Question

Compute \(\mathbf{u} \cdot \mathbf{v}\) where: a. \(\mathbf{u}=\left[\begin{array}{r}2 \\ -1 \\ 3\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-1 \\ 1 \\ 1\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}1 \\ 2 \\ -1\end{array}\right], \mathbf{v}=\mathbf{u}\) c. \(\mathbf{u}=\left[\begin{array}{r}1 \\ 1 \\ -3\end{array}\right], \mathbf{v}=\left[\begin{array}{r}2 \\ -1 \\ 1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 5\end{array}\right], \mathbf{v}=\left[\begin{array}{r}6 \\ -7 \\ -5\end{array}\right]\) e. \(\mathbf{u}=\left[\begin{array}{l}x \\ y \\ z\end{array}\right], \mathbf{v}=\left[\begin{array}{l}a \\ b \\ c\end{array}\right]\) f. \(\mathbf{u}=\left[\begin{array}{l}a \\ b \\ c\end{array}\right], \mathbf{v}=\mathbf{0}\)

Short answer

(a) 0, (b) 6, (c) -2, (d) 0, (e) \(xa + yb + zc\), (f) 0.

Step-by-step solution

Understanding the Dot Product

The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) in the same dimensional space is calculated as the sum of the products of their corresponding components: \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\). In this exercise, we will apply this formula to vector pairs given in different parts.

Compute \(\mathbf{u} \cdot \mathbf{v}\) for part (a)

For part (a), \(\mathbf{u} = \begin{bmatrix} 2 \ -1 \ 3 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} -1 \ 1 \ 1 \end{bmatrix}\). Apply the dot product formula: \(2(-1) + (-1)(1) + 3(1) = -2 - 1 + 3 = 0\). So, \(\mathbf{u} \cdot \mathbf{v} = 0\).

Compute \(\mathbf{u} \cdot \mathbf{v}\) for part (b)

For part (b), \(\mathbf{u} = \begin{bmatrix} 1 \ 2 \ -1 \end{bmatrix}\) and \(\mathbf{v} = \mathbf{u}\). For the dot product of a vector with itself, use the same formula: \(1(1) + 2(2) + (-1)(-1) = 1 + 4 + 1 = 6\). So, \(\mathbf{u} \cdot \mathbf{v} = 6\).

Compute \(\mathbf{u} \cdot \mathbf{v}\) for part (c)

For part (c), \(\mathbf{u} = \begin{bmatrix} 1 \ 1 \ -3 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} 2 \ -1 \ 1 \end{bmatrix}\). The dot product is \(1(2) + 1(-1) + (-3)(1) = 2 - 1 - 3 = -2\). So, \(\mathbf{u} \cdot \mathbf{v} = -2\).

Compute \(\mathbf{u} \cdot \mathbf{v}\) for part (d)

For part (d), \(\mathbf{u} = \begin{bmatrix} 3 \ -1 \ 5 \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} 6 \ -7 \ -5 \end{bmatrix}\). Calculate the dot product as follows: \(3(6) + (-1)(-7) + 5(-5) = 18 + 7 - 25 = 0\). Thus, \(\mathbf{u} \cdot \mathbf{v} = 0\).

Compute \(\mathbf{u} \cdot \mathbf{v}\) for part (e)

For part (e), \(\mathbf{u} = \begin{bmatrix} x \ y \ z \end{bmatrix}\) and \(\mathbf{v} = \begin{bmatrix} a \ b \ c \end{bmatrix}\). The dot product \(\mathbf{u} \cdot \mathbf{v}\) is expressed as \(xa + yb + zc\). This is the general form for any arbitrary vector components.

Compute \(\mathbf{u} \cdot \mathbf{v}\) for part (f)

For part (f), any vector \(\mathbf{u}\) with components \(\begin{bmatrix} a \ b \ c \end{bmatrix}\) and zero vector \(\mathbf{v} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\) results in a dot product of 0: \(a(0) + b(0) + c(0) = 0\). Thus, \(\mathbf{u} \cdot \mathbf{v} = 0\).

Key concepts

vector operations

Vector operations are essential in understanding many aspects of mathematics and physics, particularly in linear algebra. One key vector operation is the **dot product**. This operation involves two vectors and produces a scalar value. The formula for the dot product of two vectors \( \mathbf{u} \) and \( \mathbf{v} \), each having components \((u_1, u_2, u_3)\) and \((v_1, v_2, v_3)\) respectively, is: \(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3\). The dot product finds many applications:

• **Projection**: Finding the component of one vector along another in physics.
• **Detection of Orthogonality**: If two vectors' dot product is zero, they are orthogonal.
• **Cosine of Angle**: Calculating the angle between vectors using \( \cos\theta = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{u}\| \|\mathbf{v}\|}\), where \( \|\mathbf{u}\| \) is the magnitude of vector \( \mathbf{u} \).

Breaking down and performing these operations step by step helps in simplifying complex vector calculations.

linear algebra

Linear Algebra is a branch of mathematics focusing on vector spaces and linear mappings between these spaces. Dot products tie directly into concepts of linear algebra, representing one of the fundamental operations with vectors. In linear algebra,

• **Vectors** are objects characterized by magnitude and direction, and they can be added, scaled, or transformed.
• The **Dot Product** is used to simplify expressions and equations in vector algebra, helping to understand vector spaces' geometric nature.
• **Vector Spaces** permit different operations to be performed easily and consistently, including addition, subtraction, and multiplication operations like the dot product.

Vector operations like the dot product are foundational because they allow us to apply algebraic techniques to solve geometric problems, such as finding solutions to systems of equations, understanding transformations, and working with projections.

computational methods

Computational methods involve using algorithms and numeric calculations to solve mathematical problems, often implemented through computer programming. The dot product plays a crucial role in computational methods. Key uses and benefits include:

• **Performance**: Dot products are heavily used in graphics rendering and simulations in computer graphics. They allow for efficient computations of projections and other linear transformations.
• **Optimization**: Machine learning algorithms use dot products to compute weights and optimize functions, being core operations in gradient descent and backpropagation.
• **Parallel Processing**: Calculating dot products can be sped up significantly using parallel processing, a common technique in modern computing to handle large-scale problems.

Understanding how to compute and interpret dot products aids in developing efficient solutions and algorithms, thus forming a bridge between theoretical mathematics and practical computing applications.