/*! 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 185 Find the area of the triangle de... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 185

Find the area of the triangle determined by the points \(\mathrm{P}_{1}(2,2,0), \mathrm{P}_{2}(-1,0,1)\) and \(\mathrm{P}_{3}(0,4,3)\) by using the cross-product.

Short Answer

Expert verified
The area of the triangle determined by points \(\mathrm{P}_{1}(2,2,0)\), \(\mathrm{P}_{2}(-1,0,1)\), and \(\mathrm{P}_{3}(0,4,3)\) is \(\frac{\sqrt{116}}{2}\).

Step by step solution

01

Find the Vectors

We are given points \(\mathrm{P}_{1}(2,2,0)\), \(\mathrm{P}_{2}(-1,0,1)\), and \(\mathrm{P}_{3}(0,4,3)\). To find the vectors formed by these points, subtract the coordinates of each point as follows: $$\vec{v} = \mathrm{P}_{2} - \mathrm{P}_{1} = (-1-2, 0-2, 1-0) = (-3, -2, 1)$$ $$\vec{w} = \mathrm{P}_{3} - \mathrm{P}_{1} = (0-2, 4-2, 3-0) = (-2, 2, 3)$$
02

Compute the Cross-Product

Now we will compute the cross-product of the vectors \(\vec{v}\) and \(\vec{w}\) as follows: $$\vec{v} \times \vec{w} = \begin{bmatrix} i & j & k \\ -3 & -2 & 1 \\ -2 & 2 & 3 \end{bmatrix} = i(-2 \cdot 3 - 1 \cdot 2) - j(-3 \cdot 3 - 1 \cdot 2) + k(-3 \cdot 2 - (-2) \cdot 2)$$ $$\vec{v} \times \vec{w} = (-6,-8,4)$$
03

Calculate the Magnitude

Next, we will compute the magnitude of the cross-product: $$\Vert \vec{v} \times \vec{w} \Vert = \sqrt{(-6)^{2} + (-8)^{2} + (4)^{2}} = \sqrt{36 + 64 + 16} = \sqrt{116}$$
04

Compute the Area of the Triangle

Finally, the area of the triangle determined by the given points is half the magnitude of the cross-product: Area \(= \frac{1}{2} \Vert \vec{v} \times \vec{w} \Vert = \frac{1}{2} \sqrt{116} = \frac{\sqrt{116}}{2}\) The area of the triangle determined by points \(\mathrm{P}_{1}(2,2,0)\), \(\mathrm{P}_{2}(-1,0,1)\), and \(\mathrm{P}_{3}(0,4,3)\) is \(\frac{\sqrt{116}}{2}\).

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

Find \(\mathrm{e}^{\mathrm{A}}=\mathrm{f}(\mathrm{A})\), where \(A=\begin{array}{rrr}\mid 3 & -3 & 3 \\ \mid-1 & 5 & -2 \mid \\ -1 & 3 & 0\end{array}\)

Find a vector orthogonal to \(\mathrm{A}=(2,2,-1)\) and \(\mathrm{B}=(1,2,1)\).

By forming the augmented matrix and row reducing, determine the solutions of the following system $$ \begin{aligned} &2 x-y+3 z=4 \\ &3 x+2 z=5 \\ &-2 x+y+4 z=6 \end{aligned} $$

Reduce the following matrices to echelon form and then to row reduced echelon form. (a) \(\quad \begin{array}{rrrrr} & 0 & 1 & 3 & -2 \\ \mathrm{~A}= & 12 & 1 & -4 & 3 \\ & & 2 & 3 & 2 & -1\end{array}\) (b) \(\quad \begin{array}{rrr} & \mid 6 & 3 & -4 \\ \mathrm{~A} & =\mid-4 & 1 & -6 \\ & \mid 1 & 2 & -5\end{array} \mid\)

Evaluate det A where: \(\mathrm{A}=\begin{array}{rrr}10 & 1 & 5 \\ 3 & -6 & 9 \\ 12 & 6 & 1\end{array}\)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Probability and Statistics

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.