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Find the parametric equations of the following lines.

a) The line parallel to $\begin{bmatrix}2 \\-1 \\0\end{bmatrix}$ and passing through P(1,-1,3).

My solution $\vec{P}$ = $\begin{bmatrix}1 \\0 \\-3\end{bmatrix}$

and for the parametric equation $\begin{bmatrix}2 \\-1 \\0\end{bmatrix}$ + t$\begin{bmatrix}1 \\0 \\-3\end{bmatrix}$

Micky
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    what is $[2,-1,0]$ a vector? – Dr. Sonnhard Graubner Nov 02 '15 at 19:56
  • You set P in data, then in your solution, with a different value ! and the tangent of your line is not proportional to your guide direction. – Fabrice NEYRET Nov 02 '15 at 19:58
  • Your answer is wrong. Your line is parallel to $[1,0,-3]$ not to $[2,-1,0]$ – Emilio Novati Nov 02 '15 at 20:00
  • How would i correct it? – Micky Nov 02 '15 at 20:01
  • @Dr.SonnhardGraubner They didn't say the question is given as is – Micky Nov 02 '15 at 20:05
  • Start with the equation of a line through the origin. This is just $\vec r(t) = t\vec v$. Clearly this line is always parallel to $\vec v$ (confirm this for yourself with the definition of parallel vectors). Then shift the line so that the direction is the same but it passes through your point of interest $P$. Then your new equation becomes $\vec r(t) = t\vec v + \vec{OP}$. (Compare this to the scalar equation: $y=mx+b$). Confirm that this does pass through $P$ and the direction of the vector pointing from any point along the line to any other point along the line is parallel to $\vec v$. –  Nov 02 '15 at 20:08
  • Perhaps this answer will help you further. –  Nov 02 '15 at 20:14

1 Answers1

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Given the vector $\vec v=[2,-1,0]^T$, all the vectors of the form $[x,y,z]^T=t[2,-1,0]^T$ ( where $t$ is a real number) stay on the same line (1-dimensional subspace) passing through the origin and containing $\vec v$, so this is the equation of such a line.

You want a line parallel to this one, that does not passes through the origin but through $P=(1,-1,3)$ so, add a vector pointing to this point such that, whan $t=0$, the vector $[x,y,z]^T$ points to $P$ . This gives the equation:

$$ [x,y,z,]^T=t[2,-1,0]^T+[1,-1,3]^T $$

Emilio Novati
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