/*! 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 25 Prove that two vectors must have... [FREE SOLUTION] | 91Ó°ÊÓ

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Prove that two vectors must have equal magnitudes if their sum is perpendicular to their difference.

Short Answer

Expert verified
\( \| \mathbf{a} \| = \| \mathbf{b} \| \)

Step by step solution

01

Define Vectors and Properties

Let the two vectors be \( \mathbf{a} \) and \( \mathbf{b} \). We are to prove that if \( \mathbf{a} + \mathbf{b} \) is perpendicular to \( \mathbf{a} - \mathbf{b} \), then \( \| \mathbf{a} \| = \| \mathbf{b} \| \).
02

Use the Dot Product Property

The dot product of two perpendicular vectors is zero. Therefore, if \( \mathbf{a} + \mathbf{b} \) is perpendicular to \( \mathbf{a} - \mathbf{b} \), then i.e., \(( \mathbf{a} + \mathbf{b} ) \cdot ( \mathbf{a} - \mathbf{b} ) = 0 \).
03

Expand the Dot Product

Expand the dot product: \[ ( \mathbf{a} + \mathbf{b} ) \cdot ( \mathbf{a} - \mathbf{b} ) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} + \mathbf{b} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} \].
04

Simplify the Expression

Since the dot product is commutative (\( \mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a} \)), the expression simplifies to \[ \| \mathbf{a} \|^2 - \| \mathbf{b} \|^2 = 0 \].
05

Conclude the Solution

Rearrange the equation: \[ \| \mathbf{a} \|^2 = \| \mathbf{b} \|^2 \]. Taking the square root of both sides, we get \( \| \mathbf{a} \| = \| \mathbf{b} \| \). Thus, the magnitudes of the two vectors must be equal.

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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 algebra. It combines two vectors and outputs a scalar. The dot product of two vectors \( \textbf{a} \) and \( \textbf{b} \) is computed as: \( \textbf{a} \bullet \textbf{b} = a_1b_1 + a_2b_2 + a_3b_3 + ... + a_nb_n \). This can also be expressed as \( \textbf{a} \bullet \textbf{b} = \textbf{a} \textbf{b} \text{cos}(\theta) \), where \( \theta \) is the angle between the two vectors.

The dot product has interesting properties, such as being commutative, meaning \( \textbf{a} \bullet \textbf{b} = \textbf{b} \bullet \textbf{a} \).
In our context, dot products are instrumental in determining if two vectors are perpendicular because when vectors are perpendicular, their dot product equals zero.
Perpendicular Vectors
Two vectors are said to be perpendicular (or orthogonal) if the angle between them is 90 degrees. This orthogonality can be verified using the dot product. Specifically, if the dot product of two vectors is zero, they are perpendicular.

In mathematical terms, if \( \textbf{a} \) and \( \textbf{b} \) are perpendicular, then \( \textbf{a} \bullet \textbf{b} = 0 \). For our exercise, we utilized this property to establish that the vectors \( \textbf{a} + \textbf{b} \) and \( \textbf{a} - \textbf{b} \) are perpendicular.

This leads to the interesting conclusion that their sum's perpendicular nature to their difference can help ascertain equality in their magnitudes.
Vector Magnitude
Vector magnitude represents the length or size of a vector. It is denoted as \( \textbf{a} \| \textbf{a} \| \). The magnitude of a vector \( \textbf{a} \) is calculated as the square root of the sum of the squares of its components: \( \| \textbf{a} \| = \text{sqrt}(a_1^2 + a_2^2 + a_3^2 + ... + a_n^2) \).

In our exercise, the conclusion that \( \| \textbf{a} \| = \| \textbf{b} \| \) was achieved by leveraging the perpendicular nature of the vector sum and difference. Starting from the dot product being zero, we showed that \( \| \textbf{a} \|^2 - \| \textbf{b} \|^2 = 0 \). Upon simplifying, we concluded that the magnitudes of \( \textbf{a} \) and \( \textbf{b} \) are indeed equal.

Understanding magnitude is essential since it provides a way to compare the lengths of vectors and fortifies the basis of our equality proof in the exercise.

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Most popular questions from this chapter

Vector \(\vec{a}\) has a magnitude of \(5.0 \mathrm{~m}\) and is directed east. Vector \(\vec{b}\) has a magnitude of \(4.0 \mathrm{~m}\) and is directed \(35^{\circ}\) west of north. What are (a) the magnitude and (b) the direction of \(\vec{a}+\vec{b}\) ? What are (c) the magnitude and (d) the direction of \(\vec{b}-\vec{a}\) ? (e) Draw a vector diagram for each combination.

(a) Using unit vectors, write expressions for the four body diagonals (the straight lines from one corner to another through the center) of a cube in terms of its edges, which have length \(a\). (b) Determine the angles that the body diagonals make with the adjacent edges. (c) Determine the length of the body diagonals in terms of \(a\).

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for \(5.6 \mathrm{~km}\), but when the snow clears, he discovers that he actually traveled \(7.8 \mathrm{~km}\) at \(50^{\circ}\) north of due east. (a) How far and (b) in what direction must he now travel to reach base camp?

(a) In unit-vector notation, what is the sum of $$\vec{a}=(4.0 \mathrm{~m}) \hat{\mathrm{i}}+(3.0 \mathrm{~m}) \hat{\mathrm{j}} \text { and } \vec{b}=(-13.0 \mathrm{~m}) \hat{\mathrm{i}}+(7.0 \mathrm{~m}) \hat{\mathrm{j}} ?$$ What are (b) the magnitude and (c) the direction of \(\vec{a}+\vec{b}\) (relative to \(\hat{\mathrm{i}}\) ?

A pea leaves a pea shooter at a speed of \(5.4 \mathrm{~m} / \mathrm{s}\). It makes an angle of \(+30^{\circ}\) with respect to the horizontal. (a) Calculate the \(x\) -component of the pea's initial velocity. (b) Calculate the \(y\) -component of the pea's initial velocity. (c) Write an expression for the pea's velocity, \(\vec{v}\), using unit vectors for the \(x\) direction and the \(y\) direction.

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