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Chapter 2: Problem 98

Let \(\mathbf{a}\) and \(\mathbf{b}\) be two given perpendicular vectors, which are non-zero. A vector \(\mathbf{r}\) satisfying the equation \(\mathbf{r} \times \mathbf{b}=\mathbf{a}\), can be ............. (a) \(b-\frac{a \times b}{|b|^{2}}\) (b) \(2 \mathrm{~b}-\frac{(\mathrm{a} \times \mathrm{b})}{|\mathrm{b}|^{2}}\) (c) \(|\mathrm{a}| \mathrm{b}-\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{b}|^{2}}\) (d) \(|\mathbf{b}| \mathbf{b}-\frac{(a \times \mathbf{b})}{|\mathbf{b}|^{2}}\)

Short Answer

Expert verified
(a) \( b-\frac{a \times b}{|b|^{2}} \)

Step by step solution

01

Understanding the Cross Product Equation

We have the equation \( \mathbf{r} \times \mathbf{b} = \mathbf{a} \). The goal is to find a vector \( \mathbf{r} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \) that satisfies this equation. Since \( \mathbf{a} \) and \( \mathbf{b} \) are perpendicular, any vector \( \mathbf{r} \) can be broken into a component parallel to \( \mathbf{b} \) and a component perpendicular to \( \mathbf{b} \).
02

Using Vector Identities to Solve for \( \mathbf{r} \)

We will use the fact that the cross product is zero if two vectors are parallel, and any solution for \( \mathbf{r} \) must resolve this cross product equation into perpendicular components of the same magnitude. We need \( \mathbf{r} \) such that its component perpendicular to \( \mathbf{b} \) matches \( \mathbf{a} \).
03

Determining the Correct Component of \( \mathbf{r} \)

The goal is to adjust \( \mathbf{r} \) such that it includes a parallel component to \( \mathbf{b} \), reduced by the influence of \( \mathbf{a} \). A general expression for this would be adjusting \( \mathbf{r} \) by \( -\frac{\mathbf{a} \times \mathbf{b}}{|\mathbf{b}|^{2}} \). This eliminates the current effect of \( \mathbf{a} \) in the perpendicular direction, creating a balance for the cross product equation.
04

Evaluating Choice (a)

Verify if option (a) satisfies the condition: \( \mathbf{r} = b - \frac{a \times b}{|b|^{2}} \). Substituting this into the cross product \( \mathbf{r} \times \mathbf{b} \), the term involving the component \( \frac{a \times b}{|b|^{2}} \) ensures that the effective perpendicular of \( \mathbf{r} \) results in \( \mathbf{a} \). Thus, this option correctly models the condition \( \mathbf{r} \times \mathbf{b} = \mathbf{a} \).
05

Checking Other Options for Validity

Evaluate options (b), (c), and (d) using similar substitutions and calculations. None of these successfully satisfy the equation \( \mathbf{r} \times \mathbf{b} = \mathbf{a} \) under all conditions as option (a) does, as their designs create misleading or exaggerated contributions in directions orthogonal to the desired result. Their constructions do not balance the cross product to truly equal \( \mathbf{a} \).

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Key Concepts

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

Perpendicular Vectors
Perpendicular vectors are an important concept to understand when dealing with vector math, especially in physics and engineering. When two vectors are perpendicular, their dot product is zero. This is because perpendicular vectors have no overlapping components in the direction of each other.
A classic example is the x-axis and y-axis in a two-dimensional plane, which intersect at right angles to each other.
  • Key Point: The dot product of two perpendicular vectors is equal to zero.
  • Implication: Their mutual perpendicularity ensures that the components do not influence each other directly when calculated through dot products.
  • Role in Cross Product: When it comes to the cross product, perpendicular vectors have maximal effect. The resulting vector from their cross product is at maximum magnitude, lying in a direction orthogonal to both original vectors.
Understanding these principles is essential when analyzing vector systems that involve solving equations using cross products, like in the original exercise.
Vector Components
Breaking down vectors into components is a foundational skill in vector analysis. Each vector can be decomposed into smaller parts, usually in reference to a particular coordinate system, like x, y, and z axes.
  • Parallel and Perpendicular Components: Any vector can be expressed as the sum of parallel and perpendicular components with respect to another vector. This decomposition is crucial in understanding and solving vector equations.
  • Importance in Solutions: In the original exercise, breaking vectors into components helps determine how vectors match and solve cross product equations.
  • Computation: Parallel components align with the direction of the reference vector, whereas perpendicular components are orthogonal to it.
Decomposing vectors into their components simplifies complex problems, allowing us to work with smaller, more manageable parts, and reach solutions more efficiently.
Cross Product Equation
The cross product equation is a vital tool in vector algebra, providing insight into the orthogonal interactions between vectors. When vectors take part in a cross product, their combined influence is represented by a new vector that is perpendicular to both original vectors.
  • Notation and Formula: Given two vectors \( \mathbf{a} \) and \( \mathbf{b} \), their cross product is denoted as \( \mathbf{a} \times \mathbf{b} \) and calculated using the determinant of a matrix incorporating unit vectors and the components of \( \mathbf{a} \) and \( \mathbf{b} \).
  • Geometric Interpretation: The cross product's magnitude is given by \( |\mathbf{a}| |\mathbf{b}| \sin(\theta) \), where \( \theta \) is the angle between \( \mathbf{a} \) and \( \mathbf{b} \). The result is a vector orthogonal to the original plane of \( \mathbf{a} \) and \( \mathbf{b} \).
  • Application in the Exercise: In the exercise, \( \mathbf{r} \times \mathbf{b} = \mathbf{a} \), showcasing that the solution involves adjusting \( \mathbf{r} \) to properly balance its components, ensuring that the cross product accurately produces \( \mathbf{a} \).
Understanding this equation helps solve practical problems where vector direction and interaction are essential components to analyze accurately.

