Thread: Intersection of 2 Planes help

Hello everyone! I truly hope someone can explain this question out to help me out! It's been 2 hours and still no luck! Any help is truly appreciated!

-------------------------------------------------------------
Let A1x+B1y+C1z+D1 = 0 and A2x+B2y+C2z+D2=0 be 2 non-parallel planes in space. Show that for any fixed k,

A1x+B1y+C1z+D1+ k(A2x+B2y+C2z+D2) = 0


is the equation of the plane through the intersection of the 2 planes. As k varies, this equation generates the family of all such planes (except the second plane itself).
-------------------------------------------------------------

I'm really lost as to what the question has just explained. I am not grasping the concept that adding 2 planes will give you the equation of ANOTHER plane through the intersections of the first 2 planes.


As well, the question continues by asking me to find a scalar equation of the plane that passes through the ORIGIN and the line of intersection of the planes 3x+4y-7z-2=0 and 2x+3y-4=0.


Thank you to anyone who can help! God bless!

I'm going to have a go at explaining what I think it means.

For any two non-parallel planes, there is a line of intersection which is defined by, basically, the simultaneous solution of the two equations defining the planes in question.

To put it another way, that line can be defined as the line of points which simultaneously satisfy the two equations given.

Now as:

we also have
.

Thus if you add the two bits together, the whole thing still equals 0.

As it's an equation in x, y, and z which is actually:



This is in the form of an equation of a plane. But the coefficients are variable by the parameter .

So it's a plane, and it passes through a set of points which are common to both of the original planes. So you can think of it as a plane formed by rotation of one of those planes through that line of intersection. You can't get the second plane because there's a positive non-zeroable quantity of the first plane in the equation, but you can set k to zero, eliminating the second plane, and get an equation which just contains the first plane.

That should get you started. Not sure of a quick and easy way of proving that this is all the possible planes through that line of intersection. Someone else may need to help out.

(BTW something went horrible with the formatting when I was typing this, I think I hit a control character and it all went screwy. Hope it's readable.