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Chapter 6: Problem 21

Find coordinates for five different vectors \(\mathbf{u},\) each of which has magnitude \(5 .\)

Short Answer

Expert verified
The five vectors with a magnitude of 5 are: 1. \(\mathbf{u_1}=(3,4)\) 2. \(\mathbf{u_2}=(-3,4)\) 3. \(\mathbf{u_3}=\left(\frac{5}{\sqrt{21}}, \frac{10}{\sqrt{21}}, \frac{20}{\sqrt{21}}\right)\) 4. \(\mathbf{u_4}=\left(\frac{5}{\sqrt{21}}, -\frac{10}{\sqrt{21}}, -\frac{20}{\sqrt{21}}\right)\) 5. \(\mathbf{u_5}=\left(\frac{5}{\sqrt{21}}, -\frac{10}{\sqrt{21}}, \frac{20}{\sqrt{21}}\right)\)

Step by step solution

01

Understand the problem

We are given magnitude 5, and our task is to find the coordinates for five distinct vectors. Since there's no constraint on the coordinates, we can choose them freely as long as they satisfy the magnitude requirement.
02

Calculate the first vector

Let's start with a 2-dimensional vector. We can choose any two coordinates, say \(x=3\) and \(y=4\), and check if their magnitude equals 5. The formula to calculate the magnitude of a 2D vector is \(\sqrt{x^2 + y^2}\). So, for the given coordinate points, we have magnitude = \(\sqrt{3^2 + 4^2} = \sqrt{9+16} = \sqrt{25} = 5\). Thus, the first vector, \(\mathbf{u_1}=(3,4)\), has the required magnitude of 5.
03

Calculate the second vector

To find the second vector, we can use a slightly different approach. If we already have a vector of the required magnitude, we can easily obtain another one by negating some of its components. For example, by negating the x-component of \(\mathbf{u_1}\), we get another vector of equal magnitude but pointing in a different direction. So, the second vector, \(\mathbf{u_2}=(-3, 4)\), has the required magnitude of 5.
04

Calculate the third vector

Let's now switch to a 3-dimensional vector. We can choose any three coordinates as long as their magnitude equals 5. For example, let's choose \(x=1, y=2, z=4\). The formula to calculate the magnitude of a 3D vector is \(\sqrt{x^2 + y^2 + z^2}\). In this case, we have magnitude = \(\sqrt{1^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}\). To get a vector of magnitude 5, we can divide each component by \(\sqrt{21}\) and then multiply by the required magnitude. So, the third vector, \(\mathbf{u_3}=\left(\frac{5}{\sqrt{21}}, \frac{10}{\sqrt{21}}, \frac{20}{\sqrt{21}}\right)\), has the required magnitude of 5.
05

Calculate the fourth vector

For the fourth vector, we can simply negate two components of our third vector \(\mathbf{u_3}\). So, the fourth vector, \(\mathbf{u_4}=\left(\frac{5}{\sqrt{21}}, -\frac{10}{\sqrt{21}}, -\frac{20}{\sqrt{21}}\right)\), has the required magnitude of 5.
06

Calculate the fifth vector

Once again, we can negate one component of our third vector \(\mathbf{u_3}\) to create the fifth vector. So, the fifth vector, \(\mathbf{u_5}=\left(\frac{5}{\sqrt{21}}, -\frac{10}{\sqrt{21}}, \frac{20}{\sqrt{21}}\right)\), has the required magnitude of 5. In conclusion, here are the five vectors we found: 1. \(\mathbf{u_1}=(3,4)\) 2. \(\mathbf{u_2}=(-3,4)\) 3. \(\mathbf{u_3}=\left(\frac{5}{\sqrt{21}}, \frac{10}{\sqrt{21}}, \frac{20}{\sqrt{21}}\right)\) 4. \(\mathbf{u_4}=\left(\frac{5}{\sqrt{21}}, -\frac{10}{\sqrt{21}}, -\frac{20}{\sqrt{21}}\right)\) 5. \(\mathbf{u_5}=\left(\frac{5}{\sqrt{21}}, -\frac{10}{\sqrt{21}}, \frac{20}{\sqrt{21}}\right)\)

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