/*! 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 78 Two sides of a parallelogram are... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 12: Problem 78

Two sides of a parallelogram are formed by the vectors \(\mathbf{u}\) and \(\mathbf{v}\). Prove that the diagonals of the parallelogram are \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{u}-\mathbf{v}\)

Short Answer

Expert verified
Answer: The diagonals of the parallelogram can be expressed as follows: 1. Diagonal AC: \(\mathbf{u}+\mathbf{v}\) 2. Diagonal BD: \(\mathbf{u}-\mathbf{v}\)

Step by step solution

01

Identify properties of a parallelogram

In a parallelogram, opposite sides are parallel and equal in length. Let's denote the vertices of the parallelogram as A, B, C, and D, with AB formed by vector \(\mathbf{u}\), BC formed by vector \(\mathbf{v}\), CD formed by vector \(\mathbf{u}\), and DA formed by vector \(\mathbf{v}\). We will use these properties to find expressions for the vectors representing the diagonals.
02

Find expression for diagonal AC

To find the vector representing diagonal AC, we can go from point A to point C along two different paths: AB + BC, or AD + DC. Since both paths connect the same two points, their vectors should be the same, so we can write: $$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AD} + \overrightarrow{DC}$$ Now, substitute the given vectors \(\mathbf{u}\) and \(\mathbf{v}\): $$\overrightarrow{AC} = \mathbf{u} + \mathbf{v}$$ So the vector representing diagonal AC can be expressed as \(\mathbf{u}+\mathbf{v}\).
03

Find expression for diagonal BD

Similarly, to find the vector representing diagonal BD, we can go from point B to point D along two different paths: BC + CD, or BA + AD. Since both paths connect the same two points, their vectors should be the same, so we can write: $$\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{BA} + \overrightarrow{AD}$$ Now, substitute the given vectors \(\mathbf{u}\) and \(\mathbf{v}\) and keep in mind that \(\overrightarrow{BA} = -\mathbf{u}\): $$\overrightarrow{BD} = \mathbf{v} + \mathbf{u} = -\mathbf{u} + \mathbf{v}$$ Now, rewrite the sum to show the diagonal BD in terms of \(\mathbf{u}-\mathbf{v}\): $$\overrightarrow{BD} = \mathbf{u} - \mathbf{v}$$ So the vector representing diagonal BD can be expressed as \(\mathbf{u}-\mathbf{v}\).
04

Conclusion

In conclusion, given a parallelogram with sides formed by vectors \(\mathbf{u}\) and \(\mathbf{v}\), we have shown that its diagonals can be expressed in terms of these vectors as: 1. Diagonal AC: \(\mathbf{u}+\mathbf{v}\) 2. Diagonal BD: \(\mathbf{u}-\mathbf{v}\)

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

Parabolic trajectory In Example 7 it was shown that for the parabolic trajectory \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle, \mathbf{a}=\langle 0,2\rangle\) and \(\mathbf{a}=\frac{2}{\sqrt{1+4 t^{2}}}(\mathbf{N}+2 t \mathbf{T}) .\) Show that the second expression for \(\mathbf{a}\) reduces to the first expression.

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 1+2 t, 7-3 t, 6+t\rangle;\\\ &\mathbf{R}(s)=\langle-9+6 s, 22-9 s, 1+3 s\rangle \end{aligned}$$

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. \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|\)

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 \(d+a^{2}+b^{2}+c^{2}>0\)

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{array}{l} \mathbf{r}(t)=\langle 4+5 t,-2 t, 1+3 t\rangle ;\\\ \mathbf{R}(s)=\langle 10 s, 6+4 s, 4+6 s\rangle \end{array}$$
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

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.