/*! 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 21 Given the vector \(\overrightarr... [FREE SOLUTION] | 91Ó°ÊÓ

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Given the vector \(\overrightarrow{\mathbf{A}}=3.0 \hat{\mathbf{i}}+1.5 \hat{\mathbf{j}},\) find a vector \(\overrightarrow{\mathbf{B}}\) that is perpendicular to \(\overrightarrow{\mathbf{A}}\).

Short Answer

Expert verified
The vector \( \overrightarrow{\mathbf{B}} = \hat{\mathbf{i}} - 2\hat{\mathbf{j}} \) is perpendicular to \( \overrightarrow{\mathbf{A}} \).

Step by step solution

01

Understanding Perpendicular Vectors

Two vectors are perpendicular if their dot product is zero. Thus, if we want \( \overrightarrow{\mathbf{B}} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} \) to be perpendicular to \( \overrightarrow{\mathbf{A}} = 3.0 \hat{\mathbf{i}} + 1.5 \hat{\mathbf{j}} \), we must have \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = 0 \).
02

Setting Up the Dot Product Equation

Calculate the dot product: \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = (3.0 \cdot b_1) + (1.5 \cdot b_2) = 3.0b_1 + 1.5b_2 = 0 \).
03

Solve for the Unknowns

From the equation \( 3.0b_1 + 1.5b_2 = 0 \), express \( b_2 \) in terms of \( b_1 \): \( 1.5b_2 = -3.0b_1 \) so \( b_2 = -2b_1 \).
04

Formulate the Perpendicular Vector

Use the relationship \( b_2 = -2b_1 \) to express \( \overrightarrow{\mathbf{B}} \): Let \( b_1 = 1 \), then \( b_2 = -2 \). Therefore, a vector perpendicular to \( \overrightarrow{\mathbf{A}} \) is \( \overrightarrow{\mathbf{B}} = 1 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} \).

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Key Concepts

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

Dot Product
The dot product is an essential operation in vector mathematics, serving to measure how much two vectors align with one another. It is calculated as the sum of the products of their corresponding components. In its formulaic expression, for vectors \(\overrightarrow{\mathbf{A}}\) and \(\overrightarrow{\mathbf{B}}\), the dot product is given by:
  • \( \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = A_x B_x + A_y B_y \) for 2-dimensional vectors.
For two vectors to be perpendicular (orthogonal), their dot product should be zero. This is a pivotal concept when trying to find a vector perpendicular to a given vector. For example, in the provided exercise, we let \(\overrightarrow{\mathbf{B}} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}}\) and calculated the dot product with \(\overrightarrow{\mathbf{A}}\) to ensure it equates to zero, ensuring perpendicularity.
Vector Components
Vector components are essentially the building blocks of vectors, as they measure the vector's projection in the coordinate axes directions. Each vector can be broken down into its components, represented by coefficients alongside the unit vectors \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\). These components make it manageable to perform operations such as addition, subtraction, and taking the dot product. For instance, in the exercise, \(\overrightarrow{\mathbf{A}} = 3.0 \hat{\mathbf{i}} + 1.5 \hat{\mathbf{j}}\) has components 3.0 in the \(\hat{\mathbf{i}}\) direction and 1.5 in the \(\hat{\mathbf{j}}\) direction. Understanding how to work with these components is essential for any vector calculation, achieving efficiency and precision in mathematical or physical problems.
Vector Mathematics
Vector mathematics involves a wide array of techniques such as addition, subtraction, and scalar multiplication. It is used extensively in physics, engineering, and mathematics to describe quantities having both magnitude and direction. Consider finding a vector perpendicular to another. This requires understanding vector operations like the dot product and how leveraging components is essential for mathematical articulation of concepts. Through operations such as those demonstrated in the exercise, vectors become versatile mathematical objects allowing us to succinctly express and solve real-world spatial problems. Perpendicular vectors, in particular, have the special property of being independent in the sense that one vector does not affect the other's direction.
Solving Equations
Solving equations in vector mathematics often requires rearranging and manipulating algebraic expressions to isolate variables. In the exercise, after setting the dot product condition \(3.0b_1 + 1.5b_2 = 0\), we rearranged to express one variable in terms of another: \(b_2 = -2b_1\). This technique simplifies the process and allows for the representation of infinite solutions, each corresponding to scalars of the resulting perpendicular vector. For example, choosing \(b_1 = 1\) gives us a convenient solution for \(\overrightarrow{\mathbf{B}}\). This systematic approach—to isolate and solve variables—is crucial in finding solutions to vector problems systematically, enhancing problem-solving efficiency and accuracy.

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Most popular questions from this chapter

The net force along the linear path of a particle of mass \(480 \mathrm{~g}\) has been measured at \(10.0-\mathrm{cm}\) intervals, starting at \(x=0.0,\) to be 26.0,28.5,28.8,29.6,32.8,40.1,46.6,42.2 48.8,52.6,55.8,60.2,60.6,58.2,53.7,50.3,45.6,45.2,43.2,38.9 \(35.1,30.8,27.2,21.0,22.2,\) and \(18.6,\) all in newtons. Determine the total work done on the particle over this entire range.

(III) An elevator cable breaks when a 925 -kg elevator is 22.5\(\mathrm { m }\) above the top of a huge spring \(( k =\) \(8.00 \times 10 ^ { 4 } \mathrm { N } / \mathrm { m } )\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; \(( b )\) the speed of the elevator just betore striking the spring; \(( c )\) the amount the spring compresses (note that here work is done by both the spring and gravity.

A \(4.5-\mathrm{kg}\) object moving in two dimensions initially has a velocity \(\overrightarrow{\mathbf{v}}_{1}=(10.0 \hat{\mathbf{i}}+20.0 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} .\) A net force \(\overrightarrow{\mathbf{F}}\) then acts on the object for \(2.0 \mathrm{~s}\), after which the object's velocity is \(\vec{v}_{2}=(15.0 \hat{i}+30.0 \hat{j}) \mathrm{m} / \mathrm{s} .\) Determine the work done by \(\overline{\mathbf{F}}\) on the object.

An elevator cable breaks when a \(925-\mathrm{kg}\) elevator is \(22.5 \mathrm{~m}\) above the top of a huge spring \((k=\) \(\left.8.00 \times 10^{4} \mathrm{~N} / \mathrm{m}\right)\) at the bottom of the shaft. Calculate (a) the work done by gravity on the elevator before it hits the spring; (b) the speed of the elevator just before striking the spring; \((c)\) the amount the spring compresses (note that here work is done by both the spring and gravity).

When different masses are suspended from a spring, the spring stretches by different amounts as shown in the Table below. Masses are ±1.0 gram. \begin{tabular}{lllrrrrrrr} \hline Mass (g) & 0 & 50 & 100 & 150 & 200 & 250 & 300 & 350 & 400 \\ Stretch (cm) & 0 & 5.0 & 9.8 & 14.8 & 19.4 & 24.5 & 29.6 & 34.1 & 39.2 \\ \hline \end{tabular} (a) Graph the applied force (in Newtons) versus the stretch (in meters) of the spring, and determine the best-fit straight line. (b) Determine the spring constant ( \(\mathrm{N} / \mathrm{m}\) ) of the spring from the slope of the best-fit line. \((c)\) If the spring is stretched by \(20.0 \mathrm{~cm},\) estimate the force acting on the spring using the best-fit line.

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