Question

Compute \(\mathbf{u} \times \mathbf{v}\) where: a. \(\mathbf{u}=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right], \mathbf{v}=\left[\begin{array}{l}1 \\ 1 \\ 2\end{array}\right]\) b. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -1 \\ 0\end{array}\right], \mathbf{v}=\left[\begin{array}{r}-6 \\ 2 \\ 0\end{array}\right]\) c. \(\mathbf{u}=\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right], \mathbf{v}=\left[\begin{array}{r}1 \\ 1 \\ -1\end{array}\right]\) d. \(\mathbf{u}=\left[\begin{array}{r}2 \\ 0 \\ -1\end{array}\right], \mathbf{v}=\left[\begin{array}{l}1 \\ 4 \\ 7\end{array}\right]\)

Short answer

a. \([1, -1, 0]\), b. \([0, 0, 0]\), c. \([1, 6, 5]\), d. \([4, -15, 8]\)

Step-by-step solution

Compute Cross Product a

For \(\mathbf{u} = \left[\begin{array}{l}1 \ 2 \ 3\end{array}\right]\) and \(\mathbf{v} = \left[\begin{array}{l}1 \ 1 \ 2\end{array}\right]\), use the formula:\[\mathbf{u} \times \mathbf{v} = \left[ \begin{array}{c}u_{2} u_3 - u_3 u_2 \ u_3 u_1 - u_1 u_3 \ u_1 u_2 - u_2 u_1 \end{array} \right]\]Plug in the values:\[\mathbf{u} \times \mathbf{v} = \left[ \begin{array}{c} 1\cdot3 - 2\cdot1 \ 2\cdot1 - 1\cdot3 \ 1\cdot2 - 1\cdot2 \end{array} \right] = \left[ \begin{array}{c} 1 \ -1 \ 0 \end{array} \right]\]

Compute Cross Product b

For \(\mathbf{u} = \left[\begin{array}{r}3 \ -1 \ 0\end{array}\right]\) and \(\mathbf{v} = \left[\begin{array}{r}-6 \ 2 \ 0\end{array}\right]\), apply the cross product formula:\[\mathbf{u} \times \mathbf{v} = \left[ \begin{array}{c}-1 \cdot 0 - 2 \cdot 0 \ 0 \cdot 3 - (-6) \cdot 0 \ 3 \cdot 2 - (-1) \cdot (-6) \end{array} \right] = \left[ \begin{array}{c} 0 \ 0 \ 6 - 6 \end{array} \right] = \left[ \begin{array}{c} 0 \ 0 \ 0 \end{array} \right]\]

Compute Cross Product c

For \(\mathbf{u} = \left[\begin{array}{r}3 \ -2 \ 1\end{array}\right]\) and \(\mathbf{v} = \left[\begin{array}{r}1 \ 1 \ -1\end{array}\right]\), use the cross product formula:\[\mathbf{u} \times \mathbf{v} = \left[ \begin{array}{c}(-2) \cdot (-1) - 1 \cdot 1 \ 1 \cdot 3 - 3 \cdot (-1) \ 3 \cdot 1 - (-2) \cdot 1 \end{array} \right] = \left[ \begin{array}{c}2 - 1 \ 3 + 3 \ 3 + 2 \end{array} \right] = \left[ \begin{array}{c}1 \ 6 \ 5 \end{array} \right]\]

Compute Cross Product d

For \(\mathbf{u} = \left[\begin{array}{r}2 \ 0 \ -1\end{array}\right]\) and \(\mathbf{v} = \left[\begin{array}{l}1 \ 4 \ 7\end{array}\right]\), utilize the formula of the cross product:\[\mathbf{u} \times \mathbf{v} = \left[ \begin{array}{c}0 \cdot 7 - 4 \cdot (-1) \ (-1) \cdot 1 - 7 \cdot 2 \ 2 \cdot 4 - 0 \cdot 1 \end{array} \right] = \left[ \begin{array}{c}4 \ -1 - 14 \ 8 \end{array} \right] = \left[ \begin{array}{c}4 \ -15 \ 8 \end{array} \right]\]

Key concepts

Vector Multiplication

Vector multiplication is a fundamental operation in linear algebra, particularly when dealing with 3D vectors. While the term might suggest similar processes as basic arithmetic multiplication, in linear algebra, vector multiplication can mean either the dot product or the cross product. Here, we focus on the cross product, which is unique to three-dimensional spaces.

The cross product of two vectors, denoted \(\mathbf{u} \times \mathbf{v}\), results in another vector that is perpendicular to the plane formed by the original vectors \(\mathbf{u}\) and \(\mathbf{v}\). This orthogonal vector is determined using the formula:

\[\mathbf{u} \times \mathbf{v} = \left[ \begin{array}{c}u_{2} \, v_{3} - u_{3} \, v_{2} \ v_{3} \, u_{1} - v_{1} \, u_{3} \ u_{1} \, v_{2} - u_{2} \, v_{1} \end{array} \right]\]

Each element of the resulting vector is calculated by combining the components of the original vectors in a cyclic manner. This might sound complex, but through practice, it becomes an intuitive tool for analyzing vectors in physics and engineering.

3D Vectors

Three-dimensional vectors, commonly represented as \(\left[ x, y, z \right]\), provide a way to describe directions and positions in 3D space. Each element corresponds to a dimension: the x, y, and z axes. Vectors can represent anything from velocity in physics, position in graphics, or any directional quantity.

Working with 3D vectors involves understanding operations such as addition, scalar multiplication, and notably here, the cross product.

• The nature of 3D vectors makes them ideal for describing spatial phenomena like force and torque.
• By manipulating these vectors, we can easily transform points within a coordinate system or find perpendicular directions.
• The perpendicular resulting vector from the cross product helps us gain insights into the orientation of surfaces in space.

Understanding how to manipulate and compute with 3D vectors is crucial in fields such as robotics, where spatial orientation is key to operations.

Linear Algebra

Linear algebra is the branch of mathematics that studies vectors, matrices, and their transformations. It's a vital tool in various disciplines, such as computer graphics, engineering, and machine learning. One of its mighty operations is the cross product, which is exclusively defined for 3D vectors.

In the realm of linear algebra, understanding the cross product's properties helps in visualizing and computing complex spatial problems:

• This operation is non-commutative, meaning \(\mathbf{u} \times \mathbf{v} eq \mathbf{v} \times \mathbf{u}\) and often the two results are opposite.
• The magnitude of the cross product vector gives the area of the parallelogram formed by the two original vectors, allowing for simple calculations of area in a spatial context.
• The cross product is instrumental in determining orientation and rotational dynamics, which appear in physics and engineering applications.

Linear algebra provides the foundation for many advanced computations and applications across scientific fields, playing a crucial role in technological advancements.