$V_1 = \{(x,y,z)|\ 2x-y+z=0\ and\ x + y - z = 0\} $
$V_2 = \{(x,y,z)|\ x+2z=0\ and\ x-2y +z = 0\} $
$ V_1, V_2 \ are\ subspaces\ in\ R^{3}. $
(1) Write $V_1\ and\ V_2$ in explicit set notation.
$\quad \quad V_1 = \{(0,t,t) |\ t \in R\}$, $V_2 = \{(4s,s,-2s) | s \in R\}$
(2) Find a basis for each of $V_1\ and\ V_2$
$\quad \quad V_1 basis= \{(0,1,1)\}$, $V_2 basis = \{(4,1,-2)\}$
(3) Find the smallest subspace that contains both $V_1\ and\ V_2$ in (a) linear span, (b) explicit and implicit set notations
$\quad \quad $$(3a)\ span\{(0,1,1), (4,1,-2)\}$
$\quad \quad $$(3b)\ \{(4s,s+t,t-2s)|\ s,t \in R \}$, $ \{(4,1,-2)\}$, $\quad\ \{s(4,1,-2)+t(0,1,1)|\ s,t \in R \}$
I am having difficulty deriving the equation of the plane containing the two vectors for (3b), is that due to the fact that my answer for the previous parts are wrong, or is there a way to derive the implicit form in this case : x = 4s, y = s+t , z = t-2s?
Please correct me if I am wrong, thank you!
Take the cross product: $(4,1,-2)×(0,1,1)=(3,-4,4)$. This gives you the normal direction. Now the plane is: $3x-4y+4z=0$.