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In the line through $P(0, 0, 0)$ and is perpendicular to $x=y-5$, $z=2y-3$, when we solve the equations and get the symmetric equations in order to find the vectors $V_1$ and $V_2$, why the normal vector $N_1$ of the plane coincides with $V_1$? And what will happen if the point is $P(0, 2, 1)$?

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What do you mean "solve the equations"? What are V1 and V2? Can you make this question more easily understandable? –  anon Feb 3 '12 at 1:51
    
Hi,sorry, I mean when I convert those equations to symmetric equations (x+6) = y = (z+3)/2 and get the normal vector V1 = <1, 1, 2>; I don't know why has been said that if we go through P(0, 0, 0) and create a plane perpendicular to x=y−5, z=2y−3 the normal vector of the new plane coincides with V1? –  pablo89 Feb 3 '12 at 2:12

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The direction vector $\mathbf{v}$ points in the direction of the line $\ell$. If a plane $\pi$ is perpendicular to $\ell$ it must make right angles with the vector $\mathbf{v}$ if we place the starting point of $\mathbf{v}$ on $\pi$ itself. This ensures that wherever $\ell$ intersects $\pi$, locally it looks like a $z$ axis stood on top of an $xy$-plane (of course this scene will not necessarily take place at the origin). It doesn't matter where the plane is located in space, as long as it is perpendicular to $\ell$ we must have $\mathbf{w}\perp\mathbf{v}$ for any vector $\mathbf{w}$ affixed strictly to $\pi$ (so if we have $\mathbf{p},\mathbf{q}\in\pi$, the displacement vector $\mathbf{p}-\mathbf{q}$ that goes from $\mathbf{p}$ to $\mathbf{q}$ is stuck to $\pi$). Hence if $\mathbf{n}$ is a normal vector for the plane, $\mathbf{v}$ and $\mathbf{n}$ must point in the same direction and so are the same vector up to a scalar multiple (i.e. $\mathbf{v}=c\mathbf{n}$ means they may differ in magnitude but not direction).

This might make more sense if you tried to visualize what's going in in three-dimensional space. If there is a large, flat piece of paper floating in the room, for example, we could say this models a plane, and from there if we had a yardstick, we could by hand position the yardstick so that it makes right angles all around its contact with the piece of paper. (Don't question why the paper is floating there. Magnets, or something.) This yardstick is our line $\ell$, and the little tick marks on it tell us how we are parametrizing it. If we take a toothpick and pressed it against the paper, then the direction it points in is perpendicular to the yardstick. Similarly, if the paper has a normal vector, say a pixie stick that we're holding up to it, if we move the pixie and yardsticks together while keeping constant the direction they point, we will find they come up back-to-back and point in the very same direction (though one might be bigger than the other).

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