5

-2 -2 / 3 -1/310 / 27 -17 / 27 and Uj 14 / 27 4 / 927 ~10 ~14 /= -4 / 9point) Let U1Find vector U4 in @t such that {v, , 02, U5 , 04} is a orthonormal setAnswer;...

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-2 -2 / 3 -1/310 / 27 -17 / 27 and Uj 14 / 27 4 / 927 ~10 ~14 /= -4 / 9point) Let U1Find vector U4 in @t such that {v, , 02, U5 , 04} is a orthonormal setAnswer;

-2 -2 / 3 -1/3 10 / 27 -17 / 27 and Uj 14 / 27 4 / 9 27 ~10 ~14 /= -4 / 9 point) Let U1 Find vector U4 in @t such that {v, , 02, U5 , 04} is a orthonormal set Answer;



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Let $\mathbf{v}_{1}=(1,2,3), \mathbf{v}_{2}=(1,1,-1) .$ Determine all nonzero vectors $\mathbf{w}$ in $\mathbb{R}^{3}$ such that $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{w}\right\}$ is an orthogonal set. Hence obtain an orthonormal set of vectors in $\mathbb{R}^{3}$.

Hello there. So for this exercise we have these three vectors and we need to show that they form an author. No basis for our three. Okay so for you to do that let's first calculate. Check that the set. Um it's cool um B formed by the The 2 3 years or so. So that means that the inner product of each of the combinations of vectors of the set is equal to zero. So we want would be to Is equal to zero. The one with the three And B two B 3. So we need to check this. So indeed you will see that everyone would be too is it goes to 1/25 times minus 12 plus 12 which is imposed to zero. We won with the three. Well this is clearly zero because the we have zero zero during the Fierce two components and then the one we have during the third component. So I think there probably will be zero And B two. B 3 is the same. This game is going to be opposed to zero. So from this we have that. The this is V is smartphone. Oh, so B is so. And how to check if this is a basis for our three? Well, we need to check that disk resets. We won't be B one, B 2, B three are linearly independent. But that is clear because by the serum 6.3.1, if a set uh If any set is an orthogonal, said that that implies that the sector V is linearly independent immediately. And we have three linearly independent vectors that will span the so this pan Of three linearly independent vectors in R. three will the whole R three. So B two, B 3 Will give us our three. So that satisfied the two conditions for a basis that means that the set B is uh huh Or a seven hour Basis for our three. Then we need to extend this two or so normal basis and for that we need to see what happened with the lengths or the norms of each of the factors. So The Normal B one, it's going to be the square root 25 over 25 which is equal to forbid to this equal To again the square root of 25/25, which is the cost to one. And finally, the norm of the three that has only one element of physicals to once it. Is this what? So these three factors are unitary. That means that actually the set V is not only an orthogonal basis but actually the is right Ortho normal basis four hours three now. So we know that B in this case we have to be the set of anyone B two. B 3 is a basis for three. That means that we can Right, Director you. I think this case is two is equal to one -2 and two. So we know that we can represent any vector in this case use it particular Rectory three. So we can represent you as a new combination of the elements of the basis. But in particular this an Ortho normal basis. So we have in a specific way to represent this vector U. And it's given by the inner product of you would be one times the better if you want. Plus the inter both of you would me too B two Plus in Front of You. B three B three. And just to give you some geometric intuition, what happened with this formal education? Yeah, it's just the product of the of the victor you with each of the elements on the Euro vulnerable basis. But this is because basically what we're doing here is that the In this case V one is a unitary vector. So this represents on a projection. So let's suppose. And here we have our three. Okay, that extends here infinity. And here you have the vector for example, will be one over here. You have the two over there. You have these three form an Ortho normal basis. Okay, so the point is that here we have the best for you. And what we're doing is this In your product is calculating the projection of U. and v. one. So basically we're calculating how mhm. The amount of you in this direction. Okay then would be it is the same. So this is your problem will be the amount of you that we uh huh. Going into the direction of the two and the same or The three. So we are just projecting taking how many in each of these directions were. So you can observe that that will give you these victories. So it's just projecting on each of the basis elements of our. These are three. So that's just to give you some geometric inefficient of what's Harmony. Okay, so let's get this compute so in this case you be one is equal to -11/5. The approach of you would be too easy equals 2 -2, 5th. And finally the inner berg you would be three years they posted to. So that means the victor U. Is equal to minus 5th time's will actually is -11/25 times minus three For zero minus two. Fifth for 30 Unplug us too. 00

