$\displaystyle \vec{r}(t)= (-t+ 11/10)\vec{i}- 3t/vec{j}$

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- Sep 18th 2009, 06:12 PMquah13579i am testing the code
$\displaystyle \vec{r}(t)= (-t+ 11/10)\vec{i}- 3t/vec{j}$

- Sep 18th 2009, 06:24 PMSoroban
Hello, quah135791

I assume you're hoping for:

$\displaystyle \vec{r}(t) \;=\;\left(-t + \tfrac{11}{10}\right)\vec{i} - 3t\vec{j} $

Click on QUOTE to see how it's done.

- Sep 18th 2009, 06:40 PMquah13579
- Sep 19th 2009, 12:36 AMCaptainBlack
There are lots of ways of doing this, the one I like (because it involves remembering fewer commands that some of the others) is:

$\displaystyle \begin{array}{rcrcrcr}

x_1&-&x_2&+&x_3&=&2\\

3x_1&+&x_2&-&x_3&=&2\\

2x_1&+&x_2&-&x_3&=&k

\end{array}$

Which is produced by:

\begin{array}{rcrcrcr}

x_1&-&x_2&+&x_3&=&2\\

3x_1&+&x_2&-&x_3&=&2\\

2x_1&+&x_2&-&x_3&=&k

\end{array}

CB