/*! 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 26 (II) Show that if two nonparalle... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) Show that if two nonparallel vectors have the same magnitude, their sum must be perpendicular to their difference.

Short Answer

Expert verified
Given equal magnitudes, the sum of the vectors is perpendicular to their difference because their dot product is zero.

Step by step solution

01

Understand the Vectors

Consider two nonparallel vectors \( \mathbf{a} \) and \( \mathbf{b} \), each with the same magnitude. Denote their magnitudes as \( \|\mathbf{a}\| = \|\mathbf{b}\| \).
02

Expression for Sum and Difference

Write the expressions for the sum and the difference of the vectors: \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \).
03

Use Dot Product Property

To prove the vectors are perpendicular, their dot product should equal zero: \( (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = 0 \).
04

Calculate Dot Product

Calculate the dot product: \\((\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} \).
05

Simplify the Expression

Simplify using the properties of the dot product: \( \|\mathbf{a}\|^2 - \|\mathbf{b}\|^2 + \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{a} \). Since \( \mathbf{a}\cdot\mathbf{b} \) and \( \mathbf{b}\cdot\mathbf{a} \) are equal, they cancel out.
06

Conclude the Perpendicularity

Since \( \|\mathbf{a}\|^2 = \|\mathbf{b}\|^2 \), the terms \( \|\mathbf{a}\|^2 - \|\mathbf{b}\|^2 \) cancel, leaving the dot product as zero, proving \( \mathbf{a} + \mathbf{b} \) is perpendicular to \( \mathbf{a} - \mathbf{b} \).

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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 a fundamental operation in vector mathematics, used extensively to determine the relationship between two vectors. When we talk about the dot product, often represented as \( \mathbf{a} \cdot \mathbf{b} \) for vectors \( \mathbf{a} \) and \( \mathbf{b} \), we're essentially measuring how much of one vector goes in the direction of another.
The formula for the dot product of vectors \( \mathbf{a} = (a_1, a_2, ..., a_n) \) and \( \mathbf{b} = (b_1, b_2, ..., b_n) \) is:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + ... + a_nb_n \)
If the result is zero, it shows that the vectors are perpendicular, as none of the vector's component projects onto the other.
This property plays a crucial role in the problem of proving perpendicularity between the sum and difference of two vectors with the same magnitude.
Vector Magnitude
Vector magnitude refers to the length or size of a vector. For any vector \( \mathbf{v} = (v_1, v_2, ..., v_n) \), its magnitude is calculated using the square root of the sum of the squares of its components:
  • \( \|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + ... + v_n^2} \)
In the provided exercise, both vectors \( \mathbf{a} \) and \( \mathbf{b} \) have the same magnitude, i.e., \( \|\mathbf{a}\| = \|\mathbf{b}\| \).
This is significant because it simplifies the mathematical manipulation during the solution process, ensuring that terms like \( \|\mathbf{a}\|^2 \) and \( \|\mathbf{b}\|^2 \) cancel each other.
This ultimately leads to simplifying the dot product calculation, confirming the perpendicularity requirement.
Perpendicular Vectors
Perpendicular vectors have a special relationship: their dot product equals zero. This relationship arises because when two vectors are perpendicular, they don't "align" with each other at all, meaning one vector doesn't project onto the other.
In our problem, we are given that vectors \( \mathbf{a} + \mathbf{b} \) and \( \mathbf{a} - \mathbf{b} \) need to be shown as perpendicular
To confirm they are perpendicular, we calculate their dot product:
  • \( (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) \)
Through this computation, by simplifying and using the properties of vector magnitudes and dot products, we see that the terms cancel out, resulting in zero.
This zero result is the mathematical proof required to conclude that the vectors are indeed perpendicular.

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Most popular questions from this chapter

A \(6.10-\mathrm{kg}\) block is pushed \(9.25 \mathrm{~m}\) up a smooth \(37.0^{\circ}\) inclined plane by a horizontal force of \(75.0 \mathrm{~N}\). If the initial speed of the block is \(3.25 \mathrm{~m} / \mathrm{s}\) up the plane, calculate \((a)\) the initial kinetic energy of the block; (b) the work done by the 75.0-N force; \((c)\) the work done by gravity; \((d)\) the work done by the normal force; \((e)\) the final kinetic energy of the block.

Find a vector of unit length in the \(x y\) plane that is perpendicular to \(3.0 \hat{i}+4.0 \hat{j}\)

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In the game of paintball, players use guns powered by pressurized gas to propel \(33 -\) g gel capsules filled with paint at the opposing team. Game rules dictate that a paintball cannot leave the barrel of a gun with a speed greater than 85\(\mathrm { m } / \mathrm { s } .\) Model the shot by assuming the pressurized gas applies a constant force \(F\) to a \(33 - g\) capsule over the length of the \(32 - \mathrm { cm }\) barrel. Determine \(F ( a )\) using the work-energy principle, and \(( b )\) using the kinematic equations and Newton's second law.

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