Sign up ×  I am having difficulty understanding this question:
Graph the subspace spanned by the vectors $i = (3,-2,-1)^T$, $j
= (-2,0,-1)^T$.  You don't have to graph it, but the answer for
this problem is $z = (-3/5)x - (6/5)y$. I don't understand how
to come up with that answer. I tried setting $i=-j$ but to no
avail. Also, sorry for writing $i,j$ with $^T$, I do not know
how to format it to make it look like a column vector.
share|cite|improve this question Are you sure that's the right
answer? The points $(3,-2,-1)$ and $(-2,0,-1)$ don't seem to be
on that plane. –  yunone Sep 22 '12 at 7:27  1 Answer 1  up vote
2 down vote accepted  I think something is not quite right here.
Since we are in $\mathbb{R}^3$ we can find the normal to the
plane spanned by $i, j$ by $i \times j = (2,5,-4)^T$, hence the
equation of the plane is $\langle (x,y,z)^T, i \times j \rangle
= 2x+5y-4z =0$, which gives $z = \frac{1}{2} x + \frac{5}{4} y$.
Here is another approach that does not use the cross product:
Suppose $(x,y