Given the plane $ X = (0 , 1/2 , 0 ) + λ(1, -1/2,0) + μ(0,1,1) (λ, μ ∈ R) $ and the line $r=\left\{\begin{matrix} x-y+z=0 & \\ 2x+y-z=1 & \end{matrix}\right.$
Give their relative position
Answer: The plane contains the line
I first tried it numerically by finding the general equation for the plane by calculating the determinant of the point and the 2 vectors that define the plane. After I calculated the scalar product between the normal plane vector and the line defining vector and I got $n•v = 0$ , that says that the line is parallel to the plane or it is in the plane, after that I inserted the general points of line onto the plane equation and it resulted 0=0, that would match the answer, that the plane contais the line.
I'm always curious about the visual representation of exercises and ploted it on geogrebra, but then it diverges from the answer. I don't know whether the code is wrong, if I wrote something wrong or if my numerical steps is wrong
The error appears to be in your input to Geogebra. According to your handwritten notes, it seems you want a line parameterized by
$$ \left(\frac13,\frac13,0\right) + \gamma (0,1,1). $$
You entered this in Geogebra as $\mathsf{Line}\left(\left(\frac13,\frac13,0\right),(0,1,1) \right).$ But what that command produces in Geogebra is a line through the points $\left(\frac13,\frac13,0\right)$ and $(0,1,1).$
I think the Geogebra command for the line you wanted is $\mathsf{Line}\left(\left(\frac13,\frac13,0\right),\mathsf{Vector}((0,1,1))\right).$ And indeed this appears to give a line in the plane.