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Chapter 1: Problem 44

Normalize each vector: (a) \(u=(5,-7)\) (b) \(v=(1,2,-2,4)\) (c) \(w=\left(\frac{1}{2},-\frac{1}{3}, \frac{3}{4}\right)\)

Short Answer

Expert verified
(a) \(u = \left(\frac{5}{\sqrt{74}},\frac{-7}{\sqrt{74}}\right)\) (b) \(v = \left(\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}},\frac{-2}{21}, \frac{4}{\sqrt{21}}\right)\) (c) \(w = \left(\frac{\frac{1}{2}}{\frac{5}{4}}, \frac{-\frac{1}{3}}{\frac{5}{4}}, \frac{\frac{3}{4}}{\frac{5}{4}}\right)\)

Step by step solution

01

Calculate the magnitude of each vector

First, we will find the magnitude of each vector using the formula: Magnitude = \(\sqrt{(\text{Sum of the squares of the components})}\) (a) For vector \(u\): Magnitude of \(u = \sqrt{(5)^2 + (-7)^2}\) (b) For vector \(v\): Magnitude of \(v = \sqrt{(1)^2 + (2)^2 + (-2)^2 + (4)^2}\) (c) For vector \(w\): Magnitude of \(w = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{3}{4}\right)^2}\)
02

Divide each component of the vector by its magnitude

(a) For vector \(u\): Normalize \(u\) = \(\frac{u}{|u|} = \frac{(5,-7)}{\sqrt{(5)^2 + (-7)^2}} = \left(\frac{5}{\sqrt{74}},\frac{-7}{\sqrt{74}}\right)\) (b) For vector \(v\): Normalize \(v\) = \(\frac{v}{|v|} = \frac{(1,2,-2,4)}{\sqrt{(1)^2 + (2)^2 + (-2)^2 + (4)^2}} = \left(\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}},\frac{-2}{21}, \frac{4}{\sqrt{21}}\right)\) (c) For vector \(w\): Normalize \(w\) = \(\frac{w}{|w|} = \frac{\left(\frac{1}{2},-\frac{1}{3}, \frac{3}{4}\right)}{\sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{3}{4}\right)^2}} = \left(\frac{\frac{1}{2}}{\frac{5}{4}}, \frac{-\frac{1}{3}}{\frac{5}{4}}, \frac{\frac{3}{4}}{\frac{5}{4}}\right)\) Thus, the normalized vectors are: (a) \(u = \left(\frac{5}{\sqrt{74}},\frac{-7}{\sqrt{74}}\right)\) (b) \(v = \left(\frac{1}{\sqrt{21}}, \frac{2}{\sqrt{21}},\frac{-2}{21}, \frac{4}{\sqrt{21}}\right)\) (c) \(w = \left(\frac{\frac{1}{2}}{\frac{5}{4}}, \frac{-\frac{1}{3}}{\frac{5}{4}}, \frac{\frac{3}{4}}{\frac{5}{4}}\right)\)

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Key Concepts

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

Magnitude of a Vector
Understanding the magnitude of a vector is crucial for various applications in mathematics and physics. The magnitude, often referred to as the length or the norm of a vector, represents the distance of the vector from the origin in a vector space.

To find the magnitude of a vector, which is a Euclidean vector for our purposes, you apply the Pythagorean theorem. This theorem states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. Similarly, the magnitude of a vector is the square root of the sum of the squares of its components.

For example, the magnitude of a 2-dimensional vector \(u=(x, y)\) is given by \(\sqrt{x^2 + y^2}\), and for a 3-dimensional vector \(v=(x, y, z)\), it is \(\sqrt{x^2 + y^2 + z^2}\). This formula generalizes to higher dimensions as needed.
Normalizing Vectors
Normalizing a vector is an operation that converts it to a unit vector, which retains its direction but has a magnitude of 1. To normalize a vector, you divide each of its components by the vector's magnitude.

This process is utilized in various fields, such as computer graphics, where it helps in simplifying calculations related to lighting and shading. In physics, normalized vectors are used to express directional quantities without affecting their magnitude.

Referring to the textbook solution, after calculating the magnitude of a given vector, normalization can be accomplished by dividing each component by this calculated magnitude. The result is a vector pointing in the same direction but scaled to a length of 1.
Vector Components
A vector in mathematics and physics is defined by its components, which represent the vector's influence in various dimensions of a coordinate system. The components are values along the axis of each dimension that, when combined, give the vector itself.

For instance, in a 2D coordinate system, a vector \(v=(x, y)\) has a component \(x\) along the x-axis and a component \(y\) along the y-axis. These components can be visualized as the horizontal and vertical displacements needed to reach the end point of the vector, starting from the origin.

In application, knowing the components of a vector is essential for operations such as addition, subtraction, and multiplication by a scalar.
Euclidean Vector Space
The concept of a Euclidean vector space is fundamental in understanding vectors and their operations. It's a mathematical structure formed by a collection of vectors that can be added together and multiplied by scalars to produce another vector within the same space. It is characterized by its dimensions, also known as its rank, where each dimension corresponds to one of the vector's components.

A Euclidean vector space corresponds to the common n-dimensional Cartesian space, and vectors within the space can be visualized geometrically. Geometrically, vectors in a Euclidean space represent displacements or directions, with the vector's magnitude signifying the extent of the displacement and its direction defining where the displacement leads.

Working within a Euclidean vector space allows us to apply tools such as the dot product and cross product, and to use geometric intuition to understand vector properties and operations.

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Most popular questions from this chapter

Find \(x\) and \(y\) where: (a) \(\quad(x, y+1)=(y-2,6)\) (b) \(x(2, y)=y(1,-2)\)

Show that for complex numbers \(z\) and \(w\) : (a) \(\operatorname{Re} z=\frac{1}{2}(z+\bar{z})\) (b) \(\quad \operatorname{Im} z=\frac{1}{2}(z-\bar{z})\) (c) \(z w=0\) implies \(z=0\) or \(w=0\).

Prove Theorem \(1.3\) (Schwarz): \(|u \cdot v| \leq\|u\|\|v\|\). For any real number \(t\), and using Theorem \(1.2\), we have $$ 0 \leq(t u+v) \cdot(t u+v)=t^{2}(u \cdot u)+2 t(u \cdot v)+(v \cdot v)=\|u\|^{2} t^{2}+2(u \cdot v) t+\|v\|^{2} $$ Let \(a=\|u\|^{2}, b=2(u \cdot v), c=\|v\|^{2}\). Then, for every value of \(t, a t^{2}+b t+c \geq 0\). This means that the quadratic polynomial cannot have two real roots. This implies that the discriminant \(D=b^{2}-4 a c \leq 0\) or, equivalently, \(b^{2} \leq 4 a c\). Thus, $$ 4(\boldsymbol{u} \cdot v)^{2} \leq 4\|u\|^{2}\|v\|^{2} $$ Dividing by 4 gives us our result.

Find \(z \bar{z}\) and \(|z|\) when \(z=3+4 i\) For \(z=a+b i\), use \(z \bar{z}=a^{2}+b^{2}\) and \(z=\sqrt{z \bar{z}}=\sqrt{a^{2}+b^{2}}\). $$ z \bar{z}=9+16=25, \quad|z|=\sqrt{25}=5 $$

Find a unit vector \(u\) orthogonal to: (a) \(v=[1,2,3]\) and \(w=[1,-1,2]\); (b) \(\quad v=3 \mathbf{i}-\mathbf{j}+2 \mathbf{k}\) and \(w=4 \mathbf{i}-2 \mathbf{j}-\mathbf{k}\).
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