0
$\begingroup$

The parametric equation of the line is $x=2t+1, y=3t-1,z=t+2$

The plane it is parallel to is $x-by+2bz = 6 $

My approach so far

I know that i need to dot the equation of the normal with the equation of the line = 0

$n = <1,-b,2b>$

I would think that the equation of the line is $ L(t) = <2t+1,3t-1,t+2>$ but am not sure because it hasn't work out very well so far.

** Solve for b such that the parametric equation of the line is parallel to the plane

  • 0
    @GerryMyerson sorry i forgot to mention that the question asked to solve for$b$such that the parametric equation of the line is parallel to the plane.2012-09-30

1 Answers 1

2

Perhaps it'll be a little clearer if you write the line as

$l: (1,-1,2)+t(2,3,1)$

So now you need the direction vector $\,(2,3,1)\,$ to be perpendicular to the plane's normal $\,(1,-b,2b)\,$ :

$(2,3,1)\cdot(1,-b,2b)=0\Longrightarrow 2-3b+2b=0....$