Hello there. So for this exercise We need to find two vectors X. And the Y. In our food. That on this inner product space where the product is defined as a weighted in their products. That we multiply three by the multiplication of the first components and two on the multiplication of the second components of the vectors. So these are different definition of inner product. And we need to find two vectors X and Y. That are are tha no on this, on using their product space. But these two vectors should not be or a seminal with the usual Euclidean in their problem. So here I'm going to the note the cleaning in their product by this subscript E here. So this defined the usual inner product cleaning in our product. And even more, we need to find two vectors. This should these vectors here should be unitary vectors. Okay, so that means that the norm of the specters R. to one. Okay, so basically we can find a family of solutions that satisfied these conditions. But we we we just need to find a pair of this. So to make it simpler, we know that uh we can fix the first vector so we can choose any vector X. And then we can find the second. So Let's fix x as the simplest case in this case. one and 1 In particular, this victory is not unitary. So that's why I'm not putting here the hat. But then at the end, we're going to normalize after obtaining a vector that is orthogonal to two X. And we're going to do Fine Director. Why? As the vector? Why? one way too. So we need to find some solutions for White one. So let's take the problem in the in our space. Okay. So the inner product of these two vectors, it's going to be equals three times why? One plus two times y two. And this should be equals to zero. So from here we can obtain a relation between the components of this vector. Why? So make it simple. I'm going to find a solution of Taiwan and there's going to be minus thirds white too. So this is we have a set of solutions as I mentioned. So let's pick some choice for Y two. In this case I don't like these fractions here, so I'm going to choose why two equals 23 And that implies that why one will be close to -2. So great. We have found a vector that is orthogonal to X. So that means that or vector why Is equal to -2? Three. So if we define why in this way we have that X. Is our ethanol to why? So so far we have an orthogonal said in or in their profit space, but we need to check that these two vectors are not orthogonal in the usual and cleaning with the usual ingredients in their products. So let's check that. So by taking the inner product of X with Y. That you cleaning in their product, this is equal two -2-plus 3, which clearly is equals to one different from zero. Therefore X. In this case X is not a phono to why? And the last think that we need to do is making these two vectors unitary. So that's the last step. So I'm going to write here again the vectors. So we have the vector X equals to one one and the vector why Equals 2 -2, three. So we need to make them normal unitary. So that means that x unitary is equals to take an X divided the norm of the factor and the norm of the vector X is defined as the square root of the inner product of the vector with itself. In this case this he goes to the square root of One was one. That is equal to the square root of two. I'm sorry, this Times three and here two times of one square square. So this is the square root of five. I am going to rewrite the definition of the inner product you want B one us youtube two, two, three. And so To normalize this vector, we just need to divide it by one Over the square root of five. And that is this vector. And then we need to repeat the procedure for why? So we have that Y unitary is going to be the vector divided the north of the picture. In this case the norm is defined in the same way the square root of the inner product of why we didn't sell, and in this case this is equal to the square root oh three times for square loss, two times three quick. So this is just the the square root of 30. So we obtained this 2nd january orthogonal vector in this, In our product space, that is one over the square root of 30 minus two, three.

Hello there. So for this exercise we start with a subspace that is respond by these three vectors. We want B. two and b. three. Okay. And we need to find an Ortho normal basis for the space. So the first step is to, well the point is that we know a procedure that takes a set factors. Yeah two mm. And then it returns after playing the punishment procedure. It returns a set of orphan yeah sets the Northern Normal said. In terms also. So you start with a set of pictures and you obtain an Ortho normal set. But to apply this procedure we need to know that these vectors are linearly. Mhm. So before starting to calculate and or applying the garnishment procedure for this set of B one, B two and B. Three. What we need to do is let's remind if these vectors are Yeah. And to do that we need to check if one of them is reading as a combination of the other one. And you can observe that actually The Rector B three Is equal to be one. Yes. Beach. So you can observe that this just by inspection if you want a procedure to check this. So process or agree them to do that. What what I recommend to do is put in these factors in a matrix form. That means putting the vectors as Rolls of our matrix. That 012 -101 -113. Okay. And here then you try to reduce this matrix to the action form by applying the girls procedure and here you will observe that when you want to eliminate this. Mhm Put a zero here or here you will obtain at the end you only two pilots one and hear something like the A. B. Here. C. 00 Yeah that means that one of your rose will become full of zeros. And that means that you is enough to pick you. Only two of the vectors that you're considering this in this sense, If you obtain three pilots after applying the girls elimination Using three pilots, then your three factors are linear. But in this case what happened is you obtain something of this form or you can check based by Inspection that the effect to be three is a linear combination of the. Okay, so first you that your mind that if the set is linear independent and we observed that it is not. So it's enough to eliminate one of the factors that you have here This case I'm going to be three and we're going to use we one and two. So for this victory people for this space is enough. two. Great. That is the span of the factor we want and beat and they said expand the subspace up and they are linear independence. So now we can apply the garnishment procedure to obtain a basis. That is all for now. Okay, so let's remember that for the grant smith procedure we are going to obtain, We're going to pick this had to be one B two. And after applying the Greenwich you're gift in a set of factors Of the one Alpha 2. Both unitary vectors. To do that. We need to Fix one of the vectors here, say alpha one here. I'm used the Yeah. For unitary factor. So this vector here is not unitary. So we fixed one of the batteries in our set Either be one of you to to follow the same order. I'm going to be one. So alpha one. It's going to be B one. That means 012 Is the 1st step. So in or set. In our final set, we have already director of what Then the next step is the second vector offer to Such that this Alpha two is a phony one. And to do that we need to take the vector V two and transform Director of Alpha two Such that this vector is also an out for one. To do that we need to right offer to us the vector V two minus the projection of YouTube on all. For one When we do this we are obstructing the projection. We're eliminating the component of YouTube that is aligned with alpha one and the result will be and Better. That is Arthur 12 for one. So in this case This becomes the Alpha two is equal, two minus one, one minus. And here the projection just to remind you that the formula for the projection of a vector into another one is equal to you. The inner product of these two vectors times that picture to the one that we're projecting, two divided the norm of this square, enormous. So this projection particular case is equal to -250 times 012 And this part you can observe that corresponds to awful. So after subtracting these two pictures, We obtained that Alpha two is equal two minus one minus 2/5 And 1/15. Okay, the second vector orthogonal set. So what we have so far is a set of orthogonal factors Okay over one Equal to 0 1, two. Alfa two equal to -1 -2 and one over deep. But what we need is an Ortho normal set. So so far this is or a phone but we need say that should be normal. That means that the factors are unitary. And to do that we need just to normalize the vectors. That means that alpha one, you need to hurry Will be all for one divided the normal album. This is just One over the square root of 501 two. And for alpha tube we applied the same formula. So we normalize the vector by divider but it's not and we obtain one over The square root of 30 times minus five minus 21 These two vectors for or phone normals. Ah these vectors, so all for one, All for two, the span of these vectors, he is a subspace of you even more. We know that by theorem 6.3.1 Ortho normal set is linear. And then so that means that these two vectors are linearly independent. Therefore we call director. We will be Formed by Alpha one oh two. Is a basis is an is an orphan, success is an old on our mountain basis for done

