Question

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

Short answer

The vector \(-2\textbf{i} + 2\textbf{j} - \textbf{k}\) is perpendicular to both \(\textbf{i} + \textbf{j}\) and \(\textbf{i} - 2\textbf{k}\).

Step-by-step solution

Determine given vectors

Identify the given vectors. Let's call them \(\textbf{a}\) and \(\textbf{b}\). Thus, \(\textbf{a} = \textbf{i} + \textbf{j}\) and \(\textbf{b} = \textbf{i} - 2\textbf{k}\).

Write the vectors in component form

Convert the vectors \(\textbf{a}\) and \(\textbf{b}\) into their component forms: \(\textbf{a} = (1, 1, 0)\) and \(\textbf{b} = (1, 0, -2)\).

Compute the cross product

To find a vector perpendicular to both \(\textbf{a}\) and \(\textbf{b}\), calculate their cross product \(\textbf{a} \times \textbf{b}\). Use the determinant method for 3D vectors: \[ \textbf{a} \times \textbf{b} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 1 & 1 & 0 \ 1 & 0 & -2 \ \begin{vmatrix} \textbf{i}(1 \times -2 - 0 \times 0) - \textbf{j}(1 \times -2 - 0 \times 1) + \textbf{k}(1 \times 0 - 1 \times 1) \ = \textbf{i}(-2) - \textbf{j}(-2) + \textbf{k}(-1) \ = -2\textbf{i} + 2\textbf{j} - \textbf{k} \ \textbf{a} \times \textbf{b} = (-2, 2, -1) \].

Key concepts

vector mathematics

Vector mathematics is a branch of mathematics focusing on vectors, which are quantities defined by both magnitude and direction. Vectors are commonly represented in a coordinate system.
For example, \(\mathbf{a} = (1, 1, 0)\) and \(\mathbf{b} = (1, 0, -2)\) are vectors in 3D space. The coordinates within the parentheses represent the components of the vector in each spatial direction (x, y, z).
Vectors can undergo several operations, such as addition, subtraction, and multiplication. The cross product, which we'll discuss shortly, is a specific operation useful for finding vectors that are perpendicular to two other given vectors.
Understanding vector mathematics helps in various fields like physics, engineering, and computer graphics, as vectors can represent forces, velocities, and directions.

determinants

Determinants are mathematical constructs used in multilinear algebra to solve linear equations, find areas, and calculate volumes. In vector mathematics, determinants are crucial for computing the cross product of vectors.
The determinant of a 3x3 matrix is found using a specific rule which entails minor matrices and cofactors. For vectors \(\mathbf{a}\) and \(\mathbf{b}\), the cross product \(\mathbf{a} \times \mathbf{b}\) is computed using the determinant of a matrix formed by the unit vectors \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) and the components of \(\mathbf{a}\) and \(\mathbf{b}\):
\[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 0 \ 1 & 0 & -2 \end{vmatrix} \]
Here, you expand the matrix to find each component of the resulting vector through minors and cofactors.
If you're unfamiliar with matrix operations and determinants, practice small 2x2 and 3x3 matrices to become comfortable with expansion and the rules.

perpendicular vectors

Perpendicular vectors are vectors that meet at a right angle (90 degrees). If vectors are perpendicular, their dot product is zero. However, the cross product of two vectors yields another vector that is perpendicular to both.
When given two vectors \(\mathbf{a} = (1, 1, 0)\) and \(\mathbf{b} = (1, 0, -2)\), the cross product \(\mathbf{a} \times \mathbf{b}\) results in a vector that is perpendicular to both. In our problem, it was calculated as:
\[ \mathbf{a} \times \mathbf{b} = -2\mathbf{i} + 2\mathbf{j} - \mathbf{k} = (-2, 2, -1) \]
To verify, you can check that the dot product of \(\mathbf{a}\) with the result and \(\mathbf{b}\) with the result is zero. This confirms perpendicularity.
Understanding perpendicular vectors is essential in physics for resolving force components and in computer graphics for calculating normal vectors for surfaces.