\mathbf{r} \times \mathbf{s} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ -2 & 1 & 4 \end{vmatrix}

February 22, 2026

$\mathbf{r} \times \mathbf{s} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ -2 & 1 & 4 \end{vmatrix}$

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\boxed{\begin{pmatrix} 2 \\ 0 \\ -1 \end{pmatrix}}
An astrophysics researcher analyzes stellar motion with vectors $\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $\mathbf{s} = \begin{pmatrix} -2 \\ 1 \\ 4 \end{pmatrix}$. Find the area of the parallelogram spanned by $\mathbf{r}$ and $\mathbf{s}$.
The area is the magnitude of the cross product $\mathbf{r} \times \mathbf{s}$:
= \mathbf{i}(2 \cdot 4 - (-1) \cdot 1) - \mathbf{j}(1 \cdot 4 - (-1)(-2)) + \mathbf{k}(1 \cdot 1 - 2 \cdot (-2))
= \mathbf{i}(8 + 1) - \mathbf{j}(4 - 2) + \mathbf{k}(1 + 4)
= 9\mathbf{i} - 2\mathbf{j} + 5\mathbf{k} = \begin{pmatrix} 9 \\ -2 \\ 5 \end{pmatrix}

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