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In Exercises $7-10$ , let $H$ be the hyperplane through the listed points. (a) Find a vector $\mathbf{n}$ that is normal to the hyperplane. (b) Find a linear functional $f$ and a real number $d$ such that $H=[f : d]$$$\left[\begin{array}{l}{1} \\ {1} \\ {3}\end{array}\right],\left[\begin{array}{l}{2} \\ {4} \\ {1}\end{array}\right],\left[\begin{array}{r}{-1} \\ {-2} \\ {5}\end{array}\right]$$

a) $\vec{n}=\left[\begin{array}{l}{0} \\ {2} \\ {3}\end{array}\right]$b) $f\left(x_{1}, x_{2}, x_{3}\right)=2 x_{2}+3 x_{3}$$d=11$

Calculus 3

Chapter 8

The Geometry of Vector Spaces

Section 4

Hyperplanes

Vectors

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in this exercise, we need toe find there is a new hyper plane. The age that that passed through three points. The one on the 13 we to is a point to. For one onda 0.3, find us minus one minus 25 So we got another on hyper playing that passed through these two points on. We need to find first the normal vector, the normal vector to this plane. So technically thes is, or hyper plane age on here. In this plane lies the 3.3123 So we need to find a vector that the normal vector to these hyper plane. Okay. And then after that, we need to find the linear functional such that this age is equals to the that linear, functional f on that number. Okay, so let's start by considering the normal vector in this case on, we're going to construct in affordable way. So these three points lies on the same on the same plane. Okay, so we to the three. So we can find vectors that lies also on this hyper plane by just taking the difference between these two points so we can define here a vector Alfa, and I'm going to call this victory beat on how to construct the normal vector. We can do this in two ways. The usual way is that it's supposed that this normal vector should be perpendicular to these two vectors here. Alfa Um beater. So we just need to defined two equations and that product Alfa should be close to zero on end that product, Vita should because 20 And here we obtain our system off a linear linear system off equations on we can solve. However, there is another way to do this we can instead off considering thesis complicated could be complicated system We're going to consider geometric property of the vectors. So let's suppose that here is we one hears me to. Here is the three we find here are vectors Alfa on Vita on We know that the cross product off Alfa with Vita so we can define the normal vector Just as the cross product off these two vectors here could just a even simpler. So let's compute these vectors Alfa on a vita. So here we just need to take three to minus We won and here the three minus we won from this we obtain. This is two for one minus 113 on Dhere. Yes, minus one minus 25 minus on 13 So, Alfa, the Alfa factor is 13 minus two on de Vita Vector is minus two minus three to Okay, so we got these two vectors, and now we can compute the cross product. These two were We can use the determinant way. So that means that you were going to put the component I J k. So here is 13 minus two, minus two, minus three to on. From this we obtain that is cross product or the normal vector is 0 to 3. Okay, so this is a normal vector that we try to find for the part A. Now, now that we have defined it in normal vector, we can define the functional. So how to define functional? Just remember that the equation off hyper plane is the normal vector times a vector X equals to be We're X is on our end, depending on the dimension off the hyper plane. Okay, so this is similar. This is going to be or functional on Duh. So let's Let's compute this so the functional in this case is defined. X one x two x three Will be the normal vector times x one x two x three But that is equal to say 0 to 3 The product here x one x two x three So or functional if 6123 is two x two plus three x three Okay, now we need to find the value off the on that it seemed the is obtained by evaluating dysfunctional on some off the point B one B two or B three that lies on the hyper plane so we can choose any of this in particular. I'm going to choose the first one, the big one. So let's evaluate this. A to point B one on this is equals to 11 three, so f 11 three is equals to to times one plus three times three on this is equals to 11, so the value off the is illa so at the end, or Hypercar hyper plane is defined as a function of F D, where f is equals to two x two plus three x three. Andi is equals to 11

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