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I have some trouble following some examples in my textbook. In the examples, the book provides equations for planes, and I'm not sure how they are derived.

One example is:

Given:

$$\begin{bmatrix} x \\ y \\ z\end{bmatrix} = \alpha \begin{bmatrix} 1 \\ 1\\ 1 \end{bmatrix}e^{4t} + \beta \begin{bmatrix} -1 \\ -1 \\ 2 \end{bmatrix} e^{t} + \gamma \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix}e^{-t}$$

Any solution for which $\gamma = 0$ will lie in the plane $x - y = 0$ for all $t$.

Another example is:

Given: $$\begin{bmatrix} x \\ y\\ z \end{bmatrix} = \beta \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}e^{t} + \gamma \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} e^{4t}$$

This defines the plane $x - 3z = 0$

It's been quite some time since I had multivariable calculus, and after fumbling around for a long time, I just don't see how these plane equations are obtained. Since the book doesn't show any intermediate steps, I would greatly appreciate any help!

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Are you sure you have the second example correctly? If I understand what you have correctly, I would say that in the second example you get the plane $x-z=0$. (P.S. What is the textbook?) –  Arturo Magidin Apr 24 '12 at 16:30
    
Hi. Yes, this is as it is written in the textbook (Nonlinear Ordinary Differential Equations - An Introduction for Scientists and Engineers). However, the book contains a vast amount of typos (I've never encoutered a book with as many typos as this before), so it is quite possible that the text has got it wrong. –  Kristian Apr 24 '12 at 16:50
    
The way I would interpret this for the first problem: if $\gamma=0$, then $(x,y,z)\in\mathrm{span}\bigl((1,1,1),(-1,-1,2)\bigr) = \mathrm{span}\bigl(1,1,0),(0,0,1)\bigr)$, and that span is precisely the plane $x-y=0$. Doing the same thing with the second example, I get $x-z=0$. –  Arturo Magidin Apr 24 '12 at 16:52
    
Thanks a lot! But from where do you derive that $span((1,1,1), (-1,-1,2)) = span(1,1,0),(0,0,1)$? –  Kristian Apr 24 '12 at 17:11
    
Row-reduce. If you add the first vector to the second, you get $(1,1,1)$ and $(0,0,3)$. Divide the second by $3$ to get $(1,1,1)$ and $(0,0,1)$. Subtract the second from the first to get $(1,1,0)$ and $(0,0,1)$. None of those (elementary row) operations changes the span. –  Arturo Magidin Apr 24 '12 at 17:12
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