choose-a-set-s-of-three-points-such-that-aff-s-is-the-plane-in-mathbbr3-whose-equation-is-x_35-justi

Name: choose a set s of three points such that aff s is the plane in mathbbr3 whose equation is x_35 justi
Uploaded: 2020-07-22
Duration: 2 min 19 s
Channel: Numerade
Description: 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.

Question

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.

Michael Jacobsen · Accepted Answer

in this example, we&#x27;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&#x27;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&#x27;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

Gideon Idumah · Answer

we solve this question so we because we&#x27;re told to choose, is sets s or four dick stints points in out for you. So we do these by choosing four points from the blame Sue X one plus X two minus three X Terry. Of course it&#x27;s solved. So duds at lists we need three off them our notes Kulina. That is why we don&#x27;t we want want to pick up the street at the local inn is because if the points kulina, the uh oh would be in line I&#x27;m not a plane. So sure that those four points known is Karina just into victory. Because if you picked suit as a possibility that they are not soon maybe Kulina sold to pick the three points. So pick three points. That&#x27;s ah, holding that are not Kalinda. So, for example, we can sets x one equal to X to quart zero on dhe off. We do that off. This implies out of mine a story ex Terry recorded solve on, then extra because of minus four. So with these women, three anos cortina. So we can also said so this is one point second points you can set X one equal to X Treaty Court Zero. This would imply that&#x27;s when you plug that in. This would imply because X 10 extra is there a Thomas ext Sue Records. It solved. I&#x27;m for the foot start points because sets ext Su equal to X tree records zero. This was implied. And after two ex want records to solve and an excellent quarter to six now we need 1/4 points. Um, for the fourth points, we can just pick any points for the fourth for the fourth points. Choose whatever points you want to say. We can choose X one equal to X to the quarter one. You can pick anything you want. So this would be Sue Lost. One minor story. Ex Terry Quarter solve. This implies our ex Terry Quarts of minus three. So this implies that&#x27;s in total. The points we compete so we can choose as to be sick. Ward. So for the first point, you have x 10 extra 000 minus fall, the second point x 10 extra reserve and we got extra to be solved. Similarly, we odd 600 and then lastly, we have 11 minus

Ankit Gupta · Answer

we have given a question. Five backs Bless why? Plus these that is a quarto 15 and we need to escape the line. Is the plane represented by the medication? Then we need to list four points that lying of us. We need to find the point. Before that, we can assume something. We can assume that when, uh why is 1/4 0 and Zenden&#x27;s in Porto zero and putting this well earned the situation. So we&#x27;ll get accident Polito three. And the point we can diet here is 30 little no, then excellent quarters little and that is a world of the rose seven get overlooked by which is a sort of hefty and we can get the point, which is the role 15 a little and, uh, excellent world 00 on y is equal to zero so we can get the jobs that Mrs a quantify bill so they can died. The point here 005 No, we just need toe PLO Plourde this point The Lord, he&#x27;s to the point in three d plane. Then we&#x27;ll have a graph so we can plot, like this job and just connecting all lines from this point and we&#x27;ll have this type of plane for here. We can write possible points. Possible points will be do five little so they knows it&#x27;ll fight. Was it all 15 0? See the little little But this is on.

Petros Markopoulos · Answer

for this exercise. We&#x27;re looking at the points p with co ordinates 3 1/3 negative five que coordinates for 2/3 negative three and are recording is 201 We want to find an equation for the plane that passes through these points. So to get started, we will find two vectors that lie on that plane. Let&#x27;s find R P first and then we&#x27;ll find, aren&#x27;t you? So r. P is three minus two 1/3 10 Not gonna five minus one, which gives us 1 1/3 minus six the and are too, which will be our other vector is four minus two 2/3 line a zero negative three minus one, which is equal to two 2/3 negative for Okay, So now that we have two vectors that lie on the plane, weaken, take across product and that&#x27;s a vector that&#x27;s orthogonal to the plane efforts. A normal vector. So let&#x27;s concentrate R p Cross or que, which is given by the determinant with R J. K 1 1/3 1986 to 2/3 of it. Four. Within calculate is using, like smaller determining this 1 30 minute of six 2/3 for I minus. We have to be aware of the minus. Sign 1 96 to negative four times. J plus, we take your the determinant. 1 1/3 2 2/3 Okay, so we can further sold this by multiplying the diagonals was attracting, So we have minus 4/3. Oh, for us 12 3rd and multiplying the main diagonal and subtracting the other one times I minus negative for plus 12 J plus 2/3 minus 2/3. Okay, so solving these we get Ah, 8/3. I, uh, minus Oh, a j Excuse me, When is a J um plus zero K which were thin, right as components as 8/3 minus 80 So this is our normal vector for the plane that we&#x27;re looking for and now is use our as a point that the plane passes through unless plug in the numbers. So we get 8/3 X minus to plus, uh, excuse me, minus a timeline and zero and plus zero Z minus one equals zero. We could multiply this out. We can have 8/3 X minus 16 3rd minus a Why, this is zero, and this term is cancelled out because of the zero efficient. And we can divide by eight to make this simpler. So we have 1/3 X minus 2/3 minus y is equal to zero. So we get our final equation. 1/3 x minus Y is equal to 2/3. And this is one equation for the plane that we&#x27;re looking for. No notice that we couldn&#x27;t like scale this by any, uh, non zero constant. We can multiply this by two by three by whatever, but this is one equation that is in a simple form.

Jimmy Yao · Answer

the normal factor for the pen minus X does three. Why, Susie? Because you ate the normal nectar. Should be minus one and three and sweet. Right? So And that appointment consider t is minus one industry and fire. So this API in thesis that there should have been in there we want to permit tries appointing of here. Say it&#x27;s a so that a the permit Try stir court coordinates should be minus one minus k in three. Mine because ck and thigh plus okay. And we plug this point into very question and we can have minus minus one minus. K and three seeker speak a close the sea in street clothes five. Last week A. It&#x27;s like a 28 So therefore here is one plus K because no impasse. Nine k, this team has nine K. It&#x27;s the CO 28 So this isn&#x27;t 19 right? In your here is 25 terrify so many 70 here for here. Our k should be miners 17/19 ends Our distance should be the norm for this factor minus k in sweet K and speak a So the number of this factor is 7/19 in squared Have one because she square has square, it is ICO to 70 90 squares 20

Jason Orozco · Answer

So we have a polar equation. R equals three, and that represents a set of points in the plane that we want the sketch a graph. So will withdraw the X Y axes. It&#x27;s one and, uh, R equals three means, uh, a circle of radius. Three nights. So 123 123 123 123 Um, so we want says it&#x27;s are equal to three. No inequality. We want points on the circle and nowhere else, right? So it&#x27;s just a solid line through your with a radius of three.

Problem

Choose a set $S$ of four distinct points in $\mat…

Problem 17 Medium Difficulty

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.

Answer

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

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

Linear Algebra and Its Applications

Anna Marie V.

Kristen K.

Samuel H.

Joseph L.

Vectors Intro

Vector Basics - Example 1

David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

Anna Marie V.

Kristen K.

Samuel H.

Joseph L.

David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications