Problem 44 - Essential Calculus Early Transcendentals

Question

(a) Find all vectors \(\mathbf{v}\) such that $$\langle 1,2,1\rangle \times \mathbf{v}=\langle 3,1,-5\rangle$$ (b) Explain why there is no vector \(\mathbf{v}\) such that $$\langle 1,2,1\rangle \times \mathbf{v}=\langle 3,1,5\rangle$$

Step-by-Step Solution

Answer

(a) \(\mathbf{v} = \langle z + 1, 2z - 3, z \rangle\) for any \(z\). (b) No such \(\mathbf{v}\) exists as the system is inconsistent.

Step 1: Introduction to Cross Product Equation

We want to find vectors \(\mathbf{v}\) such that their cross product with \(\langle 1,2,1 \rangle\) gives \(\langle 3,1,-5 \rangle\). The cross product of two vectors \(\langle a,b,c \rangle\) and \(\langle x,y,z \rangle\) is \(\langle bz-cy, cx-az, ay-bx \rangle\).

Step 2: Set Up the Cross Product Components

Let \(\mathbf{v} = \langle x,y,z \rangle\). The cross product \(\langle 1,2,1 \rangle \times \langle x,y,z \rangle\) yields the vector \(\langle 2z - y, x - z, y - 2x \rangle\).

Step 3: Equate Components with Given Vector

We have the equation \(\langle 2z - y, x - z, y - 2x \rangle = \langle 3,1,-5 \rangle\). This gives us the system of equations:1. \(2z - y = 3\)2. \(x - z = 1\)3. \(y - 2x = -5\)

Step 4: Solve the System of Equations

From equation (2), \(x = z + 1\). Substitute \(x = z + 1\) into equation (3):\(y - 2(z + 1) = -5\) gives \(y = 2z - 3\). Substitute \(y = 2z - 3\) into equation (1):\(2z - (2z - 3) = 3\) simplifies to \(3 = 3\), which is always true.Thus, \(x = z + 1\) and \(y = 2z - 3\) describe the family of solutions.

Step 5: Solution for Part (a)

Given \(z\) is a parameter, the solutions are parameterized by \(z\): \(\mathbf{v} = \langle z + 1, 2z - 3, z \rangle\) for any \(z\).

Step 6: Set Up for Part (b)

Consider the equation \(\langle 1,2,1 \rangle \times \mathbf{v} = \langle 3,1,5 \rangle\). This gives us the system:1. \(2z - y = 3\)2. \(x - z = 1\)3. \(y - 2x = 5\)

Step 7: Check for Consistency in Part (b)

Following similar steps, \(x = z + 1\) gives us \(y = 5 + 2x = 5 + 2(z + 1) = 2z + 7\). Substituting into the first equation, \(2z - (2z + 7) = 3\), simplifies to \(-7 = 3\) which contradicts each other.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of unknowns. The goal is to find values for the variables that satisfy all the equations simultaneously. In the context of the cross product exercise, solving a system of equations helps determine the specific components of the vector \(\mathbf{v}\). Here, we have three equations derived from equating the cross product of two vectors to a given vector:

• \(2z - y = 3\)
• \(x - z = 1\)
• \(y - 2x = -5\)

By solving these equations, we can determine a set of possible vectors \(\mathbf{v}\) that satisfy the given conditions. Solutions to these equations can be unique, infinite, or none, depending on the relationship between the equations.

Parameterized Solutions

A parameterized solution is a way to express the solutions of an equation or system of equations in terms of one or more parameters. These parameters represent indefinite values that can give rise to multiple solutions. In our example, the solutions for \(\mathbf{v}\) were expressed in terms of the parameter \(z\). For part (a), by setting \(z\) as a parameter, we obtain the general solution:

• \(x = z + 1\)
• \(y = 2z - 3\)
• \(z\)

Thus, the solution is \(\mathbf{v} = \langle z + 1, 2z - 3, z \rangle\), which captures an infinite number of solutions by varying \(z\). Parameterized solutions are particularly useful when the system has infinitely many solutions. It elegantly presents all possible solutions in a compact form.

Vector Operations

Vector operations are calculations involving vectors, and the cross product is one such operation. The cross product of two vectors \(\mathbf{a} = \langle a_1, b_1, c_1 \rangle\) and \(\mathbf{b} = \langle a_2, b_2, c_2 \rangle\) results in a new vector \(\mathbf{c} = \langle b_1c_2 - c_1b_2, c_1a_2 - a_1c_2, a_1b_2 - b_1a_2 \rangle\). This operation is used to find vectors perpendicular to both original vectors.In our exercise, the expression \(\langle 1,2,1\rangle \times \mathbf{v}\) denotes the cross product of the vector \(\mathbf{v}\) with \(\langle 1,2,1\rangle\). For part (a), this equals to \(\langle 3,1,-5 \rangle\), hence, the components must simultaneously satisfy the cross product equations derived. In part (b), no vector \(\mathbf{v}\) satisfies the equation \(\langle 1,2,1\rangle \times \mathbf{v} = \langle 3,1,5 \rangle\) due to the inconsistency revealed during equation solving. These operations and their solutions demonstrate the fundamental properties and rules of vector mathematics.