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Point) Parameterize the plane that contains the three points (-5,-4,3), (-4, 8,-2) , and (30,35,0).r(s,t) = (Use and for the parameters in your parameterization, an...

Question

Point) Parameterize the plane that contains the three points (-5,-4,3), (-4, 8,-2) , and (30,35,0).r(s,t) = (Use and for the parameters in your parameterization, and enter your vector as a single vector with angle brackets: e.g-, as < 1 + $ + t,s-t,3-t>)

point) Parameterize the plane that contains the three points (-5,-4,3), (-4, 8,-2) , and (30,35,0). r(s,t) = (Use and for the parameters in your parameterization, and enter your vector as a single vector with angle brackets: e.g-, as < 1 + $ + t,s-t,3-t>)



Answers

Use set theoretic or vector notation or both to describe the points that lie in the given configurations. The plane spanned by $\mathbf{v}_{1}=(2,7,0)$ and $\mathbf{v}_{2}=(0,2,7)$.

So we're given three points p. Q and R. And our task is to find the vector perpendicular to the plane that contains thes three points. The way we're going to do this is first by developing our vectors P Q and P. R. So are two vectors are p Q and P. R. So peek you is going to be equal to Well, we know P is our initial point. Q. Is Air terminal point, so we have to do each of the elements of Q minus each of the elements of Pete. So that's gonna be one minus three comma to minus four, comma three minus five. And just a reminder we did it in this order because 12312123123 are all the terminal elements in this peak. You point vector calculation because Q is the terminal point in the vector. P is the initial point, which means they're the second terms that come in these subtraction components. In PR, we repeat the same process, using our instead of cute as our terminal points. So cue so are the way we're gonna use it is in the same way. Four minus three seven minus four and six minus five. Now we can solve both of these and we get our Spectres as negative two. Negative, too. I think it's two and the negative, too. And for a 2nd 1 we get 131 No. All we have to do to find the vector perpendicular to the plan containing these three points is we have to find the cross product peak. You cross PR p Q Cross P. R. So let's jog her memory on the formula for a cross product. We know that P Q Cross PR is equal to well, what's right in terms of you and be actually, for simplicity's sake, you cross V is equal to you too. VI three minus you. Three v two comma you three V one minus. You won B three and you one V two minus you, too V one. So we're gonna use the same formula to find the cross product. Except in this case, the victory you is going to be p. Q and Director V is going to be PR. So let's do the calculation Now, Peak, you cross PR is equal to you two times V three. So that's gonna be negative two times one, which is just negative, too. Minus you three times V two. So negative, too. Times three. That's just negative. Six. Next part you three times V one. So negative. Two times one is just negative, too. Minus you one times V three. So that's a negative. Two times one giving us negative, too. And third part you one times be too negative. Two times three is just negative. Six. Minus YouTube Times V one. That's negative. Two times one. Giving us a negative, too. Now, when we solve this, we get our final answer as negative. Two plus six is four negative two plus 20 and the negative six plus two is negative for. Therefore, our final answer is negative or four zero negative for.

In this problem, we are given three points and asked to find or to produce a vector that is perpendicular to the plane that contains those three points. And to do this, we're going to take the vector p Q. And take the cross product of that vector with the vector p R. So how do we get P Q and P. R. Well, to get a peek, you were going to take the point Q. And subtract the point p from it. That's because Q is the end point of the vector and P is the starting point. And so we're going to get one 23 minus three 45 And our factor is going to be the foaming we get negative, too negative, too negative, too. Okay. Now, to get the vector PR, we're going to take the point are and subtract the point p from it. And so we're going to get four seven six, minus three for five. And when we do that, we get the following factor. One, three, one And again, that is equal to the vector. PR for the one of here is equal to the vector p Q. And now we're going to take the cross product of these two doctors by running them in a matrix that has the first row of I, J and K a second row with the components of P Q, which are native to negative to negative two and 1/3 row with the components of PR, which are 13 in one. Okay, and now to take the cross product and get the component for I, we're going to cross out the first column and take the determinant of this two by two matrix when we get negative. Two times one minus native, two times three as native to minus native six United to plus six, which is for then, we always attracts the J component to get back up or not, we're going to cross at the middle column and take the determinant of this to go to Matrix. So we get negative two times one minus negative two times one her native to plus two zero and then to get the K component, we're going to cross out the third calm. You take the determinant of this two by two matrix, which is going to be negative two times three minus data two times one which is negative. Six minus negative to Grenada, six plus two or negative for and we have found our vector. It's going to be four zero negative for.

Hi. So for this exercise we have a vector that is perpendicular to the plane. Okay, so we need to find a parametric equation of this plane, this game coined pie. So how to find this? So basically we have here a condition that we should be perpendicular to this whole plane. That means that any any point X. Y. Z. That lies in the plane. Shit will satisfy that. X. Why Z In a product with 40 -5 Will be equal to zero. Okay. And It will be because 20 because this plane passed through the point through the origin. Otherwise, if this plane passed through some point, let's say V. That is a fracture, then they should be equal to be. So we have this condition to any point X. Times this vector will be professional. So we need to fine all the vectors X, Y and Z. That satisfy this condition. So here we have an equation. The question that we obtain is at four x -5 Y. is equal to zero. So we have to free variables. Here's the Wine. Sorry? Here's -5 Z. So we have to free variables. One is why that's going to be T. And Z. Then this case is going to be S. So defining this, we have the X. Is equal to 5/4 Z. So this gives us a relation of course between the the components. So we can just first some. Yeah. So yeah so the generic solution for this equation here is that the vector X. Why Z? But they find the plane is given in terms of T. M. S. Where the Point where it's passing 30. So I wouldn't put anything. And then here the two vectors that passed through. So we have this is equal to 5/4 S. T. S. And we can separate the solution in two pictures, the picture tea time 010 plus S. Times. Fight over four 01 And this will give us the equation of the plane that we need. You can rewrite actually this vector here, you can choose any multiple. So let's, for example, multiply this vector by four, and you obtain these other representation 010 plus S. Five 04. And this also will be perpendicular to the vector V.

Okay, so for this exercise we have to find the the equation diplomatic equation of the plane that passed through this point and is parallel to these two vectors. Okay, So we know that this is parallel and this zero should be contained in that plane, geometrically speaking. We're talking about the following here. We have this space X, Y and Z. And the zero define here for example. And then we have a plane as they called pie uppercase Pie that pass through this point. So how to define this at this point? Well basically we choose first zero and then we can we can use the pro factors. Because the parallel vectors will give us the permit risk will give us actually the privatization because these two vectors, this pan or any linear combination of these two vectors will define a point in this plane. Okay, so in that way we can generate this whole plane. So having that idea in mind that first which is zero and then which is any linear combination of the directors we want and V. Two is that we obtain the equation of the of the plane. The parametric equation of the plane. So the definition is that the parametric equation of the plane given by two parameters, DNS is equal to the point where this plane is passing through plus T. Times V. One plus S. V. To where these two vectors to vectors are parallel to the plane. And non cleaner if they are ca linear vectors, that means that we will define a line another plane. Okay, so we just need to replace the values of our vectors. So X0 is the .05 minus four plus T. Times 00 -5 plus S one -3 -2. And then we just need to solve this. So we have zoo in the first coordinated, we will have only S in the second one that will be five minus three S, and the third coordinate will be minus four minus four, 95 t minus two S. And this is the equation of the plane.


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