Q: Find the plane which contains the lines
$$
\langle 2,1,4\rangle + t\langle1,-2,4\rangle\text{ and }\langle2,-3,3\rangle + s\langle2,-4,8\rangle
$$
I know to get the plane I need a point and a vector normal to the plane. Since my lines are parallel if I take the cross product of the direction vectors I'll get zero. I am confused on how to get the normal vector here.
Would using the two points to get a vector, then taking the cross product of that vector and the direction of the line be the right move here?
Best Answer
take two Points, $$P_1(2;-3;3)$$ and $$P_2(2;1;4)$$ from our two lines and you can write $$[x,y,z]=[2;-3;3]+a[1;-2;4]+b[0;4;1]$$ as vector equation