Here is my vector:
$(-3,1,-4)+r(4,0,1)$
And my plane: Created from the following vectors:
$x: (3,0,1)+t(-1,1,2)$
$x: (0,2,-1)+s(2,-2,-4)$
$(3,0,1)+t(-1,1,2)+n(2,-2,-4)$
(Cartesian: $x+y-3=0$)
I've tried to do the following:
vector: $(-3+4r, 1, -4+r)$
plane: $(3-t+2n, t-2n, 1+2t-4n)$
e.g.:
1) $3-t+2n = -3+4r$
2) $t-2n = 1$
3) $1+2t-4n = -4+r$
But I didn't find the right solution...
The intersection point should be: $(1,1,-3)$(written in the math book)
What did I do wrong? Thanks in advance.
A rather straightforward way to solve this is following these 2 steps:
1) find the cartesian equation of the plane, i.e. its equation of the form $ax+by+cz+d=0$. In order to do this, just remember that the vector $(a,b,c)$ is perpendicular to the plane and observe that you can easily obtain two independent vectors $\vec v$ and $\vec w$ parallel to the plane (so that for instance $(a,b,c)=\vec v\wedge\vec w$).
2) impose that the generic point of the line satisfies the equation of the plane and solve for $t$.
I leave the details to you as an exercise.