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Chapter 7: Problem 2

Consider vectors \(u=(1,2,4), v=(2,-3,5), w=(4,2,-3)\) in \(\mathbf{R}^{3}\). Find (a) \(u \cdot v\) (c) \(\quad v \cdot w\) (d) \(\quad(u+v) \cdot w\) (e) \(\|u\|,\) (f) \(\|v\|\) (b) \(u \cdot w\) (a) Multiply corresponding components and add to get \(u \cdot v=2-6+20=16\) (b) \(u \cdot w=4+4-12=-4\) (c) \(v \cdot w=8-6-15=-13\) (d) First find \(u+v=(3,-1,9) .\) Then \((u+v) \cdot w=12-2-27=-17 .\) Alternatively, using \(\left[\mathrm{I}_{1}\right]\) \((u+v) \cdot w=u \cdot w+v \cdot w=-4-13=-17\) (e) First find \(\|u\|^{2}\) by squaring the components of \(u\) and adding: \\[\|u\|^{2}=1^{2}+2^{2}+4^{2}=1+4+16=21, \quad \text { and so } \quad\|u\|=\sqrt{21}\\] (f) \(\|v\|^{2}=4+9+25=38,\) and so \(\|v\|=\sqrt{38}\)

Short Answer

Expert verified
(a) $u \cdot v = 16$ (b) $u \cdot w = -4$ (c) $v \cdot w = -13$ (d) $(u+v) \cdot w = -17$ (e) $\|u\| = \sqrt{21}$ (f) $\|v\| = \sqrt{38}$

Step by step solution

01

Calculate Dot Products

To find the dot products, we will multiply the corresponding components of the vectors and then sum them. (a) \(u \cdot v = 1(2) + 2(-3) + 4(5) = 2 - 6 + 20 = 16\) (b) \(u \cdot w = 1(4) + 2(2) + 4(-3) = 4 + 4 - 12 = -4\) (c) \(v \cdot w = 2(4) + (-3)(2) + 5(-3) = 8 - 6 - 15 = -13\)
02

Calculate Sum of Vectors and Dot Products

To find the dot product of (u+v) and w, we first need to compute the sum of u and v. \(u + v = (1,2,4) + (2,-3,5) = (3, -1, 9)\) Now, we can find the dot product of (u+v) and w: (d) \((u + v) \cdot w = (3)(4) + (-1)(2) + (9)(-3) = 12 - 2 - 27 = -17\) Alternatively, we can use the properties of dot products: \( (u+v) \cdot w = u \cdot w + v \cdot w = -4 - 13 = -17\)
03

Calculate Magnitudes

To find the magnitudes of u and v, we will first compute the sum of the squares of the components and then take the square root. (e) \(\|u\|^2 = 1^2 + 2^2 + 4^2 = 1 + 4 + 16 = 21\) So, \(\|u\| = \sqrt{21}\) (f) \(\|v\|^2 = 2^2 + (-3)^2 + 5^2 = 4 + 9 + 25 = 38\) So, \(\|v\| = \sqrt{38}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Linear Algebra
Linear algebra is a foundational branch of mathematics focusing on vectors and matrices and how they can be used to solve systems of linear equations. In the given exercise, linear algebra principles are applied to operations involving vectors in three-dimensional space, \( \mathbf{R}^{3} \). Vectors are essentially lists of numbers that represent points or directions in this space, and they can be added together or multiplied in specific ways to yield new vectors or scalar quantities.

Vector Operations in Linear Algebra

When we talk about operations on vectors such as \(u=(1,2,4)\), \(v=(2,-3,5)\), and \(w=(4,2,-3)\), we're referring to procedures that involve their corresponding elements. Adding vectors, as shown in part (d) of the exercise, follows a component-wise method, resulting in a new vector representing the sum. Linear algebra provides powerful tools such as these to manipulate and understand multidimensional data and systems.
Magnitude of a Vector
The magnitude of a vector, often represented as \(\|u\|\) for a vector \(u\), is a measure of its length. Think of it as how far the vector reaches out from the origin point of the coordinate system. To find a vector's magnitude, each component is squared, and the results are summed up before taking the square root of that sum.

Calculating Vector Magnitude

In the exercise, calculations for \(\|u\|\) and \(\|v\|\) involve exactly this process. For example, \(\|u\|\boxed{^2} = 1^{\boxed{2}} + 2^{\boxed{2}} + 4^{\boxed{2}}\) results in \(\|u\|^2 = 21\), so the magnitude of \(u\) is \(\|u\| = \sqrt{21}\). This magnitude is essentially the 'size' of the vector in space. Understanding vector magnitude is crucial as it holds significance in various applications, including physics, where it's related to concepts such as force and velocity.
Properties of Dot Products
The dot product, also known as scalar or inner product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation is denoted by a dot (.) and is widely used to deduce various properties between vectors, such as the angle between them or their orthogonality.

