How do I determine a vector equation of a line..?

with the same x-intercept as [x,y,z]=[3,3,0]+t[3,-5,-9] and the same z-intercept as [x,y,z]=[6,-2,-3]+t[3,-1,-2]

Thanks a lot (: !

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- Oct 18th 2009, 02:59 PMbebejayHow do I determine a vector of a line..? (more inside)
**How do I determine a vector equation of a line..?**

with the same x-intercept as [x,y,z]=[3,3,0]+t[3,-5,-9] and the same z-intercept as [x,y,z]=[6,-2,-3]+t[3,-1,-2]

Thanks a lot (: ! - Oct 19th 2009, 07:23 AMGrandadMore information please
Hello bebejay

Welcome to Math Help Forum!The $\displaystyle z$-intercept on the second line is where it meets the $\displaystyle z$-axis; i.e. $\displaystyle x = y = 0$. This is where $\displaystyle t = -2$: the point whose position vector is $\displaystyle \begin{pmatrix}0\\0\\1\end{pmatrix}$.

But I'm afraid I don't understand what you mean by the $\displaystyle x$-intercept for the first line. This line doesn't intersect the $\displaystyle x$-axis. (When $\displaystyle t=0, z=0, y\ne0$.)

Are you sure you have the correct information here?

Grandad