line vector

**Faz** · December 1st 2008, 05:28 PM

Find equation of the line containing the given point (5,3,4) and parallel to the vector
-> -> ->
2 i + 5 j - 8 k

**vincisonfire** · December 1st 2008, 05:37 PM

$x=2t+5$
$y=5t+3$
$z=-8t+4$
$-\infty \leq t \leq \infty$

**Faz** · December 1st 2008, 06:32 PM

can show me the working i still don't understand??

**Plato** · December 1st 2008, 07:17 PM

Originally Posted by Faz

can show me the working i still don't understand??

I fear that one either understands or not: it is that simple.
$\begin{gathered} \ell :\left\langle {5,3,4} \right\rangle + t\left\langle {2,5, - 8} \right\rangle \hfill \\ \ell :\left\langle {5 + 2t,3 + 5t,4 - 8t} \right\rangle \hfill \\ \end{gathered}$
$\ell :\left\{ \begin{gathered} x = 5 + 2t \hfill \\ y = 3 + 5t \hfill \\ z = 4 - 8t \hfill \\ \end{gathered} \right.$

**Soroban** · December 1st 2008, 08:52 PM

Hello, Faz!

You're expected to know the formulas for this problem.

Find equation of the line containing the given point (5,3,4)
and parallel to the vector $\vec c \:=\:2\vec i + 5\vec j - 8\vec k$

Parametric equations: . $\begin{array}{ccc}x &=&5+2t \\ y&=&3+5t \\ z&=&4-8t \end{array}$

Symmetric equation: . $\frac{x-5}{2} \:=\:\frac{y-3}{5} \:=\:\frac{z-4}{-8}$

**vincisonfire** · December 2nd 2008, 03:26 AM

You set your x,y,z so that they start at (5,3,4). These are some kind of x,y,z-intercept if you want.
Then you know that in time t, x is going to increase by 2, y is going to increase by 5 and z is going to increase by -8.
You thus have "3 equations".
$ \ell :\left\{ \begin{gathered} x = 5 + 2t \hfill \\ y = 3 + 5t \hfill \\ z = 4 - 8t \hfill \\ \end{gathered} \right. $

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