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Choose a set $S$ of three points such that aff $S$ is the plane in $\mathbb{R}^{3}$ whose equation is $x_{3}=5 .$ Justify your work.

$\left(\begin{array}{l}{1} \\ {2} \\ {5}\end{array}\right),\left(\begin{array}{l}{2} \\ {4} \\ {5}\end{array}\right),\left(\begin{array}{c}{-1} \\ {8} \\ {5}\end{array}\right)$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 1

Affine Combinations

Vectors

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in this example, we're going to consider the Plain X three equals five that lives in the space are three. Our goal here is to find a set s consisting of vectors, V one, V two V three such that the a fine hole of this s will result in the same plane X three equals five. So to start out, let's begin building vectors V one and the first requirement such that the ATF fine whole generates a plane is that all the vectors must come from the plane itself, which means X three must always be equal to five. So it's right v one v two and the three such that five is the last entry. Now, whatever we put in the 1st 2 entries, at least we know that all of these vectors lie in that plane. Now we can pick just about any set of vectors in this plane. They must be distinct. Otherwise, that fine hold will generate a single point, not the plane. So I could start off with, say, 10 Then we could put in possibly a 20 and noticed that if we have a 10 and a 20 then in the plain, X three equals five thes two points will go through. These two points will go through the line in that plane. If we do say 30 all three points V one v two V three are going to lie on the same exact line in the plane. This is problematic since in this case, if all three points are Colin near the F find whole would generate a line and not a plane. So we need to be very careful with the last factor. We just need to insist that if we were to plot these on the plane, extra equals five. Not all points lie on the same line. We might try something along the lines of, say, a 10 to change it up. Now all three points are not Colin ear on the plain. X three equals five And this implies next that the I find hole of the city s where v one v two v three are as to find generates the plane. X three equals five

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