# Math Help - Equations of Planes 2

1. ## Equations of Planes 2

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

2. Hint: To find the line of intersection of the planes $x-2y-z=0$ and $x+9y-5z=0$, use the Addition Method to find the projected line on the $xy$-plane:

\begin{aligned}
5x-10y-5z &= 0 \;\;\;\;\;\;\;\;\;\;\mbox{(multiply first equation by 5)}\\
x+9y-5z &= 0 \\
4x-19y &= 0 \;\;\;\;\;\;\;\;\;\;\mbox {(subtract second from first)}\\
y &= \frac{4}{19}x
\end{aligned}

We can then find the $z$-coordinate of the line of intersection:

$z=x-2y=\frac{19}{19}x-2\cdot\frac{4}{19}x=\frac{11}{19}x.$

The equations of the line of intersection are therefore:

\begin{aligned}
y&=\frac{4}{19}x\\
z&=\frac{11}{19}x.
\end{aligned}

Now, all that remains is to find the value of $k$ that will result in these two equations of intersection for all three original equations.

3. so k=-7/19?
how would i check that it is correct?

4. can anyone please verify this? scott H?

5. can someone verify if k=-7/19 seems right

6. Originally Posted by skeske1234
so k=-7/19?
how would i check that it is correct?
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.