/*! 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 23 Find the area of the parallelogr... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 23

Find the area of the parallelogram that has two adjacent sides \(\mathbf{u}\) and \(\mathbf{v}\) $$\mathbf{u}=2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}, \mathbf{v}=3 \mathbf{i}+2 \mathbf{j}-\mathbf{k}$$

Short Answer

Expert verified
Answer: The area of the parallelogram is $\sqrt{90}$.

Step by step solution

01

Write down the given vectors

The given vectors are: $$\mathbf{u}=2 \mathbf{i}-\mathbf{j}-2 \mathbf{k}$$ $$\mathbf{v}=3 \mathbf{i}+2 \mathbf{j}-\mathbf{k}$$
02

Find the cross product of the vectors

The cross product of the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by: $$\mathbf{u} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & -2 \\ 3 & 2 & -1 \end{vmatrix}$$ To calculate the determinant, we can use the formula: $$\mathbf{u} \times \mathbf{v} = \mathbf{i}\begin{vmatrix} -1 & -2 \\ 2 & -1 \end{vmatrix} - \mathbf{j}\begin{vmatrix} 2 & -2 \\ 3 & -1 \end{vmatrix} + \mathbf{k}\begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix}$$
03

Calculate the cross product

Now calculate the cross product as follows: $$\mathbf{u} \times \mathbf{v} = \mathbf{i}((-1)(-1) - (-2)(2)) - \mathbf{j}((2)(-1) - (-2)(3)) + \mathbf{k}((2)(2) - (-1)(3))$$ $$\mathbf{u} \times \mathbf{v} = \mathbf{i}(1 + 4) - \mathbf{j}(-2 + 6) + \mathbf{k}(4 + 3)$$ $$\mathbf{u} \times \mathbf{v} = 5\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}$$
04

Find the magnitude of the cross product

The magnitude of the cross product is given by: $$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{ (5)^2 + (4)^2 + (7)^2 }$$ $$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{ 25 + 16 + 49 }$$ $$\|\mathbf{u} \times \mathbf{v}\| = \sqrt{ 90 }$$
05

Find the area of the parallelogram

The area of the parallelogram formed by the vectors \(\mathbf{u}\) and \(\mathbf{v}\) is equal to the magnitude of their cross product: $$Area = \|\mathbf{u} \times \mathbf{v}\|$$ $$Area = \sqrt{ 90 }$$ So, the area of the parallelogram is \(\sqrt{90}\).

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

Consider the motion of an object given by the position function $$\mathbf{r}(t)=f(t)\langle a, b, c\rangle+\left(x_{0}, y_{0}, z_{0}\right\rangle, \text { for } t \geq 0$$ where \(a, b, c, x_{0}, y_{0},\) and \(z_{0}\) are constants and \(f\) is a differentiable scalar function, for \(t \geq 0\) a. Explain why this function describes motion along a line. b. Find the velocity function. In general, is the velocity constant in magnitude or direction along the path?

Consider the curve described by the vector function \(\mathbf{r}(t)=\left(50 e^{-t} \cos t\right) \mathbf{i}+\left(50 e^{-t} \sin t\right) \mathbf{j}+\left(5-5 e^{-t}\right) \mathbf{k},\) for \(t \geq 0\). a. What is the initial point of the path corresponding to \(\mathbf{r}(0) ?\) b. What is \(\lim _{t \rightarrow \infty} \mathbf{r}(t) ?\) c. Sketch the curve. d. Eliminate the parameter \(t\) to show that \(z=5-r / 10\), where \(r^{2}=x^{2}+y^{2}\).

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Use the vectors \(\mathbf{u}=\langle\sqrt{a}, \sqrt{b}\rangle\) and \(\mathbf{v}=\langle\sqrt{b}, \sqrt{a}\rangle\) to show that \(\sqrt{a b} \leq(a+b) / 2,\) where \(a \geq 0\) and \(b \geq 0\).

Find the points (if they exist) at which the following planes and curves intersect. $$y=1 ; \mathbf{r}(t)=\langle 10 \cos t, 2 \sin t, 1\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Let \(\mathbf{u}=\left\langle u_{1}, u_{2}, u_{3}\right\rangle\) \(\mathbf{v}=\left\langle v_{1}, v_{2}, v_{3}\right\rangle\), and \(\mathbf{w}=\) \(\left\langle w_{1}, w_{2}, w_{3}\right\rangle\). Let \(c\) be a scalar. Prove the following vector properties. \(\mathbf{u} \cdot(\mathbf{v}+\mathbf{w})=\mathbf{u} \cdot \mathbf{v}+\mathbf{u} \cdot \mathbf{w}\)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Decision Maths

Read Explanation

Pure Maths

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.