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Chapter 21: Problem 108

Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero vectors such that no two of them are collinear and \((\vec{a} \times \vec{b}) \times \vec{c}=\frac{1}{3}|\vec{b} \| \vec{c}| \vec{a}\). If \(\theta\) is the angle between vectors \(\vec{b}\) and \(\vec{c}\), then a value of \(\sin \theta\) (A) \(\frac{-\sqrt{2}}{3}\) (B) \(\frac{2}{3}\) (C) \(\frac{-2 \sqrt{3}}{3}\) (D) \(\frac{2 \sqrt{2}}{3}\)

Short Answer

Expert verified
(D) \(\frac{2\sqrt{2}}{3}\).

Step by step solution

01

Understanding the Vector Product Identity

The problem gives us the vector triple product: \((\vec{a} \times \vec{b}) \times \vec{c} = \frac{1}{3}|\vec{b}\| \vec{c}| \vec{a}\). We start by applying the vector triple product identity: \((\vec{a} \times \vec{b}) \times \vec{c} = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{c})\vec{b}\). Notice that the given equality suggests a simplification where one term must vanish.
02

Simplify and Compare Terms

From the identity, \((\vec{c} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{c})\vec{b} = \frac{1}{3}|\vec{b}\| \vec{c}| \vec{a}\). Since multiplication by \(\vec{a}\) on both sides suggests that \((\vec{a} \cdot \vec{c})\vec{b}\) must be zero, because it cannot balance the scalar term on the right, we conclude \((\vec{a} \cdot \vec{c}) = 0\), meaning \(\vec{a}\) is perpendicular to \(\vec{c}\).
03

Use the Condition on \(\vec{b}\) and \(\vec{c}\)

Since \((\vec{a} \cdot \vec{c}) = 0\), simplify the condition further to \((\vec{c} \cdot \vec{b})\vec{a} = \frac{1}{3}|\vec{b}\| \vec{c}| \vec{a}\). Dividing both sides by \(\vec{a}\) (assuming \(\vec{a}\) is non-zero): \(\vec{c} \cdot \vec{b} = \frac{1}{3}|\vec{b}\||\vec{c}|\).
04

Relate to \(\sin \theta\)

The dot product \(\vec{c} \cdot \vec{b}\) can be expressed as \(|\vec{b}||\vec{c}|\cos \theta\). So, setting \(\vec{c} \cdot \vec{b} = \frac{1}{3}|\vec{b}||\vec{c}|\) gives us \(|\vec{b}||\vec{c}|\cos \theta = \frac{1}{3}|\vec{b}||\vec{c}|\). Simplifying gives \(\cos \theta = \frac{1}{3}\).
05

Determine \(\sin \theta\)

Use the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Substitute \(\cos \theta = \frac{1}{3}\) into this identity: \(\sin^2 \theta = 1 - \left(\frac{1}{3}\right)^2 = \frac{8}{9}\). Thus, \(\sin \theta = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3}\).
06

Select the Correct Option

Comparing the calculated value, \(\sin \theta = \pm \frac{2\sqrt{2}}{3}\), with the given choices verifies option (D) \(\frac{2\sqrt{2}}{3}\). As there are no constraints on \(\theta\) in the problem indicating it must be positive, for the problem's context assume it is the positive one directly listed.

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Key Concepts

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

Vector Triple Product
The concept of the vector triple product involves taking the cross product of two vectors and then multiplying the result with a third vector using another cross product. This operation can be described using the vector triple product identity:
  • \((\vec{a} \times \vec{b}) \times \vec{c} = (\vec{c} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{c})\vec{b}\)
This identity tells us that the product involves both dot and cross products, and it translates to creating a new vector based on a specific combination of products. In the provided exercise, this identity helps to move from a complex vector equation to a more manageable form by suggesting which terms may equal zero, thus simplifying the overall problem.
Cross Product
The cross product is an operation on two vectors in three-dimensional space that results in a third vector that is perpendicular to the plane containing the first two vectors. The magnitude of the cross product of vectors \(\vec{a}\) and \(\vec{b}\) is given by
  • \(|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}| \sin \phi\)
where \(\phi\) is the angle between \(\vec{a}\) and \(\vec{b}\).
This operation is anti-commutative, meaning \(\vec{a} \times \vec{b} = - (\vec{b} \times \vec{a})\). The direction of the resulting vector is determined by the right-hand rule. This concept is critical in the exercise as it forms the basis of manipulating the given vector equality into something that can be solved.
Dot Product
The dot product, also known as the scalar product, is a fundamental operation for vectors where you multiply two vectors to return a scalar. Its formula is
  • \(\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta\)
with \(\theta\) being the angle between the vectors \(\vec{a}\) and \(\vec{b}\). In the exercise solution, the dot product simplifies the computation by yielding a direct relationship for \(\cos \theta\), which is vital later on when we apply the Pythagorean identity. This scalar value serves as a bridge to compute trigonometric functions of the angle between two vectors.
Pythagorean Identity
The Pythagorean identity in trigonometry is critical in relating angles and sides of a triangle, especially in vector calculations. It states that for any angle \(\theta\):
  • \(\sin^2 \theta + \cos^2 \theta = 1\)
Utilizing this identity allows us to find \(\sin \theta\) when \(\cos \theta\) is known or vice versa. In the provided exercise, after calculating \(\cos \theta\) as \(\frac{1}{3}\), the Pythagorean identity helps determine \(\sin \theta\).
The key step is substituting the known \(\cos \theta\) value into the identity to solve for \(\sin^2 \theta\), further leading to the final answer of the problem. This identity simplifies dealing with angles in vector analysis by reducing the complexity of calculations.

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Most popular questions from this chapter

Given two vectors are \(\hat{i}-\hat{j}\) and \(\hat{i}+2 \hat{j}\) the unit vector coplanar with the two vectors and perpendicular to first is: (B) \(\frac{1}{\sqrt{5}}(2 \hat{i}+\hat{j})\) (C) \(\pm \frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\) (D) none of these

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A vector \(a\) has components \(2 p\) and 1 w.r.t. a rectangular cartesian system. This system is rotated through acertain angle about the origin in the counter- clockwise sense. If w.r.t. the new system, \(a\) has components \(p+1\) and 1 , then (A) \(p=0\) (B) \(p=1\) or \(p=-\frac{1}{3}\) (C) \(p=-1\) or \(p=\frac{1}{3}\) (D) \(p=1\) or \(p=-1\)

If \(\vec{u}, \vec{v}, \vec{w}\) are non-coplanar vectors and \(p, q\) are real numbers, then the equality \([3 \vec{u} p \vec{v} p \vec{w}]-[p \vec{v} \vec{w} q \vec{u}]\) \(-\left[\begin{array}{lll}2 \vec{w} & q \vec{v} & q \vec{u}\end{array}\right]=0\) holds for [2009] (A) exactly one value of \((p, q)\) (B) exactly two values of \((p, q)\) (C) more than two but not all values of \((p, q)\) (D) all values of \((p, q)\)
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