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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.

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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
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 $(x,y,z) = a i + b j = (3a-2b, -2a, -(a+b))$. Then, from $y$ we have $a = -\frac{y}{2}$, and from $x$ we have $b = -\frac{3}{4} y -\frac{1}{2} x$. Then $z = -(a+b) = \frac{1}{2} x + \frac{5}{4} y$.

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How did you get z=1/2x + 5/4y and (2,5,-4)^T ? We haven't been taught the cross product yet. – diimension Sep 22 '12 at 21:53
I added another approach. – copper.hat Sep 22 '12 at 22:00
I understand now because of your help, thank you very much copper!! – diimension Sep 22 '12 at 23:15
You are very welcome, glad to be of help. – copper.hat Sep 23 '12 at 0:30

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