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How can I convert the equation system form of the line to its corresponding vector form?

I currently have:

$x + y - z = 8$

$2x + 2y + z = 15$

I need the line description in a form of $(a_1, b_1, c_1) + \lambda (a_2, b_2, c_2)$, so that the following is true:

$(x, y, z) = (a_1 + \lambda*a_2, b_1 + \lambda*b_2, c_1 + \lambda*c_3)$

How do I do that?

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  • $\begingroup$ Special Solution + $\lambda \times$ Null Space. $\endgroup$ – Inquest Oct 18 '12 at 20:51
  • $\begingroup$ Forgive me my ignorance, but could you elaborate, please? $\endgroup$ – arik-so Oct 18 '12 at 20:54
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The two given equations represent planes, and the required line is their intersection. They can be written in vector form as $$(x,y,z)\cdot U = 8$$ $$(x,y,z)\cdot V = 15$$ where $U = (1,1,-1)$ and $V = (2,2,1)$ are vectors that are normal to the two planes.

It is obvious (I think) that the line is parallel to the cross product vector $U \times V$. So, we can use $(a_2, b_2, c_2) = U \times V$.

Next, we need some point $(a_1,b_1,c_1)$ on the line. There are many possible choices, of course. In effect, we just need to choose some third plane and intersect our line with this. In other words, we add a third linear equation to the two we already have, and solve. One easy choice is the plane $x=0$. Using this equation together with the original two gives the solution $(a_1,b_1,c_1)=(0,23/3,-1/3)$.

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