/*! 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 (II) Given the vector \(\vec { \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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

Short Answer

Expert verified
\( \vec{\mathbf{B}} = 1\hat{\mathbf{i}} - 2\hat{\mathbf{j}} \) is perpendicular to \( \vec{\mathbf{A}} \).

Step by step solution

01

Understand the Problem

We need to find a vector \( \vec{\mathbf{B}} \) such that it is perpendicular to \( \vec{\mathbf{A}} = 3.0 \hat{\mathbf{i}} + 1.5 \hat{\mathbf{j}} \). This implies that the dot product of \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \) must be zero.
02

Set Up the Dot Product Equation

If \( \vec{\mathbf{B}} = a \hat{\mathbf{i}} + b \hat{\mathbf{j}} \), then the dot product \( \vec{\mathbf{A}} \cdot \vec{\mathbf{B}} \) is given by \( 3.0a + 1.5b \). Since the vectors are perpendicular, this must equal zero: \[ 3.0a + 1.5b = 0 \]
03

Solve for the Components of Vector B

To satisfy \( 3.0a + 1.5b = 0 \), rearrange the equation to express \( b \) in terms of \( a \): \[ b = -\frac{3.0}{1.5}a = -2a \] Choose \( a = 1 \) (any non-zero value will work, but 1 keeps it simple), then \( b = -2 \), making \( \vec{\mathbf{B}} = 1\hat{\mathbf{i}} - 2\hat{\mathbf{j}} \).
04

Verify Perpendicularity

Substitute \( a = 1 \) and \( b = -2 \) back into the dot product equation:\[ 3.0(1) + 1.5(-2) = 3.0 - 3.0 = 0 \]The result is zero, confirming \( \vec{\mathbf{B}} \) is perpendicular to \( \vec{\mathbf{A}} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding the Dot Product
The dot product is a fundamental operation involving two vectors, which helps determine the angle between them. For two vectors, \( \vec{\mathbf{A}} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} \) and \( \vec{\mathbf{B}} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} \), the dot product is calculated as:\[ \vec{\mathbf{A}} \cdot \vec{\mathbf{B}} = a_1b_1 + a_2b_2 \]This operation combines components of each vector to produce a scalar value.
The outcome of the dot product provides crucial insight into the relationship between two vectors:
  • If the dot product is positive, the angle between vectors is acute.
  • If it is negative, the angle is obtuse.
  • Most importantly, if the dot product is zero, the vectors are perpendicular to each other.
In problems involving perpendicularity, finding a dot product of zero confirms that the vectors are at a right angle, providing a vital checking point for our solutions.
Determining Vector Components
Vector components break down a vector into its horizontal and vertical parts. Understanding these components is key to solving vector-related problems, like determining perpendicular vectors.
A vector, such as \( \vec{\mathbf{A}} = 3.0 \hat{\mathbf{i}} + 1.5 \hat{\mathbf{j}} \), includes components along the x-axis (\( \hat{\mathbf{i}} \)) and y-axis (\( \hat{\mathbf{j}} \)). These are the building blocks of any vector, requiring careful balancing to meet specific conditions like perpendicularity.
To adjust vector components for a condition, such as ensuring perpendicularity, set the dot product equation based on these components.
  • Use trial and error or algebraic rearrangement from a known dot product condition to find the necessary components.
  • In practice, choose one component and solve for the other to simplify calculation.
By comprehensively understanding these basic components, you can tailor any vector, like \( \vec{\mathbf{B}} \), to meet the required criteria effectively.
Understanding Perpendicular Vectors
Perpendicular vectors meet at a right angle (90 degrees). This relationship is a cornerstone of vector operations in physics and mathematics.
Communicated via the dot product, as earlier noted, vectors \( \vec{\mathbf{A}} \) and \( \vec{\mathbf{B}} \) are perpendicular if their dot product equals zero. This property is widely applicable, including in fields like engineering and game development.
  • Identifying perpendicular vectors can simplify complex vector calculations.
  • Helps in constructing orthogonal coordinate systems, which simplify component calculations.
To find a perpendicular vector, like \( \vec{\mathbf{B}} \), simply adjust components to satisfy the zero dot product.
This exercise shows choosing specific values, such as \( a = 1 \), in \( \vec{\mathbf{B}} = a\hat{\mathbf{i}} - 2a\hat{\mathbf{j}} \), to maintain simplicity while ensuring perpendicularity to \( \vec{\mathbf{A}} = 3.0 \hat{\mathbf{i}} + 1.5 \hat{\mathbf{j}} \). Such an understanding boosts your ability to manipulate and utilize vectors in practical scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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.

(II) An \(85 - \mathrm { g }\) arrow is fired from a bow whose string exerts an average force of 105\(\mathrm { N }\) on the arrow over a distance of 75\(\mathrm { cm } .\) What is the speed of the arrow as it leaves the bow?

(II) A \(1200 - \mathrm { kg }\) car rolling on a horizontal surface has speed \(v = 66 \mathrm { km } / \mathrm { h }\) when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2\(\mathrm { m }\) . What is the spring constant of the spring?

A softball having a mass of \(0.25 \mathrm{~kg}\) is pitched horizontally at \(110 \mathrm{~km} / \mathrm{h}\). By the time it reaches the plate, it may have slowed by \(10 \% .\) Neglecting gravity, estimate the average force of air resistance during a pitch, if the distance between the plate and the pitcher is about \(15 \mathrm{~m}\).

(II) A baseball \(( m = 145 \mathrm { g } )\) traveling 32\(\mathrm { m } / \mathrm { s }\) moves a fielder's glove backward 25\(\mathrm { cm }\) when the ball is caught. What was the average force exerted by the ball on the glove?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.