/*! 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 68 Find vectors parallel to \(\math... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 68

Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=\langle 3,-2,6\rangle ; \text { length }=10$$

Short Answer

Expert verified
Question: Given the vector \(\mathbf{v} = \langle 3, -2, 6 \rangle\), find a vector parallel to it with a length of \(10\). Answer: The vector parallel to \(\mathbf{v}\) with a length of \(10\) is \(\mathbf{w} = \langle \frac{30}{7}, -\frac{20}{7}, \frac{60}{7} \rangle\).

Step by step solution

01

Calculate the magnitude(length) of vector \(\mathbf{v}\)

To calculate the magnitude of \(\mathbf{v}=\langle 3,-2,6\rangle\), use the following formula: \(||\mathbf{v}||=\sqrt{x^2 + y^2 + z^2}\), where \(x\), \(y\), and \(z\) are the components of \(\mathbf{v}\). \(||\mathbf{v}||=\sqrt{3^2+(-2)^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7\)
02

Calculate the unit vector of \(\mathbf{v}\)

To find the unit vector of \(\mathbf{v}\), divide each component of \(\mathbf{v}\) by its magnitude: Unit vector of \(\mathbf{v}\) = \(\frac{\mathbf{v}}{||\mathbf{v}||}\) = \(\frac{\langle 3, -2, 6\rangle}{7} = \langle \frac{3}{7}, \frac{-2}{7}, \frac{6}{7} \rangle\)
03

Scale the unit vector to match the desired length

Now, to get the vector parallel to \(\mathbf{v}\) with the desired length of \(10\), scale the unit vector by the desired length: \(\mathbf{w} = 10 \times \langle \frac{3}{7}, \frac{-2}{7}, \frac{6}{7} \rangle= \langle 10 \times \frac{3}{7}, 10 \times \frac{-2}{7}, 10 \times \frac{6}{7} \rangle = \langle \frac{30}{7}, -\frac{20}{7}, \frac{60}{7} \rangle\) The vector parallel to \(\mathbf{v}\) with a length of 10 is \(\mathbf{w}=\langle \frac{30}{7}, -\frac{20}{7}, \frac{60}{7} \rangle\).

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

For constants \(a, b, c,\) and \(d,\) show that the equation $$x^{2}+y^{2}+z^{2}-2 a x-2 b y-2 c z=d$$ describes a sphere centered at \((a, b, c)\) with radius \(r,\) where \(r^{2}=d+a^{2}+b^{2}+c^{2},\) provided \(d+a^{2}+b^{2}+c^{2}>0\)

Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are nonzero vectors in \(\mathbb{R}^{3}\). a. Prove that the equation \(\mathbf{u} \times \mathbf{z}=\mathbf{v}\) has a nonzero solution \(\mathbf{z}\) if and only if \(\mathbf{u} \cdot \mathbf{v}=0 .\) (Hint: Take the dot product of both sides with v.) b. Explain this result geometrically.

For the given points \(P, Q,\) and \(R,\) find the approximate measurements of the angles of \(\triangle P Q R\). $$P(1,-4), Q(2,7), R(-2,2)$$

Note that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line.

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Probability and Statistics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Calculus

Read Explanation

Applied Mathematics

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.