/*! 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 32 $$(\mathbf{i} \times \mathbf{j})... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

$$(\mathbf{i} \times \mathbf{j}) \times \mathbf{i}=\mathbf{k} \times \mathbf{i}=\mathbf{j}$$

Short Answer

Expert verified
The solution is \( \mathbf{j} \).

Step by step solution

01

Understand Cross Products in Unit Vectors

The unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are part of the Cartesian coordinate system and have some known cross product relationships. Specifically, \( \mathbf{i} \times \mathbf{j} = \mathbf{k} \), \( \mathbf{j} \times \mathbf{k} = \mathbf{i} \), and \( \mathbf{k} \times \mathbf{i} = \mathbf{j} \). We will use these identities to solve the problem.
02

Simplify \((\mathbf{i} \times \mathbf{j}) \times \mathbf{i}\)

Start by finding \( \mathbf{i} \times \mathbf{j} \), which equals \( \mathbf{k} \). Then substitute \( \mathbf{k} \) into the expression for the next cross product: \( \mathbf{k} \times \mathbf{i} \).
03

Evaluate \(\mathbf{k} \times \mathbf{i}\)

According to the vector cross product identity, \( \mathbf{k} \times \mathbf{i} = \mathbf{j} \). Therefore, the expression \( (\mathbf{i} \times \mathbf{j}) \times \mathbf{i} \) simplifies to \( \mathbf{j} \).
04

Confirm Final Result

Check each step corresponds correctly with the known cross product identities. Since \( (\mathbf{i} \times \mathbf{j}) \times \mathbf{i} = \mathbf{k} \times \mathbf{i} = \mathbf{j} \), the problem statement is satisfied.

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 91Ó°ÊÓ!

Key Concepts

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

Cross Product
The cross product is a fundamental operation in vector calculus, particularly useful in physics and engineering. It helps determine a vector orthogonal to two given vectors in three dimensions. Suppose we have two vectors \( \mathbf{A} = a_1 \mathbf{i} + a_2 \mathbf{j} + a_3 \mathbf{k} \) and \( \mathbf{B} = b_1 \mathbf{i} + b_2 \mathbf{j} + b_3 \mathbf{k} \). The cross product, \( \mathbf{A} \times \mathbf{B} \), is defined as:\[\mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k}\]The result of the cross product is always a vector and follows the right-hand rule:
  • Point your index finger in the direction of the first vector.
  • Your middle finger in the direction of the second vector.
  • Your thumb will point in the direction of the cross product.
Remember, the magnitude of \( \mathbf{A} \times \mathbf{B} \) is given by:\[|\mathbf{A} \times \mathbf{B}| = |\mathbf{A}| |\mathbf{B}| \sin \theta\]where \(\theta\) is the angle between \(\mathbf{A}\) and \(\mathbf{B}\). The cross product is zero if the two vectors are parallel.
Unit Vectors
Unit vectors are vectors with a magnitude of 1. They are crucial in defining directions in space and simplify mathematical expressions. In the Cartesian coordinate system, unit vectors are represented as \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), which correspond to the directions of the x, y, and z axes respectively.
  • \( \mathbf{i} \) points in the x-direction: \( (1, 0, 0) \)
  • \( \mathbf{j} \) points in the y-direction: \( (0, 1, 0) \)
  • \( \mathbf{k} \) points in the z-direction: \( (0, 0, 1) \)
Unit vectors help break down vectors into components in vector calculations. A general vector \( \mathbf{V} \), like \( (v_x, v_y, v_z) \), can be expressed as:\[\mathbf{V} = v_x \mathbf{i} + v_y \mathbf{j} + v_z \mathbf{k}\]This breakdown enables easy computation of vector operations such as addition, scalar multiplication, and importantly, the cross product.
Cartesian Coordinate System
The Cartesian coordinate system is a foundational tool in mathematics and physics for representing and analyzing vector spaces. It utilizes three perpendicular axes: x, y, and z, intersecting at an origin.
  • Each point in space is described by three coordinates \( (x, y, z) \).
  • The system follows right-handed orientation, which affects operations like cross products.
  • It allows for easy visualization and calculation of vector relationships.
Vectors in this system are represented through their components along each axis. For instance, the position vector \( \mathbf{r} \) for a point is:\[\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\]The Cartesian coordinate system simplifies mathematical operations such as vector addition, dot product, and cross product, providing a straightforward framework to approach many physical phenomena.

