find-the-distance-between-mathbfuleftbeginarrayr0-5-2endarrayright-and-mathbfzleftbeginarrayr-4-1-8e

Question

Find the distance between $\mathbf{u}=\left[\begin{array}{r}{0} \ {-5} \ {2}\end{array}ight]$ and $\mathbf{z}=\left[\begin{array}{r}{-4} \ {-1} \ {8}\end{array}ight]$

Ashley Boni · Accepted Answer

So we want to find the distance between these two vectors. So in order to do that, we take the norm of their difference. So if we want to subtract thes two vectors, we subtract on them component wise. So we have zero minus a negative, for which is for negative five minus negative. One is negative for on and two minus eight is negative. Six. So the normal this vector is equal to the square root of four squared plus negative four squared plus negative. Six squared. So we get 16 full 16 close 36 which gives us 32 plus 36 which gives us 68. And we can simplify this a little bit. Ah, 68 breaks down into four times 17 so have yet to square it&#x27;s of 70.

Ankit Gupta · Answer

the value of the formula that distance between two points, which are excellent by one and x two by two, can be calculated by the formula Miti&#x27;s, which is X two minus x one whole square plus y tu minus y Ivan whole square. We will Denard this distance by the so in our kitchen. We are given the two points with their eight minus two I minus six minus four. We will use the above formula to calculate the distance between these two points. Therefore, we get minus six minus eat foursquare plus minus four plus two whole square minus six minus. It becomes minus 14. Who&#x27;s well placed minus rule square square off 14 is 1 96 plus square off minus two equals four. We get route over off 200 which is 14.1 poor. Therefore, the distance between two points which are given to us is 14.1 poor

Ankit Gupta · Answer

According to your own question, we have two points for common to and minus took among minus four. So the stands between these two points can wilderness route under four minus minus plus to hold squared plus two minus minus. Plus four holds great, but this is coming in six square plus six square. That means we can say that this is 36 plus 36 which is a Rule 72 and this week Undated six Route two units, So distances six row two units.

Amy Jiang · Answer

Okay, so we want to find a distance of these two points, so that&#x27;s going to be a difference of are two X points squared times the difference. We&#x27;re plus the difference of R T Y points. So this year is a negative for squared. This year&#x27;s at one squared, so we get the square root of 16 plus one, which is 17.

Matt Gibson · Answer

so I&#x27;m giving these points and I know the formula. So I just substitute in. And the distance is the square root of the second X minus, the first X squared, plus the first of the second. Why minus the first way. Just be careful with your negatives here, so my distance is the square root of negative two. Plus, this is the same as just having addition. That&#x27;s going to be positive. Three squared. This is the same is addition, also says, going to be plus negative two squared. But that always winds up being positive because you square it anyway. So my distance is the square root of nine plus four. My distance is the square root of 13. There is no factor. That&#x27;s a perfect square in 13 so that would be my answer.

Ashley Volpe · Answer

Okay, so we&#x27;re gonna do the distance, and that&#x27;s gonna be the square root. Now, we&#x27;re gonna add our X coordinates together first, so that&#x27;s gonna be one plus a negative four. So that becomes one minus four square. Then we&#x27;re gonna add a while coordinates together. And so that&#x27;s a, uh, plus a negative one that becomes five minus one. So do the subtraction There. One minus four is negative. Three squared and five minus one Makes that four squared. That becomes a square root of nine plus 16 which is the square root of 25. And since 25 has a square root, we can take it to get a final distance of our

Problem

Determine which pairs of vectors in Exercises $15…

Problem 14 Easy Difficulty

Find the distance between $\mathbf{u}=\left[\begin{array}{r}{0} \\ {-5} \\ {2}\end{array}\right]$ and $\mathbf{z}=\left[\begin{array}{r}{-4} \\ {-1} \\ {8}\end{array}\right]$

Answer

2$\sqrt{17}$

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

Linear Algebra and Its Applications

Lily A.

Catherine R.

Kayleah T.

Kristen K.

Vectors Intro

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2$\sqrt{17}$

Linear Algebra and Its Applications

Lily A.

Catherine R.

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Kristen K.

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.

Kristen K.

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

Linear Algebra and Its Applications