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Magazine Chapter 7: Problem 32
Find a vector of unit length in the \(x y\) plane that is perpendicular to \(3.0 \hat{i}+4.0 \hat{j}\)
Short Answer
Expert verified
The unit vectors are \((\frac{4}{5}, -\frac{3}{5})\) and \((-\frac{4}{5}, \frac{3}{5})\).
Step by step solution
01
Understand Perpendicular Vectors
Two vectors are perpendicular when their dot product is zero. Given the vector \((3.0 \hat{i} + 4.0 \hat{j})\), we need to find another vector \(a \hat{i} + b \hat{j}\) such that their dot product equals zero.
02
Write the Dot Product Equation
The dot product of the vectors \(3.0 \hat{i} + 4.0 \hat{j}\) and \(a \hat{i} + b \hat{j}\) is given by \(3a + 4b = 0\). This equation must hold for the vectors to be perpendicular.
03
Solve for One Variable
To solve \(3a + 4b = 0\), express one variable in terms of the other. For example, solve for \(b\): \(b = -\frac{3}{4}a\).
04
Set the Unit Length Constraint
A unit vector has a magnitude of 1. Therefore, the constraint is: \(\sqrt{a^2 + b^2} = 1\). Substitute \(b = -\frac{3}{4}a\) into the unit length equation.
05
Substitute and Solve for Magnitude
Substitute \(b = -\frac{3}{4}a\) into the unit vector equation: \((a^2 + (-\frac{3}{4}a)^2 = 1)\). Simplify to get \((a^2 + \frac{9}{16}a^2 = 1)\). This results in \((\frac{25}{16}a^2 = 1)\).
06
Calculate Magnitude of 'a'
Solve \((\frac{25}{16}a^2 = 1)\) to find \(a^2 = \frac{16}{25}\), hence \(a = \pm \frac{4}{5}\).
07
Calculate Corresponding Value of 'b'
Use \(b = -\frac{3}{4}a\). If \(a = \frac{4}{5}\), then \(b = -\frac{3}{4} \times \frac{4}{5} = -\frac{3}{5}\). If \(a = -\frac{4}{5}\), \(b = \frac{3}{5}\).
08
Identify Unit Vectors
The vectors \(\left(\frac{4}{5}, -\frac{3}{5}\right)\) and \(\left(-\frac{4}{5}, \frac{3}{5}\right)\) are both unit vectors perpendicular to \(3.0 \hat{i} + 4.0 \hat{j}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Unit Vector
A unit vector is a vector with a magnitude of 1. It simply points in a specific direction along any line or plane, but it does not have any size or strength. The main purpose of a unit vector is to describe a direction without emphasizing distance.
To find a unit vector, you typically divide a given vector by its own magnitude. This transforms any vector into one that has a length of exactly one unit. This is its normalized form.
Given a unit vector formula:
To find a unit vector, you typically divide a given vector by its own magnitude. This transforms any vector into one that has a length of exactly one unit. This is its normalized form.
Given a unit vector formula:
- If a vector is \(v = a \hat{i} + b \hat{j}\), its magnitude is \(|v| = \sqrt{a^2 + b^2}\).
- The unit vector in the direction of v is \(\hat{v} = \frac{v}{|v|}\).
Dot Product
The dot product of two vectors is a number, which tells how much of the vectors overlap. When the dot product equals zero, it implies that the vectors are perpendicular.
For two vectors \(\overrightarrow{v_1} = a_1 \hat{i} + b_1 \hat{j}\) and \(\overrightarrow{v_2} = a_2 \hat{i} + b_2 \hat{j}\), the dot product is calculated as \(a_1a_2 + b_1b_2\).
This is crucial for determining when two vectors, such as \(3.0 \hat{i} + 4.0 \hat{j}\) and some other vector, are perpendicular since their dot product should be zero.
For two vectors \(\overrightarrow{v_1} = a_1 \hat{i} + b_1 \hat{j}\) and \(\overrightarrow{v_2} = a_2 \hat{i} + b_2 \hat{j}\), the dot product is calculated as \(a_1a_2 + b_1b_2\).
This is crucial for determining when two vectors, such as \(3.0 \hat{i} + 4.0 \hat{j}\) and some other vector, are perpendicular since their dot product should be zero.
- If \(3a + 4b = 0\), then vector \(a \hat{i} + b \hat{j}\) is indeed perpendicular.
Vector Magnitude
The magnitude of a vector refers to its length or distance from its origin. For a vector \(v = a \hat{i} + b \hat{j}\), the magnitude \(|v|\) is found using the Pythagorean theorem:
\(|v| = \sqrt{a^2 + b^2}\).
This formula tells how long the vector is, regardless of its direction.
A critical part of forming unit vectors, as the concept dictates, is ensuring this magnitude is normalized to 1. When you find perpendicular vectors of unit length, like the vectors generated from an existing vector, you are essentially normalizing the vector according to its components.
\(|v| = \sqrt{a^2 + b^2}\).
This formula tells how long the vector is, regardless of its direction.
A critical part of forming unit vectors, as the concept dictates, is ensuring this magnitude is normalized to 1. When you find perpendicular vectors of unit length, like the vectors generated from an existing vector, you are essentially normalizing the vector according to its components.
- This ensures the vector remains in the same direction but has a controlled length for calculations.
Vector Components
Vector components refer to the individual parts that make up a vector, usually represented by \(a \hat{i}\) and \(b \hat{j}\) in two-dimensional space. This decomposition allows for the analysis of vectors within any coordinate system.
These components correlate to how much of the vector lies along the x and y axes, respectively.
When solving vector problems, breaking them down into components can greatly simplify solutions. For example, to determine a perpendicular vector's components:
These components correlate to how much of the vector lies along the x and y axes, respectively.
When solving vector problems, breaking them down into components can greatly simplify solutions. For example, to determine a perpendicular vector's components:
- Determine the relationship between the components using a dot product equation, such as \(3a + 4b = 0\).
- Then solve for specific values as shown in the problem, where \(b = -\frac{3}{4}a\).
