Finding intersection line question without the $z$ axis. What is the intersection line of these two planes:
$$ P_1 : 3x+4y+7z=5$$
$$P_2: 2x-5y=8$$
I will first multiply one by $4$ and the second by $5$ just to get rid of $y$:
$$15x+20y+35z -25 = 0$$
$$8x-20y-32=0$$
Adding those two, in order to get the intersection plane:
$$23x + 35z -57 = 0$$
Now we want to find $2$ points in order to get the line:
So first let's substitute a random value, say $x=1$ :
$$23 +35z -57 = 0$$
$$z = \frac{34}{35}$$
Now checking what is $y$ by checking it on our plane:
$$ 3 \cdot 1 + 4y + \frac{7 \cdot 34}{35} -5 = 0 $$
$$y = - \frac{6}{5}$$
So we have one point:
$$ (1, - \frac{6}{5}, \frac{34}{35} )$$
Now for the second point:
I substituted : $ z = 0$
$$23x = 57 \Rightarrow x = \frac{57}{23}$$
And to get $y$ I subs. it in the inters. plane:
$$3 \cdot \frac{57}{23} + 4y - 5 =0 $$
$$y = 5 - 3 \cdot \frac{53}{23} = - \frac{14}{23}$$
And so we have our second point:
$$(\frac{57}{23} , - \frac{14}{23}, 0)$$
We will now take the difference to get the direction vector:
$$ ( \frac{57}{23} - 1 , - \frac{14}{23} + \frac{6}{5} , - \frac{34}{35})  = (\frac{34}{23}, \frac{68}{115}, - \frac{34}{35})$$
To get that our intersection line is (starting at the first or second point we found) and the "direction" vector :
$$(1 , - \frac{6}{5}, \frac{34}{35}) + t ( \frac{34}{23}, \frac{68}{115}, -\frac{34}{35})$$
This is not the right answer as GeoGebra says:

This is clearly not the intersecting line...
I would appreciate your help, I really checked myself dozens of times... thanks!
 A: *

*First, find the normal vectors $n_1$ and $n_2$ of both planes (that are the vectors formed from the coefficients).


*Second, find the direction vector $d$ of the intersection line ($d = n_1 \times n_2$)


*Third, find any point that lies on both planes (hint for this step: set $z=0$, for example, and solve the system for $x$ and $y$)
After all, you will have a direction vector of a line and a point lying on this line.
Let me know if you can take it from here.
A: $P_1 : 3x+4y+7z=5$
$P_2: 2x-5y=8$
You can either find the cross-product of the normal vector of the two planes and a point on the intersection to find the equation of the line or you can do as below -
For $z = t$,
$3x+4y=5-7t$ and $2x-5y=8$
Equating the two, you get -
$x = \dfrac {57-35t}{23}, y = \dfrac{-14-14t}{23}, z = t$
The parameterized equation of the line will be,
$(x,y,z) = (\dfrac{57}{23},-\dfrac {14}{23}, 0)+(-35,-14,23)t$
If you do the cross product of the normal vectors of the planes, again you will get the direction vector of the line as ($-35,-14,23)$. You can find a point on the line assuming $z = 0$. You get the same equation.