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

In Problems 75 and 76, the cross product of the direction vector of the line with the normal vector of the given plane will be a normal vector to the desired plane. A direction vector of the line is \langle 3,5,2\rangle and a normal vector to the given plane is \(\langle 2,-4,-1\rangle .\) A normal vector to the desired plane is \(\langle-3,5,2\rangle \times\langle 2,-4,-1\rangle=\langle 3,1,2\rangle .\) A point on the line, and hence in the plane, is (2,-2,8) The equation of the plane is \(3(x-2)+(y+2)+2(z-8)=0\) or \(3 x+y+2 z=20\).

To show that the set of nonzero orthogonal vectors is linearly independent we set \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{n} \mathbf{v}_{n}=\mathbf{0}\). For \(0 \leq i \leq n,\) $$\left(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{i} \mathbf{v}_{i} \cdots+c_{n} \mathbf{v}_{n}\right) \cdot \mathbf{v}_{i}=c_{1} \mathbf{v}_{1} \cdot \mathbf{v}_{i}+c_{2} \mathbf{v}_{2} \cdot \mathbf{v}_{i}+\cdots+c_{i} \mathbf{v}_{i} \cdot \mathbf{v}_{i} \cdots+c_{n} \mathbf{v}_{n} \cdot \mathbf{v}_{i}=c_{i}\left\|\mathbf{v}_{i}\right\|^{2},$$ so \(c_{i}\left\|\mathbf{v}_{i}\right\|^{2}=0\) because $$\left(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{i} \mathbf{v}_{i} \cdots+c_{n} \mathbf{v}_{n}\right) \cdot \mathbf{v}_{i}=\mathbf{0} \cdot \mathbf{v}_{i}=0.$$ Since \(\mathbf{v}_{i}\) is a nonzero vector, \(c_{i}=0 .\) Thus, the assumption that \(c_{1} \mathbf{v}_{1}+c_{2} \mathbf{v}_{2}+\cdots+c_{n} \mathbf{v}_{n}=\mathbf{0}\) leads to \(c_{1}=c_{2}=\cdots=c_{n}=0,\) and the set is linearly independent.

$$\|\mathbf{a}+\mathbf{b}\|=\|\langle 5,12\rangle\|=\sqrt{25+144}=13 ; \quad \mathbf{u}=\frac{1}{13}\langle 5,12\rangle=\left\langle\frac{5}{13}, \frac{12}{13}\right\rangle$$

In Problems 73 and 74, the cross product of the normal vectors to the two planes will be a vector parallel to both planes, and hence a direction vector for a line parallel to the two planes. Normal vectors are \langle 1,1,-4\rangle and \(\langle 2,-1,1\rangle .\) A direction vector is \\[\langle 1,1,-4\rangle \times\langle 2,-1,1\rangle=\langle-3,-9,-3\rangle=-3\langle 1,3,1\rangle\\] Equations of the line are \(x=5+t, y=6+3 t, z=-12+t\).

$$\begin{array}{l} \overrightarrow{P_{1} P_{2}}=\mathbf{j}+2 \mathbf{k} ; \overrightarrow{P_{2} P_{3}}=2 \mathbf{i}+\mathbf{j}-2 \mathbf{k} \\ \overrightarrow{P_{1} P_{2}} \times \overrightarrow{P_{2} P_{3}}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 1 & 2 \\ 2 & 1 & -2 \end{array}\right|=\left|\begin{array}{cc} 1 & 2 \\ 1 & -2 \end{array}\right| \mathbf{i}-\left|\begin{array}{cc} 0 & 2 \\ 2 & -2 \end{array}\right| \mathbf{j}+\left|\begin{array}{cc} 0 & 1 \\ 2 & 1 \end{array}\right| \mathbf{k}=-4 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k} \\ A=\frac{1}{2}\|-4 \mathbf{i}+4 \mathbf{j}-2 \mathbf{k}\|=3 \text { sq. units } \end{array}$$

See all solutions

Recommended explanations on Physics Textbooks

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.