Question

In each case, write \(\mathbf{u}=\mathbf{u}_{1}+\mathbf{u}_{2},\) where \(\mathbf{u}_{1}\) is parallel to \(\mathbf{v}\) and \(\mathbf{u}_{2}\) is orthogonal to \(\mathbf{v}\). a. \(\mathbf{u}=\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}1 \\ -1 \\ 3\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{l}3 \\ 1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-2 \\ 1 \\ 4\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{r}2 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{r}3 \\ 1 \\ -1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-6 \\ 4 \\ -1\end{array}\right]\)

Short answer

Use the formulas for parallel and orthogonal components to decompose \(\mathbf{u}\) into \(\mathbf{u}_1\) and \(\mathbf{u}_2\) with respect to \(\mathbf{v}\).

Step-by-step solution

Understand the Problem

The task is to decompose the vector \(\mathbf{u}\) into two components: \(\mathbf{u}_1\) which is parallel to \(\mathbf{v}\), and \(\mathbf{u}_2\) which is orthogonal to \(\mathbf{v}\). This means \(\mathbf{u} = \mathbf{u}_{1} + \mathbf{u}_{2}\).

Formula for Parallel Component

The component \(\mathbf{u}_{1}\), which is parallel to \(\mathbf{v}\), is given by:\[ \mathbf{u}_{1} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \] Where \(\mathbf{u} \cdot \mathbf{v}\) is the dot product between \(\mathbf{u}\) and \(\mathbf{v}\).

Formula for Orthogonal Component

The component \(\mathbf{u}_{2}\), which is orthogonal to \(\mathbf{v}\), can be found by subtracting the parallel component from \(\mathbf{u}\):\[ \mathbf{u}_{2} = \mathbf{u} - \mathbf{u}_{1} \]

Apply Formulas to Scenario 'a'

Given \(\mathbf{u} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix}\):1. Calculate \( \mathbf{u} \cdot \mathbf{v} = 2 \times 1 + (-1) \times (-1) + 1 \times 3 = 2 + 1 + 3 = 6 \).2. Calculate \( \mathbf{v} \cdot \mathbf{v} = 1 \times 1 + (-1) \times (-1) + 3 \times 3 = 1 + 1 + 9 = 11 \).3. Compute \( \mathbf{u}_{1} = \frac{6}{11} \begin{bmatrix} 1 \ -1 \ 3 \end{bmatrix} = \begin{bmatrix} 6/11 \ -6/11 \ 18/11 \end{bmatrix} \).4. Compute \( \mathbf{u}_{2} = \mathbf{u} - \mathbf{u}_{1} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix} - \begin{bmatrix} 6/11 \ -6/11 \ 18/11 \end{bmatrix} \approx \begin{bmatrix} 16/11 \ -5/11 \ -7/11 \end{bmatrix} \).

Apply Formulas to Scenario 'b'

Given \(\mathbf{u} = \begin{bmatrix} 3 \ 1 \ 0 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -2 \ 1 \ 4 \end{bmatrix}\):1. Calculate \( \mathbf{u} \cdot \mathbf{v} = 3 \times (-2) + 1 \times 1 + 0 \times 4 = -6 + 1 + 0 = -5 \).2. Calculate \( \mathbf{v} \cdot \mathbf{v} = (-2)(-2) + 1 \times 1 + 4 \times 4 = 4 + 1 + 16 = 21 \).3. Compute \( \mathbf{u}_{1} = \frac{-5}{21} \begin{bmatrix} -2 \ 1 \ 4 \end{bmatrix} = \begin{bmatrix} 10/21 \ -5/21 \ -20/21 \end{bmatrix} \).4. Compute \( \mathbf{u}_{2} = \mathbf{u} - \mathbf{u}_{1} = \begin{bmatrix} 3 \ 1 \ 0 \end{bmatrix} - \begin{bmatrix} 10/21 \ -5/21 \ -20/21 \end{bmatrix} \approx \begin{bmatrix} 53/21 \ 26/21 \ 20/21 \end{bmatrix} \).

Apply Formulas to Scenario 'c'

Given \(\mathbf{u} = \begin{bmatrix} 2 \ -1 \ 0 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} 3 \ 1 \ -1 \end{bmatrix}\):1. Calculate \( \mathbf{u} \cdot \mathbf{v} = 2 \times 3 + (-1) \times 1 + 0 \times (-1) = 6 - 1 + 0 = 5 \).2. Calculate \( \mathbf{v} \cdot \mathbf{v} = 3 \times 3 + 1 \times 1 + (-1)(-1) = 9 + 1 + 1 = 11 \).3. Compute \( \mathbf{u}_{1} = \frac{5}{11} \begin{bmatrix} 3 \ 1 \ -1 \end{bmatrix} = \begin{bmatrix} 15/11 \ 5/11 \ -5/11 \end{bmatrix} \).4. Compute \( \mathbf{u}_{2} = \mathbf{u} - \mathbf{u}_{1} = \begin{bmatrix} 2 \ -1 \ 0 \end{bmatrix} - \begin{bmatrix} 15/11 \ 5/11 \ -5/11 \end{bmatrix} \approx \begin{bmatrix} 7/11 \ -16/11 \ 5/11 \end{bmatrix} \).