Using the Dot Product

When you calculated \(u \cdot v\), for example, you multiplied corresponding components of \(u\) and \(v\) and added the results to get 16. This is a direct application of the dot product. One of the beautiful properties of dot products seen in part (d) of the exercise is their distributive nature over vector addition: \( (u+v) \cdot w = u \cdot w + v \cdot w \). This property is frequently used to simplify calculations and show the relationships between vectors in various vector spaces. Learning how dot products behave is an essential component of mastering linear algebra and has broad applications in computer graphics, physics, and machine learning, where it is used, for instance, to calculate projections of vectors onto each other.

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Most popular questions from this chapter

Let \(V\) be a complex inner product space. Verify the relation \\[ \left\langle u, a v_{1}+b v_{2}\right\rangle=\bar{a}\left\langle u, v_{1}\right\rangle+b\left\langle u, v_{2}\right\rangle\\] Using \(\left[I_{2}^{*}\right],\left[I_{1}^{*}\right],\) and then \(\left[I_{2}^{*}\right],\) we find \\[\left\langle u, a v_{1}+b v_{2}\right\rangle=\overline{\left\langle a v_{1}+b v_{2}, u\right\rangle}=\overline{a\left\langle v_{1}, u\right\rangle+b\left\langle v_{2}, u\right\rangle}=\bar{a}\left\langle\overline{\left.v_{1}, u\right\rangle}+\overline{b\left\langle v_{2}, u\right\rangle}=\bar{a}\left(u, v_{1}\right\rangle+\bar{b}\left\langle u, v_{2}\right\rangle\right.\\]

Consider \(\mathbf{P}(t)\) with inner product \(\langle f, g\rangle=\int_{-1}^{1} f(t) g(t) d t\) and the subspace \(W=P_{3}(t)\) (a) Find an orthogonal basis for \(W\) by applying the Gram-Schmidt algorithm to \(\left\\{1, t, t^{2}, t^{3}\right\\}\) (b) Find the projection of \(f(t)=t^{5}\) onto \(W\)

Suppose \(w \neq 0 .\) Let \(v\) be any vector in \(V .\) Show that \\[c=\frac{\langle v, w\rangle}{\langle w, w\rangle}=\frac{\langle v, w\rangle}{\|w\|^{2}}\\] is the unique scalar such that \(v^{\prime}=v-c w\) is orthogonal to \(w\) In order for \(v^{\prime}\) to be orthogonal to \(w\) we must have \\[\langle v-c w, \quad w\rangle=0 \quad \text { or } \quad\langle v, w\rangle-c\langle w, w\rangle=0 \quad \text { or } \quad\langle v, w\rangle=c\langle w, w\rangle\\] Thus, \(c \frac{\langle v, w\rangle}{\langle w, w\rangle} .\) Conversely, suppose \(c=\frac{\langle v, w\rangle}{\langle w, w\rangle} .\) Then \\[\langle v-c w, w\rangle=\langle v, w\rangle-c\langle w, w\rangle=\langle v, w\rangle-\frac{\langle v, w\rangle}{\langle w, w\rangle}\langle w, w\rangle=0\\]

Consider the functions \(f(t)=5 t-t^{2}\) and \(g(t)=3 t-t^{2}\) in \(C[0,4] .\) Find (a) \(d_{\infty}(f, g)\) (b) \(d_{1}(f, g)\) (c) \(d_{2}(f, g)\)

Find an orthogonal matrix \(P\) whose first row is \(u_{1}=\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)\) First find a nonzero vector \(w_{2}=(x, y, z)\) that is orthogonal to \(u_{1}-\) that is, for which \\[0=\left\langle u_{1}, w_{2}\right\rangle=\frac{x}{3}+\frac{2 y}{3}+\frac{2 z}{3}=0 \quad \text { or } \quad x+2 y+2 z=0\\] One such solution is \(w_{2}=(0,1,-1) .\) Normalize \(w_{2}\) to obtain the second row of \(P\) \\[u_{2}=(0,1 / \sqrt{2},-1 / \sqrt{2})\\] Next find a nonzero vector \(w_{3}=(x, y, z)\) that is orthogonal to both \(u_{1}\) and \(u_{2}\) -that is, for which \\[\begin{array}{l} 0=\left\langle u_{1}, w_{3}\right\rangle=\frac{x}{3}+\frac{2 y}{3}+\frac{2 z}{3}=0 \quad \text { or } \quad x+2 y+2 z=0 \\ 0=\left\langle u_{2}, w_{3}\right\rangle=\frac{y}{\sqrt{2}}-\frac{y}{\sqrt{2}}=0 \quad \text { or } \quad y-z=0 \end{array}\\] Set \(z=-1\) and find the solution \(w_{3}=(4,-1,-1) .\) Normalize \(w_{3}\) and obtain the third row of \(P ;\) that is, \\[u_{3}=(4 / \sqrt{18},-1 / \sqrt{18},-1 / \sqrt{18})\\] Thus, \\[P=\left[\begin{array}{ccc}\frac{1}{5} & \frac{2}{3} & \frac{2}{3} \\ 0 & 1 / \sqrt{2} & -1 / \sqrt{2} \\ 4 / 3 \sqrt{2} & -1 / 3 \sqrt{2} & -1 / 3 \sqrt{2} \end{array}\right]\\] We emphasize that the above matrix \(P\) is not unique.
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