# Equation of a plane containing a point and perpendicular to a line

Find an equation of the plane containing the point $(1, 1, -1)$ and perpendicular to the line through the points $(2, 0, 1)$ and $(-1, 1, 0)$

This is what I have:

I first find the vector between the points: $\vec{n}^{\ } = (2,0,1) - (-1, 1, 0) = (3, -1, 1)$

Next I find the formula for the plane that is perpendicular to the vector by taking the dot product of the vector and the given point + another point of the plane

$(3, -1, 1) \cdot (1 + x, 1 + y, z -1 )\\ 3 + 3x -1 - y -1 -z\\ 3x - y - z +1 = 0$

I am not at all sure if I followed the correct steps

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If $(x,y,z) = (1,1,-1)$, does your plane $3x-y-z+1=0$ contain that point then? –  Han de Bruijn Nov 6 '13 at 20:26
You added the coordinates of the general point to those of the point you know should be in the plane, when you should have subtracted one from the other. Aside from that, things look good. –  dfeuer Nov 6 '13 at 20:28

The point-normal form of a plane is $\vec n \cdot(p - p_0) = 0$. If you think about the meaning of this, you will find that for any point $p$ on the plane, if you form a vector from that point and a point known to be on the plane, $p_0$, that vector will be orthogonal to your normal vector $\vec n$.

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A simple check would be to see if the point $(1,1,-1)$ satisfies the equation which it doesn't. The signs are backward as part of the dot product.

$(3,-1,1)\cdot(x-1,y-1,z+1) = 0$

$3x-y+z-3+1+1=0$

$3x-y+z-1=0$

Notice that $(1,1,-1)$ does satisfy this equation.

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@JB_King: I thought this was homework. –  Han de Bruijn Nov 6 '13 at 20:30
I see my mistake, but I do not understand why it is (x, y, z) - (1, 1, -1) as opposed (1, 1, -1) - (x, y, z) –  Leon Nov 6 '13 at 20:31
dfeuer answered my question as a comment on my original question –  Leon Nov 6 '13 at 20:32