show-that-the-orthogonal-projection-of-a-vector-y-onto-a-line-l-through-the-origin-in-mathbbr2-does-

Question

Show that the orthogonal projection of a vector $y$ onto a line $L$ through the origin in $\mathbb{R}^{2}$ does not depend on the choice of the nonzero $\mathbf{u}$ in $L$ used in the formula for $\hat{\mathbf{y}} .$ To do this, suppose $\mathbf{y}$ and $\mathbf{u}$ are given and $\hat{\mathbf{y}}$ has been computed by formula $(2)$ in this section. Replace $\mathbf{u}$ in that formula by $c \mathbf{u},$ where $c$ is an unspecified nonzero scalar. Show that the new formula gives the same $\hat{\mathbf{y}} .$

Foster Wisusik · Accepted Answer

Okay, This question wants us to prove that if we&#x27;re projecting a vector onto a line, the answer is independent of what vector along the line we choose. So we&#x27;ll say that our line l is equal to the span of some vector, which we&#x27;re gonna call you. And we&#x27;re also going to say that why is the factor that will be projecting? So if this was our Y vector and this was our l vector, this red line right here would be the projection, and we can actually find a formula for the projection. So the projection on tha does vector you of why is equal to why dotted with you all over the magnitude of you squared times you. And now we&#x27;re going to show that no matter what multiple of you we pick that spans the line, we will get the same answer. So the projection onto you of why would be equal to why dotted with see you over the magnitude of see you quantity squared times See you. Because again, we&#x27;re just taking a scaler multiple so we can account for any vector that we could possibly shoes. So I&#x27;m gonna pull a C squared out in front. So we get C squared times. Why dotted with you over Well, the magnitude of see you quantity squared which will deal with in a minute times you and what&#x27;s the magnitude of the vector? See you. Well, the magnitude of see you would just be our original magnitude scaled by sea. So by a similar argument, we just have to square, See if we&#x27;re squaring the magnitude of see you. And now we see that the C Square&#x27;s cancel giving us why dotted with you over the magnitude of you squared times you. And that&#x27;s exactly the projection onto you of Why? So we&#x27;re done.

Ashley Boni · Answer

So we&#x27;re supposing that vector y is orthogonal to both you and V So we know that. Why don&#x27;t you and why Don&#x27;t be or equal zero Nah, just by definition. And we want to show that why is our abominable toe every w in the span of human? Be so we have w in the span of you and B um, using the hit we can write this w has see once you plus C to me for some scaler So you want to see to So we want to show that why is also orthogonal to w So you want to show that why stocked up use equal to zero So let&#x27;s try to compete wide out w So we can take why dot see you on you plus c to d So this is equal Thio why dot see you want you plus why dot c to d and by the properties of um the dot product we can pull scale er&#x27;s out So we have C one at times Why don&#x27;t you plus c two times wide up b and we know that Why don&#x27;t you and wideout veer both zero? So we have zero plus zero which gives us here Ko So we shown that why and w oh, our fourth optimum

Ankit Gupta · Answer

So there is you is or toller toe B and W Then you is alternative to see the best GW we&#x27;re seeing. They are skater&#x27;s. It&#x27;s mostly use alternate to read. That implies you dot Is it going to Zito? And you don&#x27;t zero as they are both our terminals. Now we have the proof that you don&#x27;t see the plus GW is equal to zero. Now we will be taking a left turn side that is Jude or C V plus DW. This implies that you door CV plus you dot dw Now we know that CMDR scale er&#x27;s they will get outside and in the multiplication there will be C dog you taught me. Sorry. Similarly, David, go outside and and the world reply there will be issued or w Now we know that it&#x27;s you got v zero and you don&#x27;t wr zero. So we replace this by zero here and a zero here that isn&#x27;t going to Tzeitel. So we just proved there there did you dot cv plus DW is equal to zero. It implies that they are our total

Amy Jiang · Answer

okay Were asked to show that you crossed me. Is orthogonal a two you got me and you Plus we with you get starting with our right hand side We have you Not be time few plus B Bless you. It&#x27;s multiply for duty dot product or both Because we know there are a dog knows that we have you Don&#x27;t be tense. You could speak because you don&#x27;t with you cross week, okay? Did not people too. You don&#x27;t be times the got a b plus you Plus no whips write music. And it&#x27;s actually yeah, really Read this that you don&#x27;t be times you take me across you Plus you got B many times he crossed you. Plus B. Thank you. Okay. And we know from the period Ah, problem that this is actually equal to go. This is also able to go. So we have you not be times you plus B puts you, Negro. You because it you know. So we&#x27;ve shown that this is or they are vital to this

Samantha Lincroft · Answer

So if this problem were asked to show that you cross b is oh, if agonal to you plus fee and you minus C on. I like to think about this geometrically. So we have here to back dues you and the vector V and we know that by definition, the cross product you cross v is orthe agonal to you and is orthe agonal to be. And so because of that, we know that it&#x27;s orthe agonal to the plane containing U and V. So that means that as long as you plus V and you mind this fear in that plane, then we&#x27;re good and it is with agonal. And in fact, since we can define that plane as all of the points that can be reached by some linear combination of U and V, it must be orthogonal till that point. So the way that we can kind of think about this right is this factor here, for example, would be you plus B. And over here we would have you minus B. But all of these will still be in the same plane

Donald Albin · Answer

in this problem. We have an Ortho agonal set of vectors, and we&#x27;re told that these vectors are all orthogonal. And that&#x27;s how we know that there are orthogonal, um, to make them Ortho normal. We have to also make each of the vectors normal. And so, um, so if we take any of the vectors v i and we divide it by the length of the I, well, that will have the same direction as V I, but it will be, um, normal. And so that&#x27;s four. Oh, I from one tu que So now our normal set would just be Ortho normal set because they&#x27;re already orthogonal. And to make them normal, we just have to do that divide by their length. Um, then it&#x27;s just you won. You, too. Dot dot dot. You okay?

Problem

Let $\left\{\mathbf{v}_{1}, \mathbf{v}_{2}\right\…

Problem 31 Easy Difficulty

Answer

So \hat{y does not depend on the choice of nonzero u in L used in the formula. }

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

Linear Algebra and Its Applications

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Samuel H.

Michael J.

Vectors Intro

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So \hat{y does not depend on the choice of nonzero u in L used in the formula. }

Linear Algebra and Its Applications

Catherine R.

Heather Z.

Samuel H.

Michael J.

Vectors Intro

Vector Basics - Example 1

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

Catherine R.

Heather Z.

Samuel H.

Michael J.

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

Linear Algebra and Its Applications