Apply Formulas to Scenario 'd'

Given \(\mathbf{u} = \begin{bmatrix} 3 \ -2 \ 1 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -6 \ 4 \ -1 \end{bmatrix}\):1. Calculate \( \mathbf{u} \cdot \mathbf{v} = 3 \times (-6) + (-2) \times 4 + 1 \times (-1) = -18 - 8 - 1 = -27 \).2. Calculate \( \mathbf{v} \cdot \mathbf{v} = (-6)(-6) + 4 \times 4 + (-1)(-1) = 36 + 16 + 1 = 53 \).3. Compute \( \mathbf{u}_{1} = \frac{-27}{53} \begin{bmatrix} -6 \ 4 \ -1 \end{bmatrix} = \begin{bmatrix} 162/53 \ -108/53 \ 27/53 \end{bmatrix} \).4. Compute \( \mathbf{u}_{2} = \mathbf{u} - \mathbf{u}_{1} = \begin{bmatrix} 3 \ -2 \ 1 \end{bmatrix} - \begin{bmatrix} 162/53 \ -108/53 \ 27/53 \end{bmatrix} \approx \begin{bmatrix} -3/53 \ 2/53 \ 26/53 \end{bmatrix} \).

Key concepts

Dot Product

The dot product, also known as the scalar product, is a fundamental operation involving two vectors. Essentially, it provides a measure of the vectors' parallelity. To compute the dot product between two vectors \( \mathbf{u} \) and \( \mathbf{v} \), you multiply their corresponding components and sum them up. For example, if \( \mathbf{u} = \begin{bmatrix} u_1 & u_2 & u_3 \end{bmatrix} \) and \( \mathbf{v} = \begin{bmatrix} v_1 & v_2 & v_3 \end{bmatrix} \), then the dot product is given by:
\[ \mathbf{u} \cdot \mathbf{v} = u_1 v_1 + u_2 v_2 + u_3 v_3 \]

• The dot product is a single number, not a vector.
• If the dot product is zero, the vectors are orthogonal.
• The magnitude of the dot product gives an insight into how much one vector extends in the direction of the other.

The dot product plays an essential role in vector decomposition and helps determine the parallel and orthogonal components of a vector.

Parallel Vectors

Parallel vectors are vectors that have the same or opposite direction. In vector decomposition, finding a vector that is parallel to another is crucial. A vector \( \mathbf{u}_1 \) is parallel to \( \mathbf{v} \) if it can be expressed as a scalar multiple of \( \mathbf{v} \). This means:
\[ \mathbf{u}_1 = c \cdot \mathbf{v} \]where \( c \) is a scalar.

• Parallel vectors maintain the same direction ratio; they only differ in magnitude.
• The parallel component of a vector relative to another can be obtained using the dot product formula.
  \[ \mathbf{u}_1 = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \cdot \mathbf{v} \]
• This process projects \( \mathbf{u} \) onto \( \mathbf{v} \), preserving direction.

Orthogonal Vectors

Orthogonal vectors are vectors that are at a right angle to each other. This concept is especially useful in vector decomposition where one aims to break down a vector into components that are parallel and orthogonal to a reference vector. For example:
\( \mathbf{u}_2 = \mathbf{u} - \mathbf{u}_1 \).

• Two vectors are orthogonal if their dot product is zero.
• The orthogonal component \( \mathbf{u}_2 \) is what remains of a vector \( \mathbf{u} \) after subtracting the parallel component \( \mathbf{u}_1 \).
• The magnitude of \( \mathbf{u}_2 \) gives the extent to which \( \mathbf{u} \) does not align with \( \mathbf{v} \).

Understanding orthogonal vectors is key in fields such as linear algebra, computer graphics, and physics, where resolving vectors into components simplifies complex problems.

Vectors in 3D

Vectors in 3D space are pivotal for representing forces, velocities, or any directional quantities in three-dimensional space. Understanding them is essential for complex calculations and applications.

• A vector in 3D is denoted as \( \begin{bmatrix} x & y & z \end{bmatrix} \), where \( x \), \( y \), and \( z \) are its components along the respective axes.
• Operations such as the dot product, cross product, and vector decomposition are applicable in 3D space with slight modifications to account for the additional dimension.
• Decomposing vectors in 3D follows the same principles as 2D, but requires accounting for all three components.
• 3D vectors enable the representation of objects in virtual environments, simulations, and engineering solutions.

Understanding vectors in 3D helps us visualize and solve real-world problems that span across various scientific and engineering domains.