/*! 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 24 If \(a, b, c\) are three non-par... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 24

If \(a, b, c\) are three non-parallel unit vectors such that \(a \times(b \times c)=\frac{1}{2} b\), then the angles which a makes with \(b\) and \(c\) are (A) \(90^{\circ}, 60^{\circ}\) (B) \(45^{\circ}, 60^{\circ}\) (C) \(30^{\circ}, 60^{\circ}\) (D) none of these

Short Answer

Expert verified
The angles are (A) \(90^{\circ}, 60^{\circ}\).

Step by step solution

01

Understanding the Vector Triple Product

The vector triple product identity can be expressed as: \( a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \). Our given condition is \( a \times(b \times c) = \frac{1}{2} b \). Substituting this into the identity, we get the equation \( (a \cdot c)b - (a \cdot b)c = \frac{1}{2} b \).
02

Analyzing the Equation

Comparing the equation \( (a \cdot c)b - (a \cdot b)c = \frac{1}{2} b \) with \( yb \), it is clear that the only term containing \(b\) is \( (a \cdot c)b \). Thus, it requires that \( (a \cdot c) = \frac{1}{2} \) for the terms to match.
03

Understanding the Dot Product Implication

Since \( (a \cdot c) = \frac{1}{2} \), it represents the cosine of the angle \( \theta \) between vector \(a\) and vector \(c\), i.e., \( \cos(\theta) = \frac{1}{2} \). This implies that \( \theta = 60^{\circ} \).
04

Simplifying the Problem

Now consider the remaining condition fed into the equation: \(- (a \cdot b)c = 0\). As the terms are vectors, \( (a \cdot b)c \) equals the zero vector only if \( (a \cdot b) = 0 \). This implies \( cos(\phi) = 0 \), where \( \phi \) is the angle between \(a\) and \(b\).
05

Solving for the Angles

From the above steps, \( a \cdot b = 0 \) indicates that the angle \(\phi = 90^{\circ}\). Meanwhile, \( a \cdot c = 0.5\) indicates \( \theta = 60^{\circ}\). Hence, the angles between \(a\) and \(b\) is \( 90^{\circ}\), and between \(a\) and \(c\) is \( 60^{\circ}\).

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.

Vector Triple Product
The vector triple product is a formula used when evaluating the cross product of one vector with the cross product of two other vectors. It involves three vectors and can be expressed with the identity:
\( a \times (b \times c) = (a \cdot c)b - (a \cdot b)c \).
This formula helps simplify complex expressions involving vector cross and dot products.
  • The term \( (a \cdot c)b \) projects vector \(b\) in the direction scaled by \(a \cdot c\).
  • The term \( -(a \cdot b)c \) similarly projects \(c\) in the opposite direction scaled by \(a \cdot b\).
Understanding this identity is key for tackling problems that involve multiple vector products in one expression.
In this exercise, recognizing the vector triple product allows us to set up an equation using the given vectors. This equation assists in deriving relationships and solving for angles, helping us understand the geometric orientation between vectors.
Vector Cross Product
The vector cross product is a binary operation on two vectors in three-dimensional space, resulting in a vector that is perpendicular to both of the original vectors. Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the cross product \( \mathbf{a} \times \mathbf{b} \) is a vector.
This can be calculated using the determinant of a matrix that includes the unit vectors \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \), as well as the components of \( \mathbf{a} \) and \( \mathbf{b} \).
An important property of the cross product is its orthogonality.
  • The resulting vector direction can be determined by the right-hand rule.
  • The magnitude is given by \( |\mathbf{a}||\mathbf{b}| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \).
In the context of this exercise, calculating the cross product \( a \times (b \times c) \) using the vector triple product formula simplifies to analyzing directions and projecting components, helping you derive angles effectively.
Vector Dot Product
The vector dot product is a measure of how much two vectors "point" in the same direction. It is a scalar product that results in a single number. For two vectors \( \mathbf{a} \) and \( \mathbf{b} \), the dot product is calculated as:
\( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos(\phi) \),
where \( \phi \) is the angle between them.
This product tells you about the cosine of the angle between the two vectors and indicates whether the vectors are parallel, perpendicular, or neither.
  • If \( \mathbf{a} \cdot \mathbf{b} = 0 \), the vectors are perpendicular (\( 90^{\circ} \)).
  • If \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \), they are parallel and point in the same direction.
In this exercise, knowing that \( a \cdot c = \frac{1}{2} \) helped us identify an angle of \( 60^{\circ} \) between vectors \(a\) and \(c\), and \( a \cdot b = 0 \) confirmed that vectors \(a\) and \(b\) are perpendicular.

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 the following questions an Assertion (A) is given followed by a Reason (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason(R) is False (D) Assertion(A) is False, Reason(R) is True Assertion: If \(a, b, c\) are three non-coplanar, non-zero vectors, then \((a \cdot a) b \times c+(a \cdot b) c \times a+(a \cdot c) a \times b=[b c a] a\) Reason: If the vectors \(a, b, c\) are non-coplanar, then so are \(b \times c, c \times a, a \times b\)

A non-zero vector \(a\) is parallel to the line of intersection of the plane determined by the vectors \(i, i+j\) and the plane determined by the vectors \(i-j, i+k\). The angle between \(a\) and the vector \(i-2 j+2 k\) is (A) \(\frac{\pi}{3}\) (B) \(\frac{\pi}{4}\) (C) \(\frac{3 \pi}{4}\) (D) none of these

Let \(\vec{a}=\hat{i}-\hat{k}, \vec{b}=x \hat{i}+\hat{j}+(1-x) \hat{k} \quad\) and \(\vec{c}=y \hat{i}+x \hat{j}\) \(+(1+x-y) \hat{k} .\) Then \([\vec{a}, \vec{b}, \vec{c}]\) depends on \(\quad[\mathbf{2 0 0 5}]\) (A) only \(y\) (B) only \(x\) (C) both \(x\) and \(y\) (D) neither \(x\) nor \(y\)

The triangle \(A B C\) is defined by the vertices \(A(1,-2,2)\), \(B(1,4,0)\) and \(C(-4,1,1) .\) Let \(M\) be the foot of the altitude drawn from the vertex \(B\) to side \(A C\). Then, \(B M=\) (A) \((-20 / 7,-30 / 7,10 / 7)\) (B) \((-20,-30,10)\) (C) \((2,3,-1)\) (D) none of these

Let \(\nabla(x, y, z)=V_{1} i+V_{2} j+V_{3} k\) be defined and differentiable at each point \((x, y, z)\) in a certain region of space. Then, the divergence of \(V\), written \(\nabla \cdot V\) or div \(V\) is defined by $$ \begin{aligned} \nabla \cdot V &=\left(\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\right) \cdot\left(V_{1} i+V_{2} j+V_{3} k\right) \\ &=\frac{\partial V_{1}}{\partial x}+\frac{\partial V_{2}}{\partial y}+\frac{\partial V_{3}}{\partial z} \end{aligned} $$ Here, \(\nabla=\frac{\partial}{\partial x} i+\frac{\partial}{\partial y} j+\frac{\partial}{\partial z} k\) is the del operator. Note the analogy with \(A \cdot B=A_{1} B_{1}+A_{2} B_{2}+A_{3} B_{3} .\) Also, note that \(\vec{\nabla} \cdot V \neq V \cdot \nabla\). If \(r=x i+y j+z k\), then \(\nabla \cdot\left(\frac{r}{r^{3}}\right)=\) (A) 0 (B) 1 (C) \(-1\) (D) none of these
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Calculus

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.