Question

Let \(\mathbf{u}=\left[\begin{array}{r}2 \\ 0 \\ -4\end{array}\right]\) and \(\mathbf{v}=\left[\begin{array}{r}2 \\ 1 \\ -2\end{array}\right] .\) In each case find \(\mathbf{x}\) : a. \(2 \mathbf{u}-\|\mathbf{v}\| \mathbf{v}=\frac{3}{2}(\mathbf{u}-2 \mathbf{x})\) b. \(3 \mathbf{u}+7 \mathbf{v}=\|\mathbf{u}\|^{2}(2 \mathbf{x}+\mathbf{v})\)

Short answer

\( \mathbf{x} = \begin{bmatrix} -\frac{1}{3} \\ 1 \\ \frac{4}{3} \end{bmatrix} \) for (a); \( \mathbf{x} = \begin{bmatrix} -0.5 \\ -0.325 \\ -0.35 \end{bmatrix} \) for (b).

Step-by-step solution

Define Given Vectors and Problem

We start with the given vectors \( \mathbf{u} = \begin{bmatrix} 2 \ 0 \ -4 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix} \). We are asked to find vector \( \mathbf{x} \) in each problem equation.

Calculate \( \|\mathbf{v}\| \)

To solve part (a), calculate \( \|\mathbf{v}\| \), which is the magnitude of \( \mathbf{v} \). \[ \|\mathbf{v}\| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \]

Solve Equation (A)

Substitute known values into the equation for part (a):\[ 2\mathbf{u} - \|\mathbf{v}\|\mathbf{v} = \frac{3}{2}(\mathbf{u} - 2\mathbf{x}) \]This is equivalent to:\[ 2\begin{bmatrix} 2 \ 0 \ -4 \end{bmatrix} - 3\begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix} = \frac{3}{2}\left(\begin{bmatrix} 2 \ 0 \ -4 \end{bmatrix} - 2\mathbf{x}\right) \]Compute:\[ \begin{bmatrix} 4 \ 0 \ -8 \end{bmatrix} - \begin{bmatrix} 6 \ 3 \ -6 \end{bmatrix} = \frac{3}{2}\begin{bmatrix} 2 \ 0 \ -4 \end{bmatrix} - 3\mathbf{x} \]\[ \begin{bmatrix} -2 \ -3 \ -2 \end{bmatrix} = \begin{bmatrix} -3 \ 0 \ -6 \end{bmatrix} - 3 \mathbf{x} \]Solve for \( \mathbf{x} \):\[ 3 \mathbf{x} = \begin{bmatrix} -1 \ 3 \ 4 \end{bmatrix} \]\[ \mathbf{x} = \begin{bmatrix} -\frac{1}{3} \ 1 \ \frac{4}{3} \end{bmatrix} \].

Calculate \( \|\mathbf{u}\|^2 \)

For part (b), calculate \( \|\mathbf{u}\|^2 \), which is the square of the magnitude of \( \mathbf{u} \):\[ \|\mathbf{u}\|^2 = (\sqrt{2^2 + 0^2 + (-4)^2})^2 = (\sqrt{4 + 16})^2 = 20. \]

Solve Equation (B)

Substitute known values into the equation for (b):\[ 3\mathbf{u} + 7\mathbf{v} = \|\mathbf{u}\|^2(2\mathbf{x} + \mathbf{v}) \]This becomes:\[ 3\begin{bmatrix} 2 \ 0 \ -4 \end{bmatrix} + 7\begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix} = 20(2\mathbf{x} + \begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix}) \]Compute each side:\[ \begin{bmatrix} 6 \ 0 \ -12 \end{bmatrix} + \begin{bmatrix} 14 \ 7 \ -14 \end{bmatrix} = \begin{bmatrix} 20 \ 7 \ -26 \end{bmatrix} \]\[ 20(2\mathbf{x} + \begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix}) = \begin{bmatrix} 20 \ 7 \ -26 \end{bmatrix} \]Simplify:\[ 2 \mathbf{x} + \begin{bmatrix} 2 \ 1 \ -2 \end{bmatrix} = \begin{bmatrix} 1 \ 0.35 \ -1.3 \end{bmatrix} \]Solve for \( \mathbf{x} \):\[ 2\mathbf{x} = \begin{bmatrix} -1 \ -0.65 \ -0.7 \end{bmatrix} \]\[ \mathbf{x} = \begin{bmatrix} -0.5 \ -0.325 \ -0.35 \end{bmatrix} \].

Key concepts

Vector Magnitude

Understanding the magnitude of a vector is crucial. The magnitude, sometimes called the length or norm, is a measure of how long the vector is. For a vector \(\mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix}\), the magnitude is calculated using the formula:

• \(\|\mathbf{v}\| = \sqrt{v_1^2 + v_2^2 + v_3^2}\)

In the exercise, the magnitude of \(\mathbf{v} = \begin{bmatrix} 2 & 1 & -2 \end{bmatrix}\) was found to be 3. This is simply

• \(\sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3\)

This magnitude acts as a scaling factor that can alter other vectors through operations.

Vector Operations

Vector operations involve actions like addition, subtraction, and scaling. Each of these operations follows specific rules that often mirror everyday arithmetic but applied in a three-dimensional space.For example, subtracting two vectors, like \(2\mathbf{u} - \|\mathbf{v}\|\mathbf{v}\), requires

• Multiplying each component of \(\mathbf{u}\) by 2
• Scaling \(\mathbf{v}\) by its magnitude, \(3\)
• Then, subtract the two resultant vectors

Understanding these operations helps solve equations by manipulating vector quantities.

Equation Solving

Solving equations involving vectors follows a logical sequence, similar to algebra. In this exercise, we tackled two equations wherein the main goal was to isolate \(\mathbf{x}\). To do this:

• Perform operations step-by-step, adhering to the vector operation principles
• Substitute known values, then simplify as much as possible
• Solve for \(\mathbf{x}\) only after simplifying both sides of the equation

Breaking down complex vector equations into manageable parts is essential in understanding and solving them.

Linear Algebra

Linear Algebra deals with vectors and matrices, and their relationships. This exercise is rooted in concepts from linear algebra, such as linear combinations and scalar multiplication.In linear algebra:

• Vectors can be scaled by constants, like \(3\mathbf{x}\)
• Systems of vector equations represent geometric transformations or movements

It’s important to grasp these basic ideas, as they form the cornerstone of more advanced topics like vector spaces and transformations. Linear algebra provides the framework needed for analyzing and solving problems involving multiple dimensions.