parameterization

[mcwang719]

Let r(t)=<x(t),y(t),z(t)> be the position vector function which corresponds to the 3d line in space which passes through point P(-3,4,1) and is parallel to the vector v=<-2,3,6>. In particular, the most convenient parameterization has r(o)=<-3,4,1> and r(1)=<-5,7,7>
Give the standard parameterization for r(t) based on the information above.

so v=<-5+3,7-4,7-1>=<-2,3,6> so taking points(-3,4,1) equals x=-3-2t y=4+3t z=1+6t. =<-3-2t,4+3t,1+6t>. did i do this right? is that what they're asking? thanks!!!!!!!!!

 

[pka]

Look the line through \(\displaystyle (x_0 ,y_0 ,z_0 )\) parallel to the vector \(\displaystyle \left\langle {a,b,c} \right\rangle\) is:
\(\displaystyle l(t) = \left\{ \begin{eqnarray}
at + x_0 \\
bt + y_0 \\
ct + z_0 \\
\end{array} \right.\)

 

[mcwang719]

<-2t-3,3t+4,6t+1> is that right??? thanks!