# Thread: Sorry 1 more plane problem

1. ## Sorry 1 more plane problem

What is the line of intersection of the xy and xz planes?

So far this is what I have:

The planes I used are: xy=(3,4,0)+s(1,1,0)+t(2,3,0)
and xz = (2,0,1)+p(1,1,0)+q(1,0,2)

Now I need the scalar equations in the for Ax+By+Cz+D=0 so I found the cross product of the direction vectors to get the normals which were:
n1=(0,0,1) and n2=(0,-1,0) where n=(A,B,C)

Heres where my problem is, I dont see a way to plug these normals in to get a scalar equation. Did I do something wrong, or what do I do next?

2. Originally Posted by sugar_babee
What is the line of intersection of the xy and xz planes?
I do not know what the difficulty is. Draw by hand the xy-plane and the xz-plane and you shall easily see that their intersection forms the x-axis.

Are you asking how to do this problem using algebra?

3. oooh yeah that's obvious to me now, but I do need the algebraic way because I think the point is to figure that out with the question rather than just knowing it

4. Originally Posted by sugar_babee
oooh yeah that's obvious to me now, but I do need the algebraic way because I think the point is to figure that out with the question rather than just knowing it
The equation of the xy-plane is $0x+0y+1z=0$.
The equation of the xz-plane is $0x+1y+0z=0$.

I will use the following theorem that you learned.
Theorem: Two planes distinct planes are either parallel (do not intersect) or interest in a line.

Now, these two planes are distinct. Are they parallel? The answer is no because $(0,0,0)$ is a point shared by both planes. So by the theorem they intersect in a straight line. But to find the equation of a straight line we need to points (two points determine a unique straight line). By inspection we see that $(1,0,0)$ is another point.

So we need to find the equation of a line passing through $\{ (0,0,0),(1,0,0) \}$.
We need to first find a non-zero vector aligned with this line, and such a non-zero vector is $\bold{v} = (1-0)\bold{i}+(0-0)\bold{j}+(0-0)\bold{k} = \bold{i}$.
Thus, the line is descirbed parametrically as,
$\left\{ \begin{array}{c} x = 1t+0 \\ y=0t+0\\ z=0t+0 \end{array} \right\}$
Which is the x-axis.