/*! 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 70 Find vectors parallel to \(\math... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 70

Find vectors parallel to \(\mathbf{v}\) of the given length. $$\mathbf{v}=\overrightarrow{P Q} \text { with } P(1,0,1) \text { and } Q(2,-1,1) ; \text { length }=3$$

Short Answer

Expert verified
Find the vector parallel to the vector $\mathbf{v}=\overrightarrow{PQ}$, where P(1, 0, 1) and Q(2, -1, 1), with a length of 3. Solution: The vector parallel to $\mathbf{v}$ with a length of 3 is $\mathbf{w} = \langle 3\sqrt{2}/2, -3\sqrt{2}/2, 0 \rangle$.

Step by step solution

01

Find vector \(\mathbf{v}=\overrightarrow{PQ}\)

Given the coordinates of points P(1, 0, 1) and Q(2, -1, 1), we can find the vector \(\mathbf{v}\) as follows: $$\overrightarrow{PQ} = Q - P = (2 - 1, -1 - 0, 1 - 1) = (1, -1, 0)$$ So, the vector \(\mathbf{v} = \langle 1, -1, 0 \rangle\).
02

Find the magnitude of \(\mathbf{v}\)

The magnitude of a vector is calculated using the formula: $$||\mathbf{v}|| = \sqrt{x^2 + y^2 + z^2}$$ Now, we can find the magnitude of the vector \(\mathbf{v}\), which is: $$||\mathbf{v}|| = \sqrt{1^2 + (-1)^2 + 0^2} = \sqrt{1 + 1} = \sqrt{2}$$
03

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

To find the unit vector of \(\mathbf{v}\), divide the vector by its magnitude: $$\mathbf{u}_v = \frac{\mathbf{v}}{||\mathbf{v}||} = \frac{\langle 1, -1, 0 \rangle}{\sqrt{2}} = \left\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right\rangle$$
04

Find the vector parallel to \(\mathbf{v}\) with the given length

Given the length of the desired vector is 3, we can find the vector parallel to \(\mathbf{v}\) by multiplying the unit vector by the given length: $$\mathbf{w} = 3\mathbf{u}_v = 3\left\langle \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0 \right\rangle = \left\langle 3\frac{1}{\sqrt{2}}, 3(-\frac{1}{\sqrt{2}}), 3(0) \right\rangle = \langle 3\sqrt{2}/2, -3\sqrt{2}/2, 0 \rangle$$ Thus, the vector parallel to \(\mathbf{v}\) with a length of 3 is \(\mathbf{w} = \langle 3\sqrt{2}/2, -3\sqrt{2}/2, 0 \rangle\).

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Most popular questions from this chapter

An object moves along a path given by \(\mathbf{r}(t)=\langle a \cos t+b \sin t, c \cos t+d \sin t\rangle, \quad\) for \(0 \leq t \leq 2 \pi\) a. What conditions on \(a, b, c,\) and \(d\) guarantee that the path is a circle? b. What conditions on \(a, b, c,\) and \(d\) guarantee that the path is an ellipse?

Determine whether the following statements are true and give an explanation or counterexample. a. The line \(\mathbf{r}(t)=\langle 3,-1,4\rangle+t\langle 6,-2,8\rangle\) passes through the origin. b. Any two nonparallel lines in \(\mathbb{R}^{3}\) intersect. c. The curve \(\mathbf{r}(t)=\left\langle e^{-t}, \sin t,-\cos t\right\rangle\) approaches a circle as \(t \rightarrow \infty\). d. If \(\mathbf{r}(t)=e^{-t^{2}}\langle 1,1,1\rangle\) then \(\lim _{t \rightarrow \infty} \mathbf{r}(t)=\lim _{t \rightarrow-\infty} \mathbf{r}(t)\).

Evaluate the following limits. $$\lim _{t \rightarrow \pi / 2}\left(\cos 2 t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2 t}{\pi} \mathbf{k}\right)$$

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Maximum curvature Consider the "superparabolas" \(f_{n}(x)=x^{2 n},\) where \(n\) is a positive integer. a. Find the curvature function of \(f_{n},\) for \(n=1,2,\) and 3 b. Plot \(f_{n}\) and their curvature functions, for \(n=1,2,\) and 3 and check for consistency. c. At what points does the maximum curvature occur, for \(n=1,2,3 ?\) d. Let the maximum curvature for \(f_{n}\) occur at \(x=\pm z_{n} .\) Using either analytical methods or a calculator determine \(\lim _{n \rightarrow \infty} z_{n}\) Interpret your result.
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