Problem 1 - Algebra and Trigonometry

Question

The cross product of the vectors \(\mathbf{a}=\left\langle a_{1}, a_{2}, a_{3}\right\rangle\) and \(\mathbf{b}=\left\langle b_{1}, b_{2}, b_{3}\right\\}\) is the vector $$ \mathbf{a} \times \mathbf{b}=\left|\begin{array}{lll}{\mathbf{i}} & {\mathbf{j}} & {\mathbf{k}} \\ {} & {} & {}\end{array}\right| $$ = _____ \(\mathbf{i}+\) _____ \(\mathbf{j}+\) _____ \(\mathbf{k}\) So the cross product of \(\mathbf{a}=\langle 1,0,1\rangle\) and \(\mathbf{b}=\langle 2,3,0\rangle\) is a \(\times \mathbf{b}=\) _____.

Step-by-Step Solution

Answer

The cross product \( \mathbf{a} \times \mathbf{b} = -3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \).

Step 1: Understanding the Cross Product Formula

The cross product of two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is the determinant of the matrix constructed by \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \) and the components of the vectors. This can be expressed as:\[\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\]

Step 2: Setting Up the Determinant Matrix

Using the vectors \( \mathbf{a} = \langle 1, 0, 1 \rangle \) and \( \mathbf{b} = \langle 2, 3, 0 \rangle \), set up the determinant matrix:\[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 0 & 1 \ 2 & 3 & 0 \end{vmatrix}\]

Step 3: Calculating the Determinant for \( \mathbf{i} \) Component

To find the \( \mathbf{i} \) component, calculate the determinant of the 2x2 matrix obtained by removing the first column:\[\text{For } \mathbf{i}: \begin{vmatrix} 0 & 1 \ 3 & 0 \end{vmatrix} = (0 \times 0) - (1 \times 3) = -3\]

Step 4: Calculating the Determinant for \( \mathbf{j} \) Component

To find the \( \mathbf{j} \) component, calculate the determinant of the 2x2 matrix obtained by removing the second column:\[\text{For } \mathbf{j}: \begin{vmatrix} 1 & 1 \ 2 & 0 \end{vmatrix} = (1 \times 0) - (1 \times 2) = -2\]

Step 5: Calculating the Determinant for \( \mathbf{k} \) Component

To find the \( \mathbf{k} \) component, calculate the determinant of the 2x2 matrix obtained by removing the third column:\[\text{For } \mathbf{k}: \begin{vmatrix} 1 & 0 \ 2 & 3 \end{vmatrix} = (1 \times 3) - (0 \times 2) = 3\]

Step 6: Forming the Cross Product Vector

Combine all of the components found:\[\mathbf{a} \times \mathbf{b} = (-3)\mathbf{i} - (-2)\mathbf{j} + 3\mathbf{k} = -3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}\]

Key Concepts

Vector Operations

Vectors are fundamental in mathematics and physics because they express quantities that have both a magnitude and a direction. One common operation involving vectors is the **cross product**, which is particularly used in three-dimensional space to find a vector that is perpendicular to two given vectors.

• The cross product is unique to 3D space.
• It results in a vector orthogonal (at right angles) to the two input vectors.
• The direction of the resulting vector is determined by the right-hand rule.

In the exercise, the vectors \( \mathbf{a} = \langle 1, 0, 1 \rangle \) and \( \mathbf{b} = \langle 2, 3, 0 \rangle \) are used to demonstrate this operation. The result \( \mathbf{a} \times \mathbf{b} = -3\mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \) is a vector perpendicular to both \( \mathbf{a} \) and \( \mathbf{b} \). This concept is widely applied in physics, engineering, and computer graphics to calculate torque, the area of parallelograms, and normals to surfaces.

Determinant

The **determinant** is a scalar value that can be computed from the elements of a square matrix. It provides important properties of the matrix, such as whether the system has a unique solution or the volume multiplication factor of the transformation represented by the matrix. When calculating the cross product of two vectors:

• The determinant of a 3x3 matrix constructed with unit vectors \( \mathbf{i} \), \( \mathbf{j} \), \( \mathbf{k} \) and the components of the two vectors is used.
• This matrix looks like:

\[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} \] In calculations, minors and cofactors are used to find the 2x2 determinants within this matrix setup, each contributing to the \( \mathbf{i} \), \( \mathbf{j} \), and \( \mathbf{k} \) components of the resultant vector. This method links linear algebra to vector calculus, aiding in solving geometric problems involving planes and lines.

Basis Vectors

**Basis vectors** are the building blocks of vector spaces. In three-dimensional space, the standard basis vectors are \( \mathbf{i} = \langle 1, 0, 0 \rangle \), \( \mathbf{j} = \langle 0, 1, 0 \rangle \), and \( \mathbf{k} = \langle 0, 0, 1 \rangle \).

• Any vector in 3D space can be expressed as a combination of these three vectors.
• They provide a reference framework for operations like the cross product or dot product.

In the exercise, these basis vectors form the top row of the determinant matrix: \[\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \end{vmatrix}\] These vectors help in breaking down operations within three-dimensional space, allowing for complex calculations like finding normals or rotational axes. They simplify the mathematics behind spatial transformations, making it possible to visualize and compute vector relationships efficiently.

Multivariable Calculus

**Multivariable calculus** extends single variable calculus to functions with multiple inputs, which is crucial in understanding complex real-world phenomena. It helps to analyze and solve problems involving multiple dimensions. In the context of vector calculus, it introduces concepts like

• gradient, which points in the direction of the greatest increase of a function.
• the divergence and curl of vector fields.

The cross product operation stems from multivariable calculus, providing insights into how vectors interact in space. By leveraging the cross product, one can find vectors perpendicular to plane surfaces, a crucial step in deriving surface integrals. The cross product helps in the study of rotation in space, making it indispensable in physics and engineering. Understanding this leads to a greater grasp of topics like electromagnetism, fluid dynamics, and more, where multiple variables and dimensions interact fluidly.