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I want to find the intersection vectors between 2 spans, something like this: $$span\left \{ \begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix},\begin{bmatrix} 1\\ 2\\ 2 \end{bmatrix} \right \}\cap span\left \{ \begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix},\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix} \right \}$$ Visually, I know their intersection is going to be a line. But I have to find the vector of this line. I attempt to put them into an equation and solve for it this way: $$\begin{align*} a\begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix}+ b\begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} = c\begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix}+ d\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix} \\ a\begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix}+ b\begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} - c\begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix}- d\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix}=0 \\ \begin{bmatrix} 3 & 1 & -3 & -8\\ 6 & 2 & 0 & -5\\ 0 & 2 & 1& 6 \end{bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}=0 \\ \Rightarrow \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix}=t\begin{bmatrix} \frac{5}{12}\\ -\frac{25}{12}\\ -\frac{11}{6}\\ 1 \end{bmatrix}, t\in \mathbb{R}\end{align*} $$

So now, I assume I could plug in those values back into the first equation: $$\frac{5}{12}\begin{bmatrix} 3\\ 6\\ 0 \end{bmatrix}+ -\frac{25}{12}\begin{bmatrix}1\\ 2\\ 2 \end{bmatrix} = -\frac{11}{6}\begin{bmatrix} 3\\ 0\\ -1 \end{bmatrix}+ 1\begin{bmatrix} 8\\ 5\\ -6 \end{bmatrix}$$

And claim that the vector that this equation is equals to each other is the vector span of the intersection. But when I try to add them together, I get this: $$\begin{bmatrix} -\frac{5}{6}\\ -\frac{5}{3}\\ -\frac{25}{6} \end{bmatrix} \neq \begin{bmatrix} \frac{5}{2}\\ 5\\ -\frac{25}{6} \end{bmatrix}$$ They are not equal! That's weird.

How come they don't equal to each other? Is what I am doing right? How and what other ways can I use to find the intersection between the 2 vector spans?

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When you row reduced your matrix, how did you get in the end that $a = 5t/12$? –  user38268 Nov 24 '11 at 13:40
    
As Benjamin Lim says, I get $a=\frac{55}{36}t$. –  Joe Johnson 126 Nov 24 '11 at 15:05

1 Answer 1

up vote 1 down vote accepted

It seems to be a simple computation error. As already pointed out in the comments, you should get $a=\frac{55}{36}t$.

In general, although it is highly recommended to solve such problems by hand until you get the hang of it, everyone makes such computation errors and it's best not to waste time and if something goes wrong, check yourself against some computer algebra system. For example, using Sage you can find the solution in three quick rows of code:

M=MatrixSpace(QQ,3,4)
A = M.matrix([[3,1,-3,-8],[6,2,0,-5],[0,2,1,6]])
show(A.echelon_form())

So I recommend that you'll familiarize yourself even now with such algebra system - it might very well save you time in the long run. However, I repeat again that you should use such tools only to double-check yourself, and never to do the work for you while you're still a student.

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