyplication:
$\frac{x-3}{-2}=\frac{y+1}{-1} =\: \: -1(x-3)=-2(y+1)$

But I get this which is close but not it:
$-x+2y+5=0$

Although when I do this:
$\frac{y+1}{-1}=\frac{z}{3} =\: \: -z=3(y+1)$

I get something that resembles the equation in the second part:
$3y+z+3=0$

Am I doing something wrong here? How does this form a line exactly because I grew up with the y = mx+b equation and this is the first time I am seeing this.

2. Originally Posted by Bman900
Equation says this:
$\frac{x-3}{-2}=\frac{y+1}{-1}=\frac{z}{3}$
But the book tells me it can also be written like this:
$d: \left\{\begin{matrix}
x-2y-5=0\\3y+z+3=0
\end{matrix}\right.$
What the text calls $d$ is actually two planes.
The given line is the intersection of those two planes.

3. Originally Posted by Bman900
Equation says this:
$\frac{x-3}{-2}=\frac{y+1}{-1}=\frac{z}{3}$

But the book tells me it can also be written like this:
$d: \left\{\begin{matrix}
x-2y-5=0\\3y+z+3=0
\end{matrix}\right.$

I got confused on how that could be so I tried to get the second expression of a line from the first expression of the line by trying cross multyplication:
$\frac{x-3}{-2}=\frac{y+1}{-1} =\: \: -1(x-3)=-2(y+1)$

But I get this which is close but not it:
$-x+2y+5=0$
It is exactly if you multiply the equation by -1.

Although when I do this:
$\frac{y+1}{-1}=\frac{z}{3} =\: \: -z=3(y+1)$

I get something that resembles the equation in the second part:
$3y+z+3=0$
Yes, that is how they got those equations.

Am I doing something wrong here? How does this form a line exactly because I grew up with the y = mx+b equation and this is the first time I am seeing this.
y= mx+ b is the equation of a line in the plane. In three dimensions, you need 2 equations to specify a 3- 2= 1 dimensional line. A single equation in three variables determines a 3- 1= 2 dimensional surface.

You can break
$\frac{x- a}{A}= \frac{y- b}{B}= \frac{z- c}{C}$
into two independent equations just as you did. (There are other ways to form new equations but they are not independent equations.)

4. Ah I see it now, I was thinking about multiplying it by -1 but I was wondering why didn't they just leave it the way it comes out of cross multiplication. Anyway thank you for the help!