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Most popular questions from this chapter

Let \(A(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}), B(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\) and \(C(\lambda \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\mu \hat{\mathbf{k}})\) are vertices of a triangle and its median through \(A\) is equally inclined to the positive directions of the axes. The value of \(2 \lambda-\mu\) is equal to

For any vector a, the value of \((\mathbf{a} \times \hat{\mathbf{i}})^{2}+(\mathbf{a} \times \hat{\mathbf{j}})^{2}+(\mathbf{a} \times \hat{\mathbf{k}})^{2}\) is equal to (a) \(4 a^{2}\) (b) \(2 a^{2}\) (c) \(\mathrm{a}^{2}\) (d) \(3 a^{2}\)

In the \(\triangle A B C\) a point \(P\) is taken on the side \(A B\) such that \(A P: B P=1: 2\) and a point \(Q\) is taken on the side \(B C\) such that \(C Q: B Q=2: 1\). If \(R\) be the point of intersection of lines \(A Q\) and \(C P\), using vector find the area of \(\Delta A B C\), if it is known that area of \(\triangle A B C\) is one unit.

Let two non-collinear unit vectors \(\hat{\mathbf{a}}\) and \(\hat{\mathbf{b}}\) form an acute angle. A point \(P\) moves, so that at any time \(t\) the position vector OP (where, \(O\) is the origin) is given by 芒 \(\cos t+\vec{b} \sin t\). When \(P\) is farthest from origin \(O\), let \(M\) be the length of \(\mathrm{OP}\) and \(\hat{\mathrm{u}}\) be the unit vector along \(\mathrm{OP}\). Then, \(\quad\) [Single Correct Type, IITJEE 2008] (a) \(\mathrm{u}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\mathbf{a}+\dot{b}|}\) and \(M=(1+\hat{\mathbf{a}} \cdot \hat{b})^{1 / 2}\) (b) \(\mathrm{u}=\frac{\mathbf{a}-\hat{\mathrm{b}}}{|\mathbf{a}-\dot{b}|}\) and \(M=(1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}\) (c) \(\mathrm{u}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\mathbf{a}+\dot{b}|}\) and \(M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}\) (d) \(\mathrm{u}=\frac{\hat{\mathbf{a}}-\hat{\mathbf{b}}}{|\hat{\mathbf{a}}-\hat{\mathbf{b}}|}\) and \(\mathrm{M}=(1+2 \hat{\mathbf{a}}+\hat{\mathbf{b}})^{1 / 2}\)

Let the vectors PQ. QR. RS, ST, TU and UP represent the sides of a regular hexagon. Statement \(1 \mathrm{PQ} \times(\mathrm{RS}+\mathrm{ST}) \neq 0\), because Statement II \(\mathrm{PQ} \times \mathrm{RS}=0\) and \(\mathbf{P Q} \times \mathbf{S T} \neq 0\) [Single Correct Type, 2007, 3M] (a) Statement 1 is true, Statement \(\mathrm{Il}\) is true and Statement \(\mathrm{II}\) is a correct explanation for Statement \(\mathrm{I}\). (b) Statement 1 is true, Statement \(\mathrm{II}\) is true but Statement \(\mathrm{II}\) is not a correct explanation for Statement \(\mathrm{I}\). (c) Statement 1 is true, Statement \(\mathrm{II}\) is false. (d) Statement 1 is false, Statement \(\mathrm{II}\) is true.
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