For what value of k will the following set of planes intersect in a line?
x-2y-z=0
x+9y-5z=0
kx-y+z=0
Hint: To find the line of intersection of the planes and , use the Addition Method to find the projected line on the -plane:
We can then find the -coordinate of the line of intersection:
The equations of the line of intersection are therefore:
Now, all that remains is to find the value of that will result in these two equations of intersection for all three original equations.
You can check it by putting it into the original problem and seeing what happens!
If k= -7/19, then the equations become x-2y-z=0. x+9y-5z=0. and -(7/19)x-y+z=0 which we can rewrite as -7x- 19y+ 19z= 0. How many solutions does that have? If we subtract the first equation from the second, we eliminate x: (x+9y-5z)- (x-2y-z)= 11y- 4z= 0. If we multiply the first equation by 7 and add to the third, (7x-14y-7z)+(-7x-19y+19z)= -33y+ 12z= 0 or, dividing by 3, -11y+ 4z= 0, exactly the same equation as before! If z= (11/4)y ten the first equation becomes x- 2y- (11/4)y= 0 so x= (2+11/4)y= (33/4)y. That is any point on the line x= (33/4)t, y= t, z= (11/4)t lies on all three planes.