So i have to write down the equation of a plane that is parallel to this two lines:
$$p1: y=2x-1 , z=3x+2 $$
$$p2: y=-x+1, z=4x-1$$
And the plane is at the same distance from both of those lines.
I was thinking i should that the dot product of normal vector and each of those lines direction vectors should equal 0, but i really don't know how to continue, beacause i get a not so simple equation and i still have two variables when i simplify it.
Also i don't how to use information that the planes is at the same distance away from line 1 as from line 2. I would really appreciate any help in solving this problem.
Best Answer
Here is one way:
Line $p_1$ is defined by $(1,2,3)x+(0,-1,2)$, while $p_2$ is defined by $(1,-1,4)x+(0,1,-1)$.
So the direction vector for $p1$ is $(1,2,3)$ and the direction vector for $p2$ is $(1,-1,4)$. A quick way to find a vector perpendicular to both of them is to calculate their cross product, which is $(11,-1,-3)$.
Therefore any plane that is perpendicular to the vector $(11,-1,-3)$ will be parallel to or coincident with your two given lines. The equation of such a plane is
$$11x-y-3z=c$$
for some constant $c$.
If the plane is equidistant from the two lines, then the value of $c$ will be equidistant from the values of that expression when a point on each line is substituted. The value of $11x-y-3z$ for point $(0,-1,2)$ on line $p_1$ is $-5$, while the value for point $(0,1,-1)$ on line $p_2$ is $2$.
Therefore we want $c$ to be the average of $-5$ and $2$, which is $-\frac 32$. Your desired plane is
$$11x-y-3z=-\frac 32$$