/*! 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 51 Use cross products to determine ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 51

Use cross products to determine whether the points \(A, B,\) and C are collinear. $$A(-3,-2,1), B(1,4,7), \text { and } C(4,10,14)$$

Short Answer

Expert verified
Answer: No, the points A, B, and C are not collinear.

Step by step solution

01

Define the vectors

Let's define the vectors, using points A and B to create vector AB, and points B and C to create vector BC: $$\overrightarrow{AB} = B - A = (1, 4, 7) - (-3, -2, 1) = (4, 6, 6)$$ $$\overrightarrow{BC} = C - B = (4, 10, 14) - (1, 4, 7) = (3, 6, 7)$$
02

Find the cross product of AB and BC

Now, let's find the cross product of the two vectors, \(\overrightarrow{AB} \times \overrightarrow{BC}\): $$ \overrightarrow{AB} \times \overrightarrow{BC} = \begin{bmatrix} i & j & k \\ 4 & 6 & 6 \\ 3 & 6 & 7 \end{bmatrix}$$
03

Evaluate the cross product

Evaluate the determinant of the matrix in Step 2: $$ \overrightarrow{AB} \times \overrightarrow{BC} = i(6 \cdot 7 - 6 \cdot 6) - j(4 \cdot 7 - 6 \cdot 3) + k(4 \cdot 6 - 6 \cdot 3)$$ $$\overrightarrow{AB} \times \overrightarrow{BC} = 6i - 6j + 0k = (6,-6,0)$$
04

Check if the cross product is the zero vector

We check if the result of the cross product is the zero vector \((0, 0, 0)\). In this case: $$\overrightarrow{AB} \times \overrightarrow{BC} = (6, -6, 0)$$ The cross product is not the zero vector, so the points A, B, and C are not collinear.

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

An object on an inclined plane does not slide provided the component of the object's weight parallel to the plane \(\left|\mathbf{W}_{\text {par }}\right|\) is less than or equal to the magnitude of the opposing frictional force \(\left|\mathbf{F}_{\mathrm{f}}\right|\). The magnitude of the frictional force, in turn, is proportional to the component of the object's weight perpendicular to the plane \(\left|\mathbf{W}_{\text {perp }}\right|\) (see figure). The constant of proportionality is the coefficient of static friction, \(\mu\) a. Suppose a 100 -lb block rests on a plane that is tilted at an angle of \(\theta=20^{\circ}\) to the horizontal. Find \(\left|\mathbf{W}_{\text {parl }}\right|\) and \(\left|\mathbf{W}_{\text {perp }}\right|\) b. The condition for the block not sliding is \(\left|\mathbf{W}_{\mathrm{par}}\right| \leq \mu\left|\mathbf{W}_{\text {perp }}\right| .\) If \(\mu=0.65,\) does the block slide? c. What is the critical angle above which the block slides with \(\mu=0.65 ?\)

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. What conditions on \(\mathbf{u}\) and \(\mathbf{v}\) lead to equality in the Cauchy-Schwarz Inequality?

Use projections to find a general formula for the (smallest) distance between the point \(\left.P\left(x_{0}, y_{0}\right) \text { and the line } a x+b y=c . \text { (See Exercises } 62-65 .\right)\)

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle-2,2\rangle, \mathbf{v}=\langle 2,1\rangle\)

For the following vectors u and \(\mathbf{v}\) express u as the sum \(\mathbf{u}=\mathbf{p}+\mathbf{n},\) where \(\mathbf{p}\) is parallel to \(\mathbf{v}\) and \(\mathbf{n}\) is orthogonal to \(\mathbf{v}\). \(\mathbf{u}=\langle 4,3,0\rangle, \mathbf{v}=\langle 1,1,1\rangle\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Probability and Statistics

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.