I need help with this exercise. It says:
Find the symmetric equation for the line that is defined by the intersection of the planes $x-2y+4z=2$ and $x+y-2z=5$
Using my book as a guide. I see that I have to get the cross product of the normal vectors and that I need a point in the line $L$.
For the point, I set in both equations $z=0$. So I got the system of equations:
\begin{cases} x-2y=2 \\ x+y=5\end{cases}
The solution for this is $(4,1,0)$.
The normal vectors are:
$$n_1=<1,-2,4>$$ $$n_2=<1,1,-2>$$
The cross-product of the normal vectors is:
$$v=n_1×n_2=<0,6,3>$$
I'm using this formula for the symmetric equations:
$$\frac{x-x_0}{a}=\frac{y-y_0}{a}=\frac{z-z_0}{a}$$
So here it is where I have my doubts:
$$\frac{x-4}{0}=\frac{y-1}{6}=\frac{z-0}{3}$$ $$\frac{x-4}{0}=\frac{y-1}{6}=\frac{z}{3}$$
I have that division between zero that is making me wonder if I'm doing something wrong or if not what happens next. In the book there is just one example, and doesn't happen this.
Please help!
A member like
$$\frac{x-x_0}{0}$$ symbolically means that $x=x_0$, the line is contained in a plane of constant $x$.
Indeed, by adding twice the second equation to the first, $$3x=12.$$
The remaining members can be seen as the implicit equation of a 2D straight line in a plane,
$$\frac{y-1}{6}=\frac z3\iff y-2z-1=0.$$
The notation with a zero at the denominator is usually accepted (in this special context).