/*! 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 8 Let \(u=(5,4,1), v=(3,-4,1), w=(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 8

Let \(u=(5,4,1), v=(3,-4,1), w=(1,-2,3) .\) Which pair of vectors, if any, are perpendicular (orthogonal)?

Short Answer

Expert verified
The vectors \(u\) and \(v\) and the vectors \(u\) and \(w\) are both orthogonal because their dot products are 0. The vectors \(v\) and \(w\) are not orthogonal because their dot product is 14.

Step by step solution

01

Compute Dot Product of u and v

To compute the dot product of u and v: \(u \cdot v = (5)(3)+(4)(-4)+(1)(1)\).
02

Evaluate Dot Product of u and v

\( u \cdot v = 15 - 16 + 1= 0 \)
03

Compute Dot Product of u and w

To compute the dot product of u and w: \(u \cdot w = (5)(1)+(4)(-2)+(1)(3)\).
04

Evaluate Dot Product of u and w

\( u \cdot w = 5 - 8 + 3 = 0 \)
05

Compute Dot Product of v and w

To compute the dot product of v and w: \(v \cdot w = (3)(1)+(-4)(-2)+(1)(3)\).
06

Evaluate Dot Product of v and w

\( v \cdot w = 3 + 8 + 3 = 14 \) Vectors u and v have a dot product of 0, so they are orthogonal. Vectors u and w also have a dot product of 0, so they are orthogonal too. Therefore, the pairs of vectors \(u, v\) and \(u, w\) are perpendicular (orthogonal). The dot product of the vectors v and w is 14, which is not equal to zero, so they are not orthogonal.

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 91影视!

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

Show that if \(S_{1}\) and \(S_{2}\) are subsets of a vector space \(\mathrm{V}\) such that \(S_{1} \subseteq S_{2}\), then $\operatorname{span}\left(S_{1}\right) \subseteq \operatorname{span}\left(S_{2}\right) .\( In particular, if \)S_{1} \subseteq S_{2}\( and \)\operatorname{span}\left(S_{1}\right)=\mathrm{V}$, deduce that span \(\left(S_{2}\right)=\mathrm{V}\). Visit goo.gl/Fi8Epr for a solution.

Is the set of all differentiable real-valued functions defined on \(R\) a subspace of \(C(R)\) ? Justify your answer.

Let \(u, v\), and \(w\) be distinct vectors of a vector space \(\mathrm{V}\). Show that if \(\\{u, v, w\\}\) is a basis for \(\mathrm{V}\), then $\\{u+v+w, v+w, w\\}\( is also a basis for \)\mathrm{V}$.

Let $\mathrm{V}=\left\\{\left(a_{1}, a_{2}\right): a_{1}, a_{2} \in F\right\\}\(, where \)F\( is a field. Define addition of elements of \)\mathrm{V}$ coordinatewise, and for \(c \in F\) and $\left(a_{1}, a_{2}\right) \in \mathrm{V}$, define $$ c\left(a_{1}, a_{2}\right)=\left(a_{1}, 0\right) . $$ Is \(\mathrm{V}\) a vector space over \(F\) with these operations? Justify your answer.

Given \(u=3 \mathbf{i}-4 \mathbf{j}+2 \mathbf{k}, \quad v=2 \mathbf{i}+5 \mathbf{j}-3 \mathbf{k}, \quad w=4 \mathbf{i}+7 \mathbf{j}+2 \mathbf{k} . \quad\) Find: (a) \(2 u-3 v\) (b) \(3 u+4 v-2 w\) \((\mathrm{c}) \quad u \cdot v, \quad u \cdot w, \quad v \cdot w\) \((\mathrm{d}) \quad\|u\|,\|v\|,\|w\|\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation
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.