/*! 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 27 Find a unit vector \(u\) orthogo... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 27

Find a unit vector \(u\) orthogonal to \(v=[1,3,4]\) and \(w=[2,-6,-5].\)

Short Answer

Expert verified
The unit vector orthogonal to both v and w is: \(u = \frac{[39, 13, -12]}{\sqrt{1834}}\).

Step by step solution

01

Calculate the cross product of v and w

To find a vector orthogonal to both v and w, we compute the cross product: \[ v \times w = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & 4 \\ 2 & -6 & -5 \end{vmatrix} \]
02

Expand the determinants

Expanding the determinants, we have: \[ v \times w = \hat{i} \begin{vmatrix} 3 & 4 \\ -6 & -5 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 4 \\ 2 & -5 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 3 \\ 2 & -6 \end{vmatrix} \]
03

Calculate the values of the determinants

Calculating the values of the determinants, we have: \[ v \times w = \hat{i} (3*(-5)-4*(-6)) - \hat{j} (1*(-5)-4*2) + \hat{k} (1*(-6)-3*2) \]
04

Simplify the Cartesian components

Simplifying the Cartesian components, we have: \[ v \times w = \hat{i} (15+24) - \hat{j} (-5-8) + \hat{k} (-6-6) = 39\hat{i} + 13\hat{j} - 12\hat{k} \] So, the orthogonal vector is \([39, 13, -12]\).
05

Calculate the magnitude of the orthogonal vector

Now, find the magnitude of the orthogonal vector: \[ \|v \times w\| = \sqrt{(39)^2 + (13)^2 + (-12)^2} \]
06

Simplify the magnitude expression

Simplifying the expression, we get: \[ \|v \times w\| = \sqrt{1521 + 169 + 144} = \sqrt{1834} \]
07

Normalize the orthogonal vector

To find a unit vector, we need to normalize the orthogonal vector: \[ u = \frac{v \times w}{\|v \times w\|} = \frac{[39, 13, -12]}{\sqrt{1834}} \] The unit vector orthogonal to both v and w is: \(u = \frac{[39, 13, -12]}{\sqrt{1834}}\).

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross Product
The cross product is a mathematical operation used in linear algebra to determine a vector that is orthogonal to two given vectors in three-dimensional space. To compute the cross product of two vectors, such as \( v = [1,3,4] \) and \( w = [2,-6,-5] \), you can utilize the determinant of a matrix.The cross product \( v \times w \) can be calculated by creating a 3x3 matrix, where the first row contains the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) (representing the x, y, and z axes respectively), and the second and third rows contain the components of vectors \( v \) and \( w \).
  • First, arrange these vectors in a matrix.
  • Then, expand and compute the determinants for each component.
  • Simplify the resulting expressions to find the cross product.
For example, the calculation results in the vector \([39, 13, -12]\), which is orthogonal to vectors \( v \) and \( w \). This orthogonal vector is important in various applications, including physics and computer graphics.
Orthogonal Vector
In geometry and linear algebra, a vector is orthogonal to another if they meet at a right angle. This means their dot product is zero. The cross product of two vectors gives a third vector which is orthogonal to the first two.For instance, given vectors \( v = [1,3,4] \) and \( w = [2,-6,-5] \), their cross product, \([39, 13, -12]\), is orthogonal to both. This is verified using the dot product:
  • The dot product of \( v \) and the orthogonal vector is: \((1 \times 39) + (3 \times 13) + (4 \times -12) = 0\).
  • The dot product of \( w \) and the orthogonal vector is: \((2 \times 39) + (-6 \times 13) + (-5 \times -12) = 0\).
Both results are zero, proving orthogonality. Orthogonal vectors are used in numerous calculations, especially when finding normal vectors, in projections, and in constructing coordinate systems.
Unit Vector
A unit vector is a vector with a magnitude of one. To convert any vector into a unit vector, you divide the vector by its magnitude. This process is known as normalization. For the orthogonal vector \([39, 13, -12]\) obtained from the cross product, the magnitude is calculated as:\[\sqrt{39^2 + 13^2 + (-12)^2} = \sqrt{1834}\]To normalize the vector, divide each component by the magnitude:
  • \( x \) component: \( \frac{39}{\sqrt{1834}} \)
  • \( y \) component: \( \frac{13}{\sqrt{1834}} \)
  • \( z \) component: \( \frac{-12}{\sqrt{1834}} \)
The resulting vector, \( u = \frac{[39, 13, -12]}{\sqrt{1834}} \), is a unit vector orthogonal to both \( v \) and \( w \). Unit vectors are fundamentally important because they maintain direction but simplify calculations by having a uniform magnitude of one.

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

Consider the following curve \(C\) in \(\mathbf{R}^{3}\) where \(0 \leq t \leq 5\) : $$ F(t)=t^{3} \mathbf{i}-t^{2} \mathbf{j}+(2 t-3) \mathbf{k} $$ (a) Find the point \(P\) on \(C\) corresponding to \(t=2\). (b) Find the initial point \(Q\) and the terminal point \(Q^{\prime}\). (c) Find the unit tangent vector \(\mathbf{T}\) to the curve \(C\) when \(t=2\).

