let-mathbfyleftbeginarrayl2-6endarrayright-and-mathbfuleftbeginarrayl7-1endarrayright-write-mathbfy-

Name: let mathbfyleftbeginarrayl2 6endarrayright and mathbfuleftbeginarrayl7 1endarrayright write mathbfy
Uploaded: 2019-08-20
Duration: 2 min
Channel: Numerade
Description: Let $\mathbf{y}=\left[\begin{array}{l}{2} \\ {6}\end{array}\right]$ and $\mathbf{u}=\left[\begin{array}{l}{7} \\ {1}\end{array}\right] .$ Write $\mathbf{y}$ as the sum of a vector in $\operatorname{Span}\{\mathbf{u}\}$ and a vector orthogonal to $\mathbf{u} .$

Question

Let $\mathbf{y}=\left[\begin{array}{l}{2} \ {6}\end{array}ight]$ and $\mathbf{u}=\left[\begin{array}{l}{7} \ {1}\end{array}ight] .$ Write $\mathbf{y}$ as the sum of a vector in $\operatorname{Span}\{\mathbf{u}\}$ and a vector orthogonal to $\mathbf{u} .$

Ashley Boni · Accepted Answer

so first to find the part that is in the span of you, we want to take the orthogonal projection of why onto you. So why hat is our projection And we know this is equal to white out you over you dot you times you. So why don&#x27;t you is 14 plus six? Uh, you dot you is seven times seven is 49 plus one and we want to multiply this coefficient into the vector or seven. So those coefficient comes out to be 20 over 50 which is 2/5 to fist times seven hiss, 14 over five, and then we get to over five on the bottom. So this is the first part. So now when you define, uh, the other part, that is orthogonal. So we take why minus y hat our Why waas to six union. We subtract our 14 5th in 2/5 to get negative 4/5 and 28 5th So therefore we can, right? Why? As a song of 14 over 52 over five, plus negative for over five. 25

Foster Wisusik · Answer

Okay, This question wants us to write the vector. Why, as some of vectors perpendicular and parallel to you, So we&#x27;ll start by finding the parallel component. And that&#x27;s the same thing. Is asking What&#x27;s the component of why along you which we get straight from the projection formula. So we&#x27;re projecting Why on to you? So that would be Why don&#x27;t you divided by the magnitude of you squared distributed into you and the dot product would be two times for minus 21. So eight minus 21 and eight minus 21 is just negative 13. So that&#x27;s negative. 13 The magnitude of you squared would be 16 plus 49 giving 65 and then our vector you is just for negative seven And this vector is the same thing as negative 1/5 times four negative seven. When we reduce, so are parallel Vector would just be negative. 4/5 and 7/5. And now, since we know that why is formed of parallel and perpendicular components, all we have to dio to find the perpendicular component is subtract this parallel component from the original factor. So our original vector Y is 23 and are parallel Vector is negative. 4. 57 50. So why perpendicular would be tu minus 4/5 giving us 14 5th and then three minus 7/5 which would be 8/5. And that is our perpendicular vector. So we have our two answers right here.

Shaza Hammoud · Answer

And that&#x27;s a question given the following two vectors, and we have to find the victor projection off you onto V and then write you a song off two or three of Victor&#x27;s one off, which is the projection off you on Tau V. So you should be written as, or the composer&#x27;s tow you one plus you to where these two vectors are orthogonal. You want on you too and you want, since it says that one of them should be the projection off you on Tau V. So I&#x27;ll take you want as this projection you on TV and the formula off the production off you on TV and given a right here, which is deducted product over the magnitude off V squared multiplied by the component for off the victor V. So first, let&#x27;s find the Notre product off U and V, which is equal to the first component off you three. Multiply by the first component of the minus two plus the second component off you minus seven, multiplied by second component of the minus six is equal to minus six in US 42 as mine seven multiplied by minus six is 42 on dhe. This is equals toe 30 six. Now let&#x27;s find the magnitude of a vector V. The magnitude off vector V is the square root off the first component squared, which is minus two squared plus minus six squared. So this is four plus 3600 square root. In other words, it&#x27;s 40 100 square roots. Now let&#x27;s substitute the values inside the projection youand V D formula, which is the daughter product. You got a product off you and Fi as given by 36 the magnitude off we as square root off 40. And it&#x27;s also so the part off too so did pour off to will be cancelled with the square root multiplied by the multiplied by the component for Victor V, which is minus two minus six. So this is is by multiplying thesis Keller number 3 36 over 40. Multiplied by minds to and minus six, this becomes minus 1.8 minus 5.4. So this is the component from off you one. Now you, too from this far from this equation you as you want plus you to than you two. It was tow you minus you want you as three minus seven minus You won, which is minus 1.8 minus 5.4. So this is a three plus 1.8, which is 4.8 then minus seven minus minus 5.4. Just minus seven. Plus 5.4 on This is minus minus 1.6. So this is you too, Andi, this is you want. Now the final think is to just write you in terms off you want and you two just substitute devalues off you and you two inside the equation you want is minus 1.8 minus 5.4 plus 4.8 minus 1.6. So this is D answer O r the d a. You as a some off two orthogonal vector&#x27;s you you want. And you too on this is di Father solution off this question

