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Chapter 2: Problem 78

Let \(\hat{\mathbf{r}}\) be a unit vector satisfying \(\hat{\mathbf{r}} \times \mathbf{a}=\mathbf{b}\), where \(|\mathbf{a}|=\sqrt{3}\) and \(|\mathbf{b}|=\sqrt{2}\). Then, (a) \(\hat{\mathbf{r}}=\frac{2}{3}(\mathbf{a}+\mathbf{a} \times \mathbf{b})\) (b) \(\hat{r}=\frac{1}{3}(a+a \times b)\) (c) \(\hat{r}=\frac{2}{3}(a-a \times b)\) (d) \(\hat{r}=\frac{1}{3}(-a+a \times b)\)

Short Answer

Expert verified
The correct answer is (a) \(\hat{\mathbf{r}} = \frac{2}{3}(\mathbf{a} + \mathbf{a} \times \mathbf{b})\).

Step by step solution

01

Understand the Cross Product Formula

The cross product of vectors is given by \( \mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin(\theta)\hat{\mathbf{n}} \), where \( \hat{\mathbf{n}} \) is a unit vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). In this problem, \( \hat{\mathbf{r}} \times \mathbf{a} = \mathbf{b} \). This implies that \( \mathbf{b} \) is perpendicular to \( \hat{\mathbf{r}} \) and \( \mathbf{a} \).
02

Use Magnitudes and Cross Product Properties

Given the magnitude of vectors \( |\mathbf{a}| = \sqrt{3} \) and \( |\mathbf{b}| = \sqrt{2} \), along with the unit vector \( \hat{\mathbf{r}} \), use the formula: \( \hat{\mathbf{r}} \times \mathbf{a} = \mathbf{b} \). Therefore, we have: \( |\mathbf{b}| = |\mathbf{a}|\cdot|\hat{\mathbf{r}}|\cdot\sin(\theta) \). Since \( \hat{\mathbf{r}} \) is a unit vector, \(|\hat{\mathbf{r}}| = 1 \).
03

Calculate the Sine Component

Substitute the known magnitudes into the equation: \( \sqrt{2} = \sqrt{3} \cdot 1 \cdot \sin(\theta) \). Thus, \( \sin(\theta) = \frac{\sqrt{2}}{\sqrt{3}} \), implying \( \theta = \arcsin\left(\frac{\sqrt{2}}{\sqrt{3}}\right)\).
04

Determine the Relationship for \(\hat{\mathbf{r}}\)

Given the perpendicularity and from the cross product definition, we infer that \( \hat{\mathbf{r}} \) depends on both \( \mathbf{a} \) and \( \mathbf{b} \). Look for a linear combination: \( \hat{\mathbf{r}} = c ( \mathbf{a} + \mathbf{a} \times \mathbf{b} ) \). The task is to find the constant \( c \).
05

Check Consistency with the Unit Magnitude

The final expression must yield a unit vector as \( \hat{\mathbf{r}} \), thus \( 1 = c \cdot | \mathbf{a} + \mathbf{a} \times \mathbf{b} | \). Using vector identities, knowing that \( \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 \), find the magnitude for different options. Substitute possible options and solve for \( c \).
06

Validate Options Considering Orthogonality

Calculate for given options each potential form: (a) \( c = \frac{2}{3} \). Confirm the condition \( |c(\mathbf{a}+\mathbf{a}\times\mathbf{b})|=1 \), showing orthogonality and unit property together with given vector conditions: (\( \mathbf{b} = \hat{\mathbf{r}} \times \mathbf{a} \).

