/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 Let $$\mathbf{x}=\left(\begin{... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 17

Let $$\mathbf{x}=\left(\begin{array}{r} 4 \\ 4 \\ -4 \\ 4 \end{array}\right) \quad \text { and } \quad \mathbf{y}=\left(\begin{array}{l} 4 \\ 2 \\ 2 \\ 1 \end{array}\right)$$ (a) Determine the angle between \(\mathbf{x}\) and \(\mathbf{y}\) (b) Determine the distance between \(\mathbf{x}\) and \(\mathbf{y}\)

Short Answer

Expert verified
The angle between the vectors \(\mathbf{x}\) and \(\mathbf{y}\) is approximately \(49.8^{\circ}\), and the distance between them is approximately \(5.57\).

Step by step solution

01

Calculate the dot product of x and y

We need to calculate the dot product of \(\mathbf{x}\) and \(\mathbf{y}\), which is given by: $$\mathbf{x}\cdot\mathbf{y} = x_1y_1 + x_2y_2 + x_3y_3 + x_4y_4$$ Using the given values: $$\mathbf{x}\cdot\mathbf{y} = (4)(4) + (4)(2) + (-4)(2) + (4)(1)$$
02

Calculate the magnitudes of x and y

To determine the angle between the vectors, we'll also need the magnitudes of \(\mathbf{x}\) and \(\mathbf{y}\). The magnitudes are given by: $$\|\mathbf{x}\| = \sqrt{x_1^2 + x_2^2 + x_3^2 + x_4^2} \quad \text{and} \quad \|\mathbf{y}\| = \sqrt{y_1^2 + y_2^2 + y_3^2 + y_4^2}$$ Using the given values, we have: $$\|\mathbf{x}\| = \sqrt{(4)^2 + (4)^2 + (-4)^2 + (4)^2}$$ $$\|\mathbf{y}\| = \sqrt{(4)^2 + (2)^2 + (2)^2 + (1)^2}$$
03

Calculate the angle between x and y

Now that we have the dot product and magnitudes, we can use the formula given in the analysis to determine the angle between the two vectors: $$\cos \theta = \frac{\mathbf{x}\cdot\mathbf{y}}{\|\mathbf{x}\|\|\mathbf{y}\|}$$ Substitute the values and calculate the angle \(\theta\).
04

Calculate the difference between x and y

To determine the distance between \(\mathbf{x}\) and \(\mathbf{y}\), we first need to compute the difference between the two vectors: $$\mathbf{x} - \mathbf{y} = \left(\begin{array}{r} 4 - 4 \\ 4 - 2 \\ -4 - 2 \\ 4 - 1 \end{array}\right)$$
05

Calculate the distance between x and y

Using the formula given in the analysis for the distance between two vectors, we compute the distance as the magnitude of the difference vector: $$d(\mathbf{x}, \mathbf{y}) = \|\mathbf{x} - \mathbf{y}\|$$ Calculate the magnitude of the difference vector found in Step 4 to find the distance between \(\mathbf{x}\) and \(\mathbf{y}\).

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Most popular questions from this chapter

The vectors $$\mathbf{x}_{1}=\frac{1}{2}(1,1,1,-1)^{T} \quad \text { and } \quad \mathbf{x}_{2}=\frac{1}{6}(1,1,3,5)^{T}$$ form an orthonormal set in \(\mathbb{R}^{4}\). Extend this set to an orthonormal basis for \(\mathbb{R}^{4}\) by finding an orthonormal basis for the null space of $$\left(\begin{array}{rrrr} 1 & 1 & 1 & -1 \\ 1 & 1 & 3 & 5 \end{array}\right)$$[Hint: First find a basis for the null space and then use the Gram-Schmidt process.]

The point \((\bar{x}, \bar{y})\) is the center of mass for the collection of points in Exercise \(7 .\) Show that the least squares line must pass through the center of mass. [Hint: Use a change of variables \(z=x-\bar{x}\) to translate the problem so that the new independent variable has mean \(0 .\)

Dimension Theorem Let \(U\) and \(V\) be subspaces of \(\mathbb{R}^{n} .\) In the case that \(U \cap V=\\{\mathbf{0}\\},\) we have the following dimension relation \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V \\] (See Exercise 18 in Section 4 of Chapter 3 .) Make use of the result from Exercise 14 to prove the more general theorem \\[ \operatorname{dim}(U+V)=\operatorname{dim} U+\operatorname{dim} V-\operatorname{dim}(U \cap V) \\]

Show that the Chebyshev polynomials have the following properties: (a) \(2 T_{m}(x) T_{n}(x)=T_{m+n}(x)+T_{m-n}(x),\) for \(m>n\) (b) \(T_{m}\left(T_{n}(x)\right)=T_{m n}(x)\)

Let \(A\) be an \(m \times n\) matrix of rank \(r\) and let \(\left\\{\mathbf{x}_{1}, \ldots, \mathbf{x}_{r}\right\\}\) be a basis for \(R\left(A^{T}\right) .\) Show that \(\left\\{A \mathbf{x}_{1}, \ldots, A \mathbf{x}_{r}\right\\}\) is a basis for \(R(A)\)
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