# Math Help - Line of intersection of two planes

1. ## Line of intersection of two planes

2x+3y-5z=0
3x-y+2z=0

I have found the direction vector using the vector product. The answer in the book says the lines goes through point (2,1,1). I think the 0's are a misprint. How do I find that the line goes though that point?

2. ## Re: Line of intersection of two planes

Originally Posted by Stuck Man
2x+3y-5z=0
3x-y+2z=0
I have found the direction vector using the vector product. The answer in the book says the lines goes through point (2,1,1). I think the 0's are a misprint. How do I find that the line goes though that point?
That is indeed a misprint. The line must be a subset of both planes. That point is on neither plane.
The obvious point to pick is $(0,0,0)$.

3. ## Re: Line of intersection of two planes

Is it possible to figure out what the constants should be?

4. ## Re: Line of intersection of two planes

Originally Posted by Stuck Man
Is it possible to figure out what the constants should be?
I have no idea what "the constants should be" could mean.
The line is $\left\{ \begin{gathered} x(t) = t \hfill \\ y(t) = - 19t \hfill \\ z(t) = - 11t \hfill \\ \end{gathered} \right.$

5. ## Re: Line of intersection of two planes

The zeroes. If I put the point (2,1,1) into the plane equations I get -4 for the first constant and 7 for the other.

6. ## Re: Line of intersection of two planes

Originally Posted by Stuck Man
The zeroes. If I put the point (2,1,1) into the plane equations I get -4 <-- that should be 2
for the first constant and 7 for the other.
The planes $p_1: 2x+3y-z=2$ and $p_2: 3x-y+2z = 7$ have the line of interception

$l:\left \{ \begin{array}{l}x=-t+\frac{23}{11} \\ y=19t - \frac8{11} \\ z= 11t\end{array}\right.$ With $t=\frac1{11}$ you'll get the point $(2,1,1)$

Yes, thanks.