# Intersection of 2 planes

1. Feb 7, 2009

### gtfitzpatrick

find the general equation of a plane that passed through the line of intersection (PQ) of 2x-7y+5z+1=0 and x+47-3z=0

find the equation of the particular plane through PQ which is parralel to the line
x/-1 = (y-1)/3 = (z-3)/13

ok i think there is a couple of ways of doing this,this is the way i went with...

2 eq 3 unknowns
solved get y = (11z+1)/15
sub back in x = (t+4)/15

then let z=t where t is any real number.

so then general eq (x,y,z) = ((t+4)/15 , (11t+1)/15 , t)

2. Feb 7, 2009

### slider142

According to the defining equations of the line, x = -(y - 1)/3, but your y = (11t+1)/15, which means -(y - 1)/3 = (-11t + 14)/45, which is not equivalent to your x = (t+4)/15, so something went wrong somewhere (Always check back to make sure your solution actually satisfies the original equation).
Since 3 and 13 have nothing in common, the simplest move for me would be to rewrite the equations as x = (1 - y)/3 = (3 - z)/13 and let x = t. Then we have L(t) = (t, 1 - 3t, 3 - 13t). One avenue of attack is to note that if we are spinning a plane around the previous line, and we want it to be parallel to another line, the normal to the plane is coincident with the vector denoting the (shortest) distance between the two lines.

3. Feb 7, 2009

### gtfitzpatrick

yes, i checked it out i had a sign wrong it should have been y = (11t-1)/15 thanks

which i put back in and everything seem fine.
as for the second part i'm not sure what is happening

Last edited: Feb 7, 2009
4. Feb 7, 2009

### gtfitzpatrick

ok i think i see...

When i did out the first part i let z = t but then you let x = t, will this compute? should i re do the first part this time letting x = t?

5. Feb 7, 2009

### slider142

Never mind, I thought you were parametrizing the second line. The sign was correct the first time.
As for finding the vector that points directly from one line to the other (the normal to the plane we're looking for), note that this vector will also be normal to the slope vectors for both lines (Draw a picture).
Thus, our first line has a slope vector of (1/15, 11/15, 1) and the line they give you has a slope vector of (1, -3, -13). Have you covered the cross product?

6. Feb 8, 2009

### gtfitzpatrick

i have covered cross product so i cross the slope of the 2 vectors?

7. Feb 8, 2009

### slider142

Right. That will give you a vector normal to both lines and thus normal to the plane we're looking for.

8. Feb 8, 2009

### gtfitzpatrick

thanks a millions for all the help,

i crossed the 2 of them and got (98/15, -28/15, 14/15) which is the normal vector right?

9. Feb 8, 2009

### gtfitzpatrick

(98/15, -28/15, 14/15) . (x-0, y-1, z-3) = 0

which gives

(98/15)x - (28/15)y + (14/15)z = 14/15