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Let $\pi$ be the plane that contains the points $(0,1,2)$, $(1,1,2)$, $(2,2,2)$. Find the equation of $\pi$.

My guess was to define two vectors by using the three given points, find a vector orthogonal to both, and then define the plane as span(u, v). The prof says I was wrong because the plane isn't spanned by two vectors, but instead is made of three points. The problem was that I had set the equation of the plane to equal zero, as I had assumed that the plane went through the origin, but in this case it apparently does not.

So my question is, how can I tell which methods to use and why are they not ultimately the same?

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You've given the reason: Unless you know tha tthe plane contains the origin,, it is not the span of two vectors. However, it is the translate of such a span. – Hagen von Eitzen Jun 25 '13 at 20:53
If you found a vector orthogonal to two vectors with both star and end points on this plane (some obvious choices!), then you are more or less done. The plane won't necessarily go through the origin, but you know three out four coefficients in the equation of the plane, and can solve for the last one simply by plugging in one of the points (you can also plug in the other two to catch an eventual error as a bonus). – Jyrki Lahtonen Jun 25 '13 at 22:04
See 2.3 in Plane: two vectors, lying on the plane, are not enough. – Tony Piccolo Jun 25 '13 at 22:30
up vote 1 down vote accepted

So just as the professor told you, you found the equation of the plane which passes through the origin. However, that is only the case when you are either told that, or your points happen to be given such that the plane defined by the three points contains the origin. If it helps, imagine that by assuming your plane went through the origin, you essentially created a parallel plane to the plane defined through the three points, and your parallel plane was created to pass through the origin, but have the same 'tilt' as the plane which passes through the three points.

Explicitly, you should have done something like: form the vectors $(1,0, 0)$ from the first two, and $1, 1, 0$ from the second two. Then their cross product is $(0,0,1)$ to give you a normal vector. Using this normal vector, we can use a point on the line and a normal vector to write the equation of a plane as: our normal vector: $(n1, n2, n3) = (0, 0, 1)$, and our point (we could use any of them, lets just choose the first) $(p1, p2, p3) = (0,1,2)$. Then our plane will be $n1(x-p1) + n2(y-p2) + n3(z-p3)$ which in this case is $0 + 0 + 1(z-2) =0 $ or $ z=2$. You can check to make sure that this works for our three points by just seeing that all of their third coordinates are in fact $2$. Hope that answers what you were still uncertain about.

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