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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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Read off the normal vector from the equation of the plane, and read off the direction vector of the line from the parametrization. – Arturo Magidin Feb 15 '11 at 4:04
Oh well I already knew the normal vector of the plane, $<2,3,5>$, and now the normal vector of the line (if I can call it that) is just $<1,0,2>$? – maq Feb 15 '11 at 4:05
@Arturo any idea about the parametrization of the plane? – maq Feb 15 '11 at 5:03
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. – Arturo Magidin Feb 15 '11 at 5:08
up vote 6 down vote accepted

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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So it'd be $<2,3,5>X<-4,-12,18>$? – maq Feb 15 '11 at 4:18
@mohabitar: Yes, that's the cross-product you want to find, to see if it's the zero vector. – Isaac Feb 15 '11 at 4:21
and what about for finding the parametrization of the plane? – maq Feb 15 '11 at 4:37
@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. – Isaac Feb 15 '11 at 4:45

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