# Thread: Finding parametric scalar equations of lines which are parallel to a plane

1. ## Finding parametric scalar equations of lines which are parallel to a plane

The question: Let pi be the plane 2x-3y-6z=6 and M be the point of intersection of the plane with the y axis. Find parametric scalar equations of the line m which lies on the plane pi, passes through the point M and m is parallel to the xz plane.
My attempt:
M must be (o,-2,0)
And i know the eqn of a parametric vector eqn of a line is r=r0+tv

Where c will not have a y component. Am i right in then solving the line x+Z-2=0
with 2x-3y-6z to find the vector? Then the line will be r=-2j=t(v)

2. Ref http://www.mathhelpforum.com/math-he...wo-planes.html

• Line is parallel to plane x-z, then Line perpendicular y-axis, $\displaystyle \vec{e}_2$
• Line lies on plane P, then Line perpendicular $\displaystyle \vec{v}_p\begin{bmatrix}2\\-3\\-6\end{bmatrix}$
• M(0,-2,0)
• Line normal vector $\displaystyle \vec{v}=\ker[\vec{e}_2,\vec{v}_p]=\ker\begin{bmatrix}2&-3&-6\\0 &1 &0\end{bmatrix}=\begin{bmatrix}3\\0\\1\end{bmatrix }$
• Line equation $\displaystyle \frac{x}{3}=\frac{z}{1},~y=-2$
.
• Or $\displaystyle \vec{x} = t\vec{v}+\vec{m}$