compute-the-quantities-in-exercises-1-8-using-the-vectors-mathbfuleftbeginarrayr-1-2endarrayright--5

Question

Compute the quantities in Exercises $1-8$ using the vectors
$$
\mathbf{u}=\left[\begin{array}{r}{-1} \ {2}\end{array}ight], \quad \mathbf{v}=\left[\begin{array}{l}{4} \ {6}\end{array}ight], \quad \mathbf{w}=\left[\begin{array}{r}{3} \ {-1} \ {-5}\end{array}ight], \quad \mathbf{x}=\left[\begin{array}{r}{6} \ {-2} \ {3}\end{array}ight]
$$
$\left(\frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}}ight) \mathbf{v}$

Ryan Sander · Accepted Answer

Hello. We&#x27;re going to continue working with dot products, so the quantity of interest this time is going to be given by the dot product You times you dot p divided by V v times the vector V and we&#x27;ll just give the definitions. Here we have you is equal to it. Negative 12 and V is equal to 46 These are both two d vectors. And so you might be asking, What is this quantity you dot divided by V v times via. So this first term in the Prentice&#x27;s is ah, ratio of dot products. This will be a scaler divided by a scaler which gives you a scaler. And the second quantity here. This V is a vector, and so will we end up with is scaler times a vector which gives us a vector so we can expect a factor in our final result on So now let&#x27;s actually go through and derive this. So we have UW divided by V dive. Whoever multiply that by V. So, um, this is gonna be equal to you dot v is going to be negative one to ah, well, read it alphabetically. First, If we remember the DOT product is, uh, you don&#x27;t b is equal toe u transpose v and feed obviously goingto be transposed fee. That was V. And then this is equal to negative one to transpose times 46 and then be transposed. V is going to be equal to 46 transpose for six and then times 46 of the output. So now we can evaluate these. So this will be a negative 12 and then for 46 and then 46 times 46 And now that the dimensions airline, we can apply the that product so we&#x27;ll have this is equal to negative. One times four plus two times six and the numerator and then on the bottom will have four squared four times four a six squared six top six. And this will give us, um oh, and then 46 here on the up. And this will give us, um, negative floor plus 12/16 plus 36. I&#x27;m 46 of the output and then this gives us eight divided by 52 times the vector for six, which gives us to over 13 times for six or 8/13 and then 12/13. And this is our final output, which, as we said before, is a vector. Thank you.

Ryan Sander · Answer

Hello. We&#x27;re going to continue working with the vectors defined as U and V, which are vectors, and Judy. So we have you is equal toe negative 12 visas and go to for six and W is equal to three negative 15 negative. Eso these two vectors air in two D and this one&#x27;s in three D in the Thea promised we&#x27;re going to compute in this case is one over w dot w times w And this is an interesting one because this w w quantity is actually w squared are the magnitude of w squared. So this is equivalent to W over w squared. And so Ah, this w dot w quantity on the bottom Here, this is going to be a sense. It&#x27;s a dot product or inner product. It&#x27;ll be a scaler, right? That here. Yeah, but w as we have a periods of vector and so we&#x27;re effectively end up with Over here is a vector divided by a scaler, which is going to be a vector, so we can expect a factor in our final results. And now if we compute the dot product of W W, that&#x27;s going to be equal to W transposed W, which is three negative one negative. Five transpose three. Sorry. You sure of that? Three Negative. One negative five. Then we get, uh, this column vector turned sideways to become a row vector after its trance post. And then three negative one negative five. And what this amounts to is three squared plots negative one squared, plus negative five squared, which is equal to nine plus one plus 25 which is equal to 35. So this is w dot w. And now we just, uh, divide our vector. W divided by this magnitude. Squared. So then just bringing this over here, we have w over W. I will use the what they ask for w W. And this is equal to three negative one negative five. Divided by 35 which just gives us a vector. 3/35. Negative one divided by 35 negative five. Divided by 35. And this is our answer. Thank you.

Ashley Boni · Answer

So we&#x27;re giving the vector you and we want to compute. Um, the following. Ah. So we first when you find you, don&#x27;t you? So you got you. We just take the corresponding first entries. Multiply on, and I have the corresponding second entries. So we have negative one times. Negative one, just one plus two times two is four. So that gives us five. So this quandary on the outside here is just, uh, won over five. So we want to take 1/5 of the factory. So when we do that, we just divide each entry by five. So we get negative. 1/5 and 2/5. Not sorry, sir.

Ashley Boni · Answer

So we&#x27;re giving the vectors x and W And we want to puke this quantity here. So starting off, we&#x27;re just gonna take x dot W So we have six times three is 18 two times one is two and negative. Three earth are negative. Five times three is minus 15 80 miles. 15 is three plus two is five s. So now we want to compute x dot Exe. So six squared is 36 plus negative. Two squared is four plus three squared is nine. So we have 36 plus, uh, 13 which gives us 49. So ah, ex dot w over x dot Exe is equal to five over 49 and we want to multiply the vector X. So we have five over 49 times six negative, 23 So we just multiply this coefficient into each, uh into each number. So six times five is 30 over 49 negative. Two times five is negative. 10 over 49 and three times five is 15 over 49. That&#x27;s our final vectors.

Ryan Sander · Answer

Hello. In this problem, we want to compute the magnitude of W where w is a three dimensional vector given by three negative 15 on these arts three components Magnitude of W is going to be equal to, um, the square root of each of its components squared, which will be three squared plus negative, one squared plus five squared. And, uh, this is going to be equal to nine plus one plus 25. And this is equal to the square root of 35. And this is our magnitude of the vector W thank you.

Ashley Boni · Answer

So we&#x27;re giving the Vector X and we want to compute the norm of X. So if you remember, the definition of a norm is the square root of the dot product of the vector with itself. So first we want to take x dot exe and square rooted. So we have the square root of six squared, which is 36 plus negative two squared, which is four plus three squared, which is nine. So if we add these things up, 36 plus 13 gives us square root of 49 which is seven.

Problem

Compute the quantities in Exercises $1-8$ using t…

Problem 5 Easy Difficulty

Answer

$\left[\begin{array}{c}{\frac{8}{13}} \\ {\frac{12}{13}}\end{array}\right]$

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

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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$\left[\begin{array}{c}{\frac{8}{13}} \\ {\frac{12}{13}}\end{array}\right]$

Linear Algebra and Its Applications

Lily A.

Anna Marie V.

Caleb E.

Samuel H.

Vectors Intro

Vector Basics - Example 1

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

Lily A.

Anna Marie V.

Caleb E.

Samuel H.

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

Linear Algebra and Its Applications