Hello. Hope you're doing well. So we have our two vectors here. We need to find a unit vector. That's orthogonal to both of these factors. So to do that, the first thing we're gonna do is we're gonna find the cross product of U cross V. So I'm just gonna give a refresher on how to do that. So, for take the cross product, you're going to create a three by three matrix. The first road is going to be your unit vectors. I, j and K second row is going to be the components of your first vector. So in this case, that's you X u Y music. The third row is going to be the components of your second factor, which is V X Y v c. So to find the cross product, you're going to take your first vector I and multiply it by the determinant of this matrix. Then you're going to subtract Jay time to the determinant of this matrix here. Then you're gonna add K times the determinant of this matrix here and then once you do that, you will simplify it and you'll end up with a vector. That's the result of this cross product. So I mentioned that you'll have to find the determinant of some matrices. So to find the determinant of a two by two matrix, let's stay with Elements A, B, C and D. It's going to be called eight times D minus B times C, so you can use that and you'll find your cross product. So once you get your cross product of U cross V, you'll get a vector that's orthogonal to U and V. It won't necessarily be unit factor, so you have to convert it into your ineffective. So to do that, you're gonna find the length of your vector. That's the result of the cross products. That's just gonna be the square root, the sum of the squares of each of your vector competitive that will get you really like. Then you want to divide your vector. That's the vector that you get from you. Cross. If you want to divide it by its length, that will result in a unit factor that's also orthogonal to your vectors. U and V. So we're gonna go and do that then. So we're gonna start by taking the cross product of you cross be so to do that, we're going to do the create our three by three matrix with first row the unit vectors i, J and K Second row has the components of our you vector, which is going to be two minus 13 The third row is going to be the components of a vector which is zero for 10 minus two. So Thio so then, to continue solving this, we've got I times the determinant this matrix here which is minus 130 minus two when we have these vertical bars on either side of the Matrix, that means that we're taking the determinant of the matrix and we're going to subtract a unit vector J times the determinant of this matrix. Here it's 231 minus two. We're gonna add the K unit vector times the determinant of this matrix here. So it's two minus 110 Okay, so remember from before we said that determinant determinant of the two by two matrix is going to be eight times d minus B times c. So keeping that in mind, we've got I times eight times he is minus three times minus one, which is to minus B times C. It's three times zero. That's just zero. We got minus J times eight times D is two times minus two, which is minus four minus a few times. See, just three times one that's three plus k times eight times d is two times zero, which is zero Uh, sorry, my three times C, which is one times negative one this negative one because simplifying this to minus year, that's to I minus four times minus students minus seven. Got a minus. Sign here. So that's seven J and then zero minus minus one. That's plus one is plus K. So writing that in bracketed for me and up to 71 So this factor here is orthogonal to U and V. But it's not a unit factor going to turn it into Unit Vector. So do that. We want to find the length of this unit vectors. So to do that, we're gonna take the square root, the squares of our components, the sum of the squares of our components. It's gonna be two squared. Well, seven squared plus one squared is equal to the square of four plus 49 plus one. It's equal to the square of 54. So now we want to divide this vector here by its length, which is the square to 54 to make it into a unit vector with the length of what? Some things that we end up with one over square to 54 times the vector 271 And just to simplify this a little bit, we're going thio, make sure we don't have a square root on the denominator. We're going to multiply the numerator and denominator by square to 54. So when we do that, we end up with square to 54/54 times the unit vector to or not the unit vector, but times the Vector 271 And this is our final answer right here. This is a vector. That's a unit vector. And it's orthogonal vector's you and all right. Well, thanks. And I hope that halt


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