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?
The equation of the xz-plane is .
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 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 is another point.
So we need to find the equation of a line passing through .
We need to first find a non-zero vector aligned with this line, and such a non-zero vector is .
Thus, the line is descirbed parametrically as,
Which is the x-axis.