# Thread: Find the intersection of two lines in 3d space.

1. ## Find the intersection of two lines in 3d space.

$-2x-5y=3, -8y+4z=-7$

How do you go about finding (x,y,z)? I can't find this in my textbook.

2. Those are planes, not lines. Are you trying to find the equation of the line of intersection?

3. You're right. I'm trying to find the intersection of two planes

4. So, set up a system of two equations in three unknowns and start solving it. What do you get?

5. First find the cross product of the two normals. That is the direction vector of your line.
This point is on both planes: $\left( {\frac{{ - 3}}{2},0,\frac{{ - 7}}{4}} \right)$.

6. I would prefer Ackbeet's method. Since z does not appear in the first equation and x does not appear in the second, it is easy to solve the two equations for x and z in terms of y. Then you can use "y" as parameter. Once you have x= f(y) and z= g(y), your parametric equations for the line of intersection are x= f(t), y= t, z= g(t).

7. $\left\{\begin{array}{ccc}
x&=-\displaystyle\frac{5y+3}{2} \\ \\
z&=\displaystyle\frac{8y-7}{4}
\end{array}\right.\rightarrow
\left[\begin{array}{ccc}
x\\y\\z
\end{array}\right]=
\left[\begin{array}{ccc}
-\displaystyle\frac{5y+3}{2} \\ \\
y \\ \\
\displaystyle\frac{8y-7}{4}
\end{array}\right]=
\left[\begin{array}{ccc}
-\displaystyle\frac{5}{2}y-\frac{3}{2} \\ \\
y \\ \\
\displaystyle\frac{8}{4}y-\frac{7}{4}
\end{array}\right]=
y\left[\begin{array}{ccc}
-\displaystyle\frac{5}{2}\\ \\
1 \\ \\ 2
\end{array}\right]+\left[\begin{array}{ccc}
-\displaystyle\frac{3}{2} \\ \\
0 \\ \\
-\displaystyle\frac{7}{4}
\end{array}\right]
$

Line pass through Point $(-\displaystyle\frac{3}{2},0,-\frac{7}{4})$ with normal vector $(-\displaystyle\frac{5}{2},1,2)^T$

8. Reply to math2009:

That's fairly slick, too. I would say, though, that you have a direction vector there, not a normal vector.