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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. –  Daryl Sep 30 '12 at 3:36
    
oh sorry that was a typo there –  40Plot Sep 30 '12 at 3:42
    
do i just dot it with <2t+1, 3t-1, t+2> ? –  40Plot Sep 30 '12 at 3:43
    
You want the dot product with a vector parallel to the line. –  Daryl Sep 30 '12 at 3:44
    
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? –  Gerry Myerson Sep 30 '12 at 3:48

1 Answer 1

up vote 2 down vote accepted

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

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

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