Equation for line connecting 2 planes

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In this question:
http://imageshack.us/m/705/917/questionn.png
I know how to get the normal vector and the tangent plane but I have no idea how to get the equation for a line connecting the tangent plane to a x - y = 0 plane.
 
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What do you need to come up with an equation of a line?
You need a point and a slope.

A point P is given in the problem, and you can think of your normal vector as a slope. Now its just a matter of putting these things together in a multivariable point slope formula to get the equation of a plane.
 
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For the second part of the question (equation of line from the intersection of the planes), take the cross product of the normal vector from the 1st and 2nd planes (this will give you a new normal vector that is perpendicular to the normal vectors of the two planes, or in other words this new vector is parallel to the planes' line of intersection.) Then you're going to need a point where the planes intersect (just set them equal and solve for a point.)
So now you have a point and a vector "slope" that you can put together in an equation of a line.
 
magicarpet512 said:
For the second part of the question (equation of line from the intersection of the planes), take the cross product of the normal vector from the 1st and 2nd planes (this will give you a new normal vector that is perpendicular to the normal vectors of the two planes, or in other words this new vector is parallel to the planes' line of intersection.) Then you're going to need a point where the planes intersect (just set them equal and solve for a point.)
So now you have a point and a vector "slope" that you can put together in an equation of a line.


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Another way to do this:
The equation of the first plane is Ax+ By+ Cz= D and the equation of the other plane is x- y= 0, which is the same as y= x. Putting that into the first equation, Ax+ Bx+ Cz= D so we can solve for z: z= (D- Ax- Bx)/C. Letting x= t, we have the parametric equations x= t, y= t, z= (D- (A+B)t)/C for the line of intersection.