Find the (parametric) equation of the line \(L\) : (a) through the point \(P(2,5,-3)\) and in the direction of \(v=4 \mathbf{i}-5 \mathbf{j}+7 \mathbf{k}\); (b) perpendicular to the plane \(2 x-3 y+7 z=4\) and containing \(P(1,-5,7)\).

Prove Theorem 1.2: For any \(u, v, w\) in \(\mathbf{R}^{n}\) and \(k\) in \(\mathbf{R}\) : (i) \((u+v) \cdot w=u \cdot w+v \cdot w\), (ii) \((k u) \cdot v=k(u \cdot v)\), (iii) \(u \cdot v=v \cdot u\), (iv) \(u \cdot u \geq 0\), and \(u \cdot u=0\) iff \(u=0\). Let \(u=\left(u_{1}, u_{2}, \ldots, u_{n}\right), v=\left(v_{1}, v_{2}, \ldots, v_{n}\right), w=\left(w_{1}, w_{2}, \ldots, w_{n}\right)\) (i) Because \(u+v=\left(u_{1}+v_{1}, u_{2}+v_{2}, \ldots, u_{n}+v_{n}\right)\), $$ \begin{aligned} (u+v) \cdot w &=\left(u_{1}+v_{1}\right) w_{1}+\left(u_{2}+v_{2}\right) w_{2}+\cdots+\left(u_{n}+v_{n}\right) w_{n} \\ &=u_{1} w_{1}+v_{1} w_{1}+u_{2} w_{2}+\cdots+u_{n} w_{n}+v_{n} w_{n} \\ &=\left(u_{1} w_{1}+u_{2} w_{2}+\cdots+u_{n} w_{n}\right)+\left(v_{1} w_{1}+v_{2} w_{2}+\cdots+v_{n} w_{n}\right) \\ &=u \cdot w+v \cdot w \end{aligned} $$ (ii) Because \(k u=\left(k u_{1}, k u_{2}, \ldots, k u_{n}\right)\), $$ (k u) \cdot v=k u_{1} v_{1}+k u_{2} v_{2}+\cdots+k u_{n} v_{n}=k\left(u_{1} v_{1}+u_{2} v_{2}+\cdots+u_{n} v_{n}\right)=k(u \cdot v) $$ (iii) \(u \cdot v=u_{1} v_{1}+u_{2} v_{2}+\cdots+u_{n} v_{n}=v_{1} u_{1}+v_{2} u_{2}+\cdots+v_{n} u_{n}=v \cdot u\) (iv) Because \(u_{i}^{2}\) is nonnegative for each \(i\), and because the sum of nonnegative real numbers is nonnegative, $$ u \cdot u=u_{1}^{2}+u_{2}^{2}+\cdots+u_{n}^{2} \geq 0 $$ Furthermore, \(u \cdot u=0\) iff \(u_{i}=0\) for each \(i\), that is, iff \(u=0\)

Find the dot products \(u \cdot v\) and \(v \cdot u\) where: (a) \(u=(1-2 i, 3+i), v=(4+2 i, 5-6 i)\), (b) \(u=(3-2 i, 4 i, 1+6 i), v=(5+i, 2-3 i, 7+2 i)\). Recall that conjugates of the second vector appear in the dot product $$ \left(z_{1}, \ldots, z_{n}\right) \cdot\left(w_{1}, \ldots, w_{n}\right)=z_{1} \bar{w}_{1}+\cdots+z_{n} \bar{w}_{n} $$ $$ \text { (a) } \begin{aligned} u \cdot v &=(1-2 i)(\overline{4+2 i})+(3+i)(\overline{5-6 i}) \\ &=(1-2 i)(4-2 i)+(3+i)(5+6 i)=-10 i+9+23 i=9+13 i \\ v \cdot u &=(4+2 i)(1-2 i)+(5-6 i)(\overline{3+i}) \\ &=(4+2 i)(1+2 i)+(5-6 i)(3-i)=10 i+9-23 i=9-13 i \end{aligned} $$ $$ \text { (b) } \begin{aligned} u \cdot v &=(3-2 i)(\overline{5+i})+(4 i)(2-3 i)+(1+6 i)(\overline{7+2 i}) \\ &=(3-2 i)(5-i)+(4 i)(2+3 i)+(1+6 i)(7-2 i)=20+35 i \\ v \cdot u &=(5+i)(3-2 i)+(2-3 i)(4 i)+(7+2 i)(\overline{1+6 i}) \\ &=(5+i)(3+2 i)+(2-3 i)(-4 i)+(7+2 i)(1-6 i)=20-35 i \end{aligned} $$ In both cases, \(v \cdot u=\overline{u \cdot v}\). This holds true in general, as seen in Problem 1.40.

Find \(x\) and \(y\) where: (a) \(\quad(x, y+1)=(y-2,6)\) (b) \(x(2, y)=y(1,-2)\)
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Probability and Statistics

Read Explanation

Geometry

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.