Intersection of 2 sets of vectors? How do you find the intersection of these two sub-spaces of (the set of real numbers)^3?
$S=\{  a(0, -1, 1)^T + b(-2, 0, 0)^T \}$, $T=\{  c(1, -3, 1)^T + d(0, 2, 2)^T \}$?
I set the general vectors equal to each other and get:
$-2b=c$,
$-a=-3c+2d$,
$a=c+2d$
From this, I don't see how to obtain the intersection.
Note: a,b,c and d are real numbers.
 A: I think there's a mistake in your third equation; the system should be:
$$-2b=c$$
$$-a=-3c+2d$$
$$a=c+2d$$
Adding the last two:
$$0=-2c+4d\implies 0=-c+2d\implies c=2d$$
and, from the first equation, $b=-d$. Can you take it from here?
Also, here's another approach: Find vectors $v$ and $w$ orthogonal to $S$ and $T$, respectively, and note that $v\times w$ generates the intersection (the intersection of two distinct planes through the origin is a line).
A: First method:
You have a system of 3 equations with 4 unknowns.
Solve it by keeping the 3 (arbitrary) "main" variables $a,b,c$ only in the LHS and considering $d$ as parameter and obtain:
$$\pmatrix{a\\b\\c}=\pmatrix{4d\\-d \\2d }$$
It suffices to use the expressions of $a$ and $b$ to obtain:
$$S=\{  4d(0, -1, 1)^T + -d(-2, 0, 0)^T = d(2 ,  -4, 4)^T\}.$$
Thus the intersecting line passes through the origin with directing vector $(2 ,  -4, 4)^T $ or any non zero vector proportional to it like $(1,-2,2)^T$...
Second method:
Let $U,V$ be the cross products :
$$U=(0, -1, 1)^T \times (-2, 0, 0)^T \ \ \ V=(1, -3, 1)^T \times (0, 2, 2)^T.$$
$U,V$ are normal vectors to the planes under consideration. Thus their intersecting has $U \times V=(8,16,-16)$ for a directing vector. This vector is in fact proportional to the one we just have found by the first method. 
