Find a vector parallel to the intersection of the planes $2x-3y+5z=2$ and $4x+y-3z=7$

Question

The solution is $(4,26,14)$. I know how to find the intersection of the planes, but not a parallel vector.

Answer 1

Here is a figure of two arbitrary planes.

A vector parallel to the intersection of the planes is the same as a vector perpendicular to one of the normal vectors.

Let's take $2x-3y+5z=2$. The normal vector of this plane is $(2,-3,5)$. We need to know the vector $(a,b,c)$ for which the vectors are perpendicular, so we solve the following equation: $2a-3b+5c=0$, to which the solution is $(4,26,14)$.

Answer 2

Solving the equations simultaneously we get: $$x=\frac{3+13z}{7},~y=\frac{23+4z}{14},~z=z$$ This means that the intersection can be described as $$(x,y,z)=(23/14,3/7,0)+\vec{u}z,~~z\in\mathbb R$$ wherein $\vec{u}=(2/7,13/7,1)$. This is what you're looking for.