Question

Find a vector perpendicular to both \(\mathbf{i}-3 \mathrm{j}+2 \mathrm{k}\) and \(5 \mathrm{i}-\mathrm{j}-\mathrm{k}\).

Short answer

The vector perpendicular to both is \(5 \textbf{i} + 11 \textbf{j} + 14 \textbf{k}\).

Step-by-step solution

Identify Given Vectors

The given vectors are \(\textbf{A} = \textbf{i} - 3 \textbf{j} + 2 \textbf{k}\) and \(\textbf{B} = 5 \textbf{i} - \textbf{j} - \textbf{k}\).

Set Up the Cross Product Formula

To find a vector perpendicular to both \(\textbf{A}\) and \(\textbf{B}\), use the cross product \(\textbf{A} \times \textbf{B}\). For vectors in component form, the cross product is \[\textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 1 & -3 & 2 \ 5 & -1 & -1 \ \end{vmatrix} \]

Calculate the Determinant

Evaluate the determinant: \[\textbf{A} \times \textbf{B} = \textbf{i} \begin{vmatrix} -3 & 2 \ -1 & -1 \end{vmatrix} - \textbf{j} \begin{vmatrix} 1 & 2 \ 5 & -1 \end{vmatrix} + \textbf{k} \begin{vmatrix} 1 & -3 \ 5 & -1 \end{vmatrix} \]

Evaluate Minor Determinants

Calculate each minor determinant: \[ \textbf{i}((-3 \times -1) - (2 \times -1)) = \textbf{i}(3 + 2) = 5 \textbf{i} \] \[ -\textbf{j}((1 \times -1) - (2 \times 5)) = -\textbf{j}(-1 - 10) = -\textbf{j}(-11) = 11 \textbf{j} \] \[ \textbf{k}((1 \times -1) - (-3 \times 5)) = \textbf{k}(-1 + 15) = 14 \textbf{k} \]

Combine Results

Combine the results of the minor determinants to form the perpendicular vector: \(\textbf{A} \times \textbf{B} = 5 \textbf{i} + 11 \textbf{j} + 14 \textbf{k}\).

Key concepts

Vector Perpendicularity

To understand the concept of **vector perpendicularity**, it's essential to grasp what it means for vectors to be perpendicular. Two vectors are perpendicular (or orthogonal) if their dot product is zero. However, when we need a vector that is perpendicular to two given vectors, the **cross product** becomes very useful. The cross product of two vectors results in a third vector that is perpendicular to both of the original vectors. For example, in our exercise, the cross product of \( \textbf{i} - 3\textbf{j} + 2\textbf{k} \) and \( 5\textbf{i} - \textbf{j} - \textbf{k} \) gives us a vector that is orthogonal to both.

Determinants in Linear Algebra

Determinants are a crucial concept in linear algebra. They are used to perform various calculations, including finding the cross product of vectors. For two vectors, the cross product can be visualized as the determinant of a 3x3 matrix. Understanding determinants helps in evaluating such a matrix.
In our cross product calculation, we set up the determinant as follows: \[ \textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 1 & -3 & 2 \ 5 & -1 & -1 \ \end{vmatrix} \] Each minor (2x2 determinant) is then calculated step-by-step, leading us to the final perpendicular vector. These determinants break the original matrix into smaller parts, simplifying our problem.

Vector Components

Vectors are often broken down into components to simplify calculations and improve understanding. Each vector has three components corresponding to the x, y, and z-axes: \textbf{i}, \textbf{j}, and \textbf{k}. For example, the vector \( \textbf{A} = \textbf{i} - 3\textbf{j} + 2\textbf{k} \) has components of 1, -3, and 2, respectively.
When performing a cross product, we treat each of these components as elements of our matrix. By understanding how these components interact in operations like addition, subtraction, and cross products, you can more easily visualize and solve problems in vector analysis.

• The x-component is aligned with \textbf{i}
• The y-component is aligned with \textbf{j}
• The z-component is aligned with \textbf{k}

Breaking vectors into components allows us to handle complex tasks in vector mathematics more efficiently. This concept is reinforced by the exercise, where we've used vector components to systematically find the cross product and derive a perpendicular vector.