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Consider the plane defined by $2x+3y+5z=30$. Is the line parametrized by $l(s):=(-1,0,2)+s(-4,-12,18)$ for all s orthogonal to the plane?

So I know this is easy, but I'm missing one key piece. If the normal vector of the plane cross the line is equal to zero, then this confirms the line is indeed orthogonal. But now how would I cross the normal vector of the plane with this line?

And the question also asks to find a parametrization of the plane. How would I go about doing this? There are no similar examples in the textbook..

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    It's not the "normal vector of the line", it's the *direction* vector of the line. As for parametrizing a plane, take a point $\mathbf{p}$ on the plane, and two non-parallel vectors $\mathbf{u}$ and $\mathbf{v}$ such that $\mathbf{p}+\mathbf{u}$ and $\mathbf{p}+\mathbf{v}$ are in the plane. Then you get all the points of the plane as $\mathbf{p}+r\mathbf{u}+s\mathbf{v}$. Think about it geometrically.2011-02-15

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As you said in your comment, the normal vector of the plane is $\langle 2, 3, 5\rangle$; the line contains the point $(-1,0,2)$ and is in the direction $\langle -4,-12,18\rangle$. You want to see if the cross-product of the normal vector to the plane and the direction vector of the line is zero.

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    @mohabitar: That I'm less sure about. More or less, if you let two of the three variables be parameters, the third variable is determined. For example, if $x=s$ and $y=t$, then $z=\frac{1}{5}(30-2s-3t)$, so that $(x,y,z)=(s,t,\frac{1}{5}(30-2s-3t))$ is a parameterization of the plane. I'm just not sure that's what the question intends.2011-02-15