Dr. Michael Lewchuk · Answer

So the question 25 we want to find the projection of you want to ve given these coordinates. And so we sold for you times the first negative eight times Negative six is 48. And from to that we add three times negative two, which is negative. Six positive 48 negative six is 42 and then the magnitude of the squared is 36 plus four is 40 and we multiply that by our value v negative six negative too. And then we come up with a V with a you one value of negative 6.3 comma. Negative 2.1 Now to thought for you to we seem to take u minus you one. So you two is equal to our your value. Negative A comma three and we subtract from that negative 6.3 and two point negative 2.1 and we end up with a U two of negative 1.7 five point one and worth

Ashley Boni · Answer

So you want to write the vector? Why? As a sum, um, a factor And w and a vector orthogonal to w or W is a span off the to use. I was So to do this when you look at, uh, the earth orginal projection of why onto this band of you wanted you two to get the vectors in W s o. The formula for that is a vector. No more protection. Why hat? We project onto one. It&#x27;s a top product and we add a projection onto you too. So first, let&#x27;s find this vector. Eso Let&#x27;s find over dot products you have. Why dot you won Does he call to negative one plus four plus three. So we get six. Uh, you want a product with itself is just gonna be one plus one plus one, 43 So those first coefficient six over three, which gives us two, uh, for the next one. Why dot product with you too, is one plus 12 minus six, which gives us 30 minus six or seven. So you two dot product with itself one plus nine plus four. That gives us 14. It&#x27;s our second coefficient is seven over 14 or one. So let&#x27;s go ahead and multiplies coefficients in. So to find you one gives us 2 to 2 and 1/2 multiplied by a factor. You too gives us minus 1/2 three halfs and get up on so adding use together to minus 1/2 iss Very halfs. Four halfs and three half&#x27;s gives us seven halfs and two minus one is So this is director. Why hat and so to find her other vector, we just need us attract. So we&#x27;re going to take, uh, why Ann&#x27;s a track. Why hat? So we get what we need to add. So why I had to get back to the vector y. So why minus why hat? So we have negative one minus the halfs. Four minus and a halfs and three minus one, which gives us negative five house more minus 7/2 sis 1/2 and three minus Want us to so we can write Why, as the son of this vector and why had

Jose Hannan · Answer

Hello, everybody. In this video, I&#x27;ll be showing you how to solve exercise 44 in chapter 13 Section four of calculus Early. Transcendental. No, this problem. They want us to find a vector which I&#x27;m the note here by V, which is orthogonal to both of the vectors with components 80 and four as well as negative 8 to 1. Now to solve this problem or they really want us to do is find a cross product of these two vectors as by definition across product between any two vectors is orthogonal to both of the original two vectors. So let&#x27;s compute this cross product. The vector V is equal the vector eight zero four cross with I think you ate 21 in this Elvis cross product will be using the determinant method from here on 13 points A and this method we find the determinant of a three by three matrix whose top row as a three standard unit vectors i, J and K, your second row as the components of the first vector and his last through. That&#x27;s the components of the secondary in the cross product. Now let&#x27;s break down the determinant of this matrix. It&#x27;s determine it is equal toe I times the determinant of the two by two sub matrix formed by this bottom rate corner here which is zero for to in one detracted. From that we have Jay times, it&#x27;s sub matrix, which is he&#x27;s made by these two sub columns on the left and right here we have eight or negative eight and one times J. Added to that, we have a similar quantity for K, which is K times the determinant of the sub matrix and this bottom left here zero negative eight and to So now let&#x27;s compute the determinants of these 32 but two major sees in the first term we have zero times one which is zero minus two times for which is eight. We have minus eight times I subtracted from that we have eight times one which is a minus negative eight times four, which is negative. 32. So we add 32 times J and then added to this we have eight times two, which is 16 minus negative. Eight time zero, which is zero. This just gives us 18 times k. And now let&#x27;s just simplify the middle term and parentheses. We have minus AI minus eight plus 32 which is 40. We have minus 40 times J and to that we add 16 K This cross product is just equal to the vector with components negative. Eight, they get a 40 in 16. So we have here a vector that is orthogonal to both of the original vectors given to us that the lesson to know is that any scaler multiple of the spectre is also a valid answer. As it is, any skill in multiple is parallel to this factor and remains orthogonal to both of the vectors given, However, for the purposes of this problem, finding the cross product of the original two vectors is the easiest way to arrive. A vector that answers the question. So that is how you solve problems. 44

Problem

Let $\mathbf{y}=\left[\begin{array}{l}{3} \\ {1}\…

Problem 14 Medium Difficulty

Let $\mathbf{y}=\left[\begin{array}{l}{2} \\ {6}\end{array}\right]$ and $\mathbf{u}=\left[\begin{array}{l}{7} \\ {1}\end{array}\right] .$ Write $\mathbf{y}$ as the sum of a vector in $\operatorname{Span}\{\mathbf{u}\}$ and a vector orthogonal to $\mathbf{u} .$

Answer

$\left[\begin{array}{c}{14 / 5} \\ {2 / 5}\end{array}\right]+\left[\begin{array}{c}{-4 / 5} \\ {28 / 5}\end{array}\right]$

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Linear Algebra and Its Applications

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$\left[\begin{array}{c}{14 / 5} \\ {2 / 5}\end{array}\right]+\left[\begin{array}{c}{-4 / 5} \\ {28 / 5}\end{array}\right]$

Linear Algebra and Its Applications

Lily A.

Catherine R.

Kayleah T.

Joseph L.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Catherine R.

Kayleah T.

Joseph L.

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

Linear Algebra and Its Applications