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Chapter 13: Problem 18

Find the length of the vectors. $$\vec{v}=7.2 \vec{i}-1.5 \vec{j}+2.1 \vec{k}$$

Short Answer

Expert verified
The vector's length is approximately 7.65.

Step by step solution

01

Identify the Components

To find the length of a vector, start by identifying its components. The vector \(\vec{v} = 7.2 \vec{i} - 1.5 \vec{j} + 2.1 \vec{k}\) has components: \(v_x = 7.2\), \(v_y = -1.5\), and \(v_z = 2.1\).
02

Formula for Vector Length

Use the formula for the length (or magnitude) of a vector in three-dimensional space: \(\|\vec{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}\).
03

Calculate Each Component's Square

Compute the squares of each of the vector's components: - \(v_x^2 = (7.2)^2 = 51.84\)- \(v_y^2 = (-1.5)^2 = 2.25\)- \(v_z^2 = (2.1)^2 = 4.41\)
04

Sum the Squared Components

Add the squared components together: \(51.84 + 2.25 + 4.41 = 58.5\).
05

Take the Square Root

Finally, take the square root of the sum: \(\|\vec{v}\| = \sqrt{58.5} \approx 7.65 \).

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Key Concepts

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

Vector Components
Understanding the components of a vector is crucial when dealing with vectors in three-dimensional space. Each vector can be thought of as having separate parts, or components, that lie along the x, y, and z axes.
These components help in clearly breaking down the overall vector into manageable parts:
  • The x-component in the vector points in the direction of the x-axis and is denoted by a scalar value multiplied by the unit vector \(\vec{i}\).
  • The y-component points along the y-axis and is indicated with \(\vec{j}\), also being a scalar multiple.
  • The z-component runs along the z-axis and involves the unit vector \(\vec{k}\), once again multiplied by a scalar.
In the vector \(\vec{v} = 7.2 \vec{i} - 1.5 \vec{j} + 2.1 \vec{k}\), these components are \(v_x = 7.2\), \(v_y = -1.5\), and \(v_z = 2.1\).
Identifying these components is the first step in many vector-related calculations, including finding a vector's length.
Magnitude Calculation
The magnitude, or length, of a vector is a measure of how long the vector is when the total distance is considered in three dimensions.
The formula for calculating the magnitude of a vector \(\vec{v}\) with components \(v_x\), \(v_y\), and \(v_z\) is given by:\[\|\vec{v}\| = \sqrt{v_x^2 + v_y^2 + v_z^2}\]To calculate this:
  • Square each component: \(v_x^2 = 51.84\), \(v_y^2 = 2.25\), and \(v_z^2 = 4.41\).
  • Add the squares: \(51.84 + 2.25 + 4.41 = 58.5\).
  • Take the square root of the result to find the magnitude: \(\|\vec{v}\| = \sqrt{58.5} \approx 7.65\).
This calculation process reveals how the separate components contribute to the vector鈥檚 overall size.
Three-Dimensional Vectors
Vectors in three-dimensional space offer a rich view of motion and forces, extending beyond the flat, two-dimensional plains.
These vectors are often represented in terms of unit vectors, \(\vec{i}\), \(\vec{j}\), and \(\vec{k}\), for each respective axis.With three-dimensional vectors, you can model scenarios that involve movement or forces acting in three directions at once. For instance:
  • Physical forces operating in space, such as gravity and electromagnetism, are traditionally expressed in three dimensions.
  • Displacements or velocities in navigation tasks, such as aviation or maritime journey paths, are described with three-dimensional vectors.
Understanding these vectors allows you to not only calculate magnitudes but also perform operations like addition, subtraction, and finding the dot or cross product with other vectors. This versatility makes three-dimensional vectors indispensable in physics and engineering.

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Most popular questions from this chapter

Give reasons for your answer. If \(\vec{u} \cdot \vec{v}=\|\vec{u}\|\|\vec{v}\|\) then \(\|\vec{u}+\vec{v}\|=\|\vec{u}\|+\|\vec{v}\|\)

Perform the following operations on the given 3 -dimensional vectors.$$\vec{a}=2 \vec{j}+\vec{k} \quad \vec{b}=-3 \vec{i}+5 \vec{j}+4 \vec{k} \quad \vec{c}=\vec{i}+6 \vec{j}$$ $$\vec{y}=4 \vec{i}-7 \vec{j} \quad \vec{z}=\vec{i}-3 \vec{j}-\vec{k}$$ $$((\vec{c} \cdot \vec{c}) \vec{a}) \cdot \vec{a}$$

Evaluate \(\vec{u} \cdot \vec{w}\) \(\|\vec{u}\|=10,\|\vec{w}\|=20 ;\) the angle between \(\vec{u}\) and \(\vec{w}\) is \(120^{\circ}\)

Are the statements true or false? Give reasons for your answer. Given a nonzero vector \(\vec{v}\) in 3 -space, there is a nonzero vector \(\vec{w}\) such that \(\vec{v} \times \vec{w}=\overrightarrow{0}\)

For two-dimensional vectors \(\vec{a}\) and \(\vec{b},\) if \(\|\vec{a}\|=2\) and \(\|\vec{b}\|=4,\) find \(\|\vec{a}+\vec{b}\|\) for the given \(\vec{a} \cdot \vec{b}\) $$\vec{a} \cdot \vec{b}=8$$
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