# Find all vectors in an intersection of 2 vector spans

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

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