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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

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    Two hints. 1. $n$ should be $[1,-b,2b]$. 2. $n$ should be perpendicular to the line.2012-09-30
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    oh sorry that was a typo there2012-09-30
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    do i just dot it with <2t+1, 3t-1, t+2> ?2012-09-30
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    You want the dot product with a vector **parallel** to the line.2012-09-30
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    The question is not clear. You give the parametric equations for the line in your first sentence. Then you rewrite those same equations in the last sentence, and ask whether they are correct. Well, if your first sentence is correct, then of course your last sentence is, too. Is it possible that what you really want to know is the value of $b$? Or that you really want to know whether your first sentence is correct, given the second sentence?2012-09-30
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    so I solve for vector v in: L(t) = r0 + tv?2012-09-30
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    @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

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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 $\,(1,-b,2b)\,$ :

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