The equations of two different planes that intersect are the following: (1)---> x-3y+5z+8=0 (2)---> 5x+y-2z+7=0
It says that I must create a third plane that passes through the line of intersection of the original two and which is parallel to the x-axis.
So far, I've found a point on the line that intersects, in this case it's [0,51,29]. I thought that since the plane had to be parallel to the line of intersection, that the normal of the x-axis would be the same as the normal of the plane. Is this correct? Plus, doesn't the normal of an axis have more than one answer to it?
I'm not too sure how to continue with finding the normal of the third plane.
To find a plane, you need to vectors parallel to the plane.
And a point on the plane.
$ $
You have one parallel to the plane, that is $\hat{\textbf{ i }}$
$ $
Another vector parallel to the plane is a vector which is perpendicular to normals of both the planes.
That is $<1, -3, 5>\times <5, 1, -2>$
$ $
Now you need a point that lies on the intersection of the two planes, find it, and you win.