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The unknown vector \(\mathbf{v}\) satisfies \(\mathbf{b} \cdot \mathbf{v}=\lambda\) and \(\mathbf{b} \times \mathbf{v}=\mathbf{c},\) where \(\lambda, \mathbf{b},\) and \(\mathbf{c}\) are fixed and known. Find \(\mathbf{v}\) in terms of \(\lambda, \mathbf{b},\) and \(\mathbf{c}\)

Short Answer

Expert verified
\( \mathbf{v} = \frac{\lambda}{||\mathbf{b}||^2} \mathbf{b} + \mathbf{x} \), where \( \mathbf{x} = \mathbf{c} \times \frac{\mathbf{b}}{||\mathbf{b}||^2} \).

Step by step solution

01

Understand the Given Conditions

We are given two equations involving the unknown vector \(\mathbf{v}\):1. The dot product equation \( \mathbf{b} \cdot \mathbf{v} = \lambda \).2. The cross product equation \( \mathbf{b} \times \mathbf{v} = \mathbf{c} \).Here, \(\mathbf{b}\) and \(\mathbf{c}\) are known vectors, and \(\lambda\) is a known scalar.
02

Use the Dot Product Equation

The dot product equation \( \mathbf{b} \cdot \mathbf{v} = \lambda \) implies that the component of \( \mathbf{v} \) in the direction of \( \mathbf{b} \) is \( \frac{\lambda}{||\mathbf{b}||} \). Therefore, one component of \( \mathbf{v} \) is parallel to \(\mathbf{b}\).
03

Use the Cross Product Equation

The cross product equation \( \mathbf{b} \times \mathbf{v} = \mathbf{c} \) implies that \( \mathbf{v} \) must have components perpendicular to \( \mathbf{b} \), resulting in the vector \( \mathbf{c} \).
04

Construct the Vector v

Since \( \mathbf{v} \) has a component \( a\mathbf{b} \) parallel to \( \mathbf{b} \) and a component that results in \( \mathbf{c} \) when crossed with \( \mathbf{b} \), we propose \( \mathbf{v} = a\mathbf{b} + \mathbf{x} \) where \( \mathbf{x} \) is perpendicular to \( \mathbf{b} \).
05

Solve for the Scalar a

To ensure the dot product condition, substitute \( \mathbf{v} = a\mathbf{b} + \mathbf{x} \) into the dot product equation:\[ \mathbf{b} \cdot (a\mathbf{b} + \mathbf{x}) = \lambda \]Since \( \mathbf{x} \) is perpendicular to \( \mathbf{b} \), \( \mathbf{b} \cdot \mathbf{x} = 0 \). This simplifies to:\[ a||\mathbf{b}||^2 = \lambda \]Thus, \( a = \frac{\lambda}{||\mathbf{b}||^2} \).
06

Determine Vector x

We need \( \mathbf{b} \times (a\mathbf{b} + \mathbf{x}) = \mathbf{c} \). Since \( \mathbf{b} \times a\mathbf{b} = \mathbf{0} \), it follows that \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \). Here \( \mathbf{x} \) is any vector that satisfies this equation, which can be solved using known methods to find a vector perpendicular to \( \mathbf{b} \).
07

Complete Solution for v

\( \mathbf{v} = a\mathbf{b} + \mathbf{x} \), where \( a = \frac{\lambda}{||\mathbf{b}||^2} \) and \( \mathbf{x} \) satisfies \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \) and is perpendicular to \( \mathbf{b} \). Use \( \mathbf{x} = \mathbf{c} \times \frac{\mathbf{b}}{||\mathbf{b}||^2} \) to satisfy the cross product condition.

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Key Concepts

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

Dot Product
The dot product is a fundamental operation in vector algebra. It involves two vectors, say \( \mathbf{b} \) and \( \mathbf{v} \), and produces a scalar. The dot product is calculated as:
  • \( \mathbf{b} \cdot \mathbf{v} = ||\mathbf{b}|| ||\mathbf{v}|| \cos(\theta) \)
where \( ||\mathbf{b}|| \) and \( ||\mathbf{v}|| \) are the magnitudes of the vectors, and \( \theta \) is the angle between them.
The dot product tells us how much of one vector goes in the direction of the other. In our exercise, when \( \mathbf{b} \cdot \mathbf{v} = \lambda \), it tells us that the projection of \( \mathbf{v} \) onto \( \mathbf{b} \) is \( \lambda \). This can be visualized as the component of \( \mathbf{v} \) that points in the direction of \( \mathbf{b} \). The condition creates a parallel component in \( \mathbf{v} \), represented as \( a\mathbf{b} \), where \( a = \frac{\lambda}{||\mathbf{b}||^2} \).
This parallel part is just one piece of the full vector \( \mathbf{v} \). The solution process requires us to balance this with the cross product condition to fully define \( \mathbf{v} \).
Cross Product
The cross product is another essential operation in vector algebra, which gives us a vector instead of a scalar, unlike the dot product. When you take the cross product of two vectors, say \( \mathbf{b} \) and \( \mathbf{v} \), it results in:
  • \( \mathbf{b} \times \mathbf{v} \)
This product is a vector perpendicular to both \( \mathbf{b} \) and \( \mathbf{v} \). The direction can be determined using the right-hand rule, and its magnitude is calculated as:
  • \( ||\mathbf{b} \times \mathbf{v}|| = ||\mathbf{b}|| ||\mathbf{v}|| \sin(\theta) \)
where \( \theta \) is the angle between the vectors.
In the exercise we're considering, \( \mathbf{b} \times \mathbf{v} = \mathbf{c} \) implies that the component of \( \mathbf{v} \) that is normal to \( \mathbf{b} \) results in the vector \( \mathbf{c} \).
This leads to the solution part where we need to find a vector \( \mathbf{x} \) such that \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \). This \( \mathbf{x} \) is the orthogonal component needed to make \( \mathbf{v} \) satisfy both the dot and cross product conditions.
Perpendicular Vectors
The concept of perpendicular vectors is an important one in vector algebra. Two vectors are perpendicular when their dot product is zero:
  • \( \mathbf{b} \cdot \mathbf{x} = 0 \)
This means there is no component of \( \mathbf{x} \) in the direction of \( \mathbf{b} \).
In the context of our exercise, finding a vector \( \mathbf{x} \), perpendicular to \( \mathbf{b} \), is crucial to satisfying the cross product condition. If \( \mathbf{x} \) is perpendicular to \( \mathbf{b} \), \( \mathbf{x} \) does not contribute to \( \mathbf{b} \cdot \mathbf{v} \), thus fulfilling the requirement \( \mathbf{b} \times \mathbf{x} = \mathbf{c} \).
This forms an orthogonal component \( \mathbf{x} \) that combines with the parallel part \( a\mathbf{b} \) to give the final vector \( \mathbf{v} \). Finding \( \mathbf{x} \) often involves solving equations that respect these perpendicularity conditions, and various approaches like using cross products help in determining such vectors.

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Most popular questions from this chapter

One of the many uses of the scalar product is to find the angle between two given vectors. Find the angle between the vectors \(\mathbf{b}=(1,2,4)\) and \(\mathbf{c}=(4,2,1)\) by evaluating their scalar product.

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