# S6.802.12.5.5 Find a vector equation and parametric equations

• MHB
Gold Member
MHB
see post 3

Last edited:

Gold Member
MHB
$\tiny{s6.802.12.5.5}$
$\textsf{ Find a vector equation and parametric equations for:}\\$
$\textsf{The line through the point$(1,0,6)$}\\$
$\textsf{and perpendicular to the plane$x=-1+2t, y=6-3 t, z=3+9t$}$

$\textit{looking at some examples but? }$

What you have posted is not a plane, but a line. There are an infinite number of planes that that line could go through...

Gold Member
MHB
$\textsf{sorry copied problem incorrectly it should read}\\$

$\textsf{Find a vector equation and parametric equations for the line.}\\$

$\textsf{through the point$(1,0.6)$and perpendicular to the plane$x+3y+z=5$!.}$

$\textit{the book answer to this was}\\$
$r=(i+6k)+t(i+3j+k)\\$
$x=1+t, y=3t, z=6+t$
$\textit{but don't know how it was derived!}$

Last edited:
Gold Member
MHB
$\textsf{sorry copied problem incorrectly it should read}\\$

$\textsf{Find a vector equation and parametric equations for the line.}\\$

$\textsf{through the point$(1,0.6)$and perpendicular to the plane$x+3y+z=5$!.}$

$\textit{the book answer to this was}\\$
$r=(i+6k)+t(i+3j+k)\\$
$x=1+t, y=3t, z=6+t$
$\textit{but don't know how it was derived!}$

A plane's normal vector always has the same coefficients as the plane, so the normal vector to the plane is (1, 3, 1). To make it infinitely long, multiply by a parameter (t) which can take on any real number, so t(1, 3, 1) = (t, 3t, t). Then position it so it can go through the point (1, 0, 6), giving (t + 1, 3t + 0, t + 6). Thus x = t + 1, y = 3t and z = t + 6.