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Key Concepts

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

Cross Product
The cross product is a vector operation used in three-dimensional space that outputs a vector perpendicular to the two input vectors. This is expressed as \( \mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin(\theta)\hat{\mathbf{n}} \). The important aspects of the cross product include:
  • The direction of the cross product is determined by the right-hand rule.
  • The magnitude of the cross product is equal to the product of the magnitudes of the vectors and the sine of the angle between them.
  • In this problem, \( \hat{\mathbf{r}} \times \mathbf{a} = \mathbf{b} \), meaning \( \mathbf{b} \) is perpendicular to both \( \hat{\mathbf{r}} \) and \( \mathbf{a} \).
Understanding these properties is imperative when solving problems involving cross products, especially when determining relationships between different vectors.
Unit Vector
A unit vector has a magnitude of one and is often used to specify a direction. It is represented by a vector with a hat, like \( \hat{\mathbf{r}} \). Here are some key points about unit vectors:
  • The magnitude of a unit vector is always 1.
  • They are usually defined by dividing a vector by its magnitude.
  • In this context, \( \hat{\mathbf{r}} \) satisfies the condition \( \hat{\mathbf{r}} \times \mathbf{a} = \mathbf{b} \).
  • We ensure our solution yields a unit vector by keeping its magnitude equal to 1.
In the problem, the task was to express \( \hat{\mathbf{r}} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \) while maintaining its unit vector property.
Vector Magnitudes
The magnitude of a vector provides information about the length of the vector, calculated as the square root of the sum of the squares of its components. Important notes about vector magnitudes include:
  • For a vector \( \mathbf{a} \), its magnitude is denoted \( |\mathbf{a}| \).
  • In the exercise, \( |\mathbf{a}| = \sqrt{3} \) and \( |\mathbf{b}| = \sqrt{2} \).
  • Calculations often involve substituting these magnitudes into formulas to solve for other unknowns such as the sine of the angle between two vectors, as shown by \( \sin(\theta) = \frac{|\mathbf{b}|}{|\mathbf{a}|} \).
These magnitudes are crucial for performing operations like the cross product and validating conditions like the unit vector property.
Orthogonality
Orthogonality describes the perpendicularity between vectors, resulting in a dot product of zero. This is critical in vector analysis:
  • Two vectors are orthogonal if their dot product is zero.
  • In the context of the cross product, orthogonality helps define the relationship \( \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0 \).
  • The cross product itself ensures orthogonality as its result is always perpendicular to both input vectors.
In the problem, orthogonality confirms \( \mathbf{b} \) is perpendicular to both \( \hat{\mathbf{r}} \) and \( \mathbf{a} \), providing part of the framework for constructing expressions like \( \hat{\mathbf{r}} = c(\mathbf{a} + \mathbf{a} \times \mathbf{b}) \). This hints at checking unit and perpendicular properties in suggested solutions.

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Most popular questions from this chapter

The distance between the line \(\mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})\) and the plane \(\mathbf{r} \cdot(\hat{\mathrm{i}}+5 \hat{\mathrm{j}}+\overrightarrow{\mathbf{k}})=5\) is \(\quad\) [AIEEE 2005] (a) \(\frac{10}{3}\) (b) \(\frac{3}{10}\) (c) \(\frac{10}{3 \sqrt{3}}\) (d) \(\frac{10}{9}\)

Two points \(P\) and \(Q\) are given in the rectangular cartesian coordinates in the curve \(y=2^{x+2}\), such that OP. \(\hat{\mathbf{i}}=-1\) and \(\mathbf{O Q} \cdot \hat{\mathbf{i}}=2\), where \(\hat{\mathbf{i}}\) is a unit vector along the \(X\)-axis. Find the magnitude of \(\mathrm{OQ}-4 \mathrm{OP}\).

If the triangle \(P Q R\) varies, then the minimum value of \(\cos (P+Q)+\cos (Q+R)+\cos (R+P)\) is (a) \(-\frac{3}{2}\) (b) \(\frac{3}{2}\) (c) \(\frac{5}{3}\) (d) \(-\frac{5}{3}\)

If the vectors \(\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\mathbf{c}=\lambda \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mu \hat{\mathbf{k}}\) are mutually orthogonal, then \((\lambda, \mu)\) is equal to [AJEEE 2010] (a) \((-3,2)\) (b) \((2,-3)\) (c) \((-2,3)\) (d) \((3,-2)\)

The unit vector which is orthogonal to the vector \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and is coplanar with the vectors \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\overrightarrow{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathbf{k}}\) is [Single Correct Type, IIT JEE 2004] (a) \(\frac{2 \hat{i}-6 \hat{j}+\hat{\mathbf{k}}}{\sqrt{41}}\) (b) \(\frac{2 \hat{i}-3 \hat{j}}{\sqrt{13}}\) (c) \(\frac{3 \hat{j}-\hat{\mathbf{k}}}{\sqrt{10}}\) (d) \(\frac{4 \hat{\mathbf{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathbf{k}}}{\sqrt{34}}\)
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