Problem on computing a cross product
$\newcommand{\bfA}{\mathbf{A}}$ $\newcommand{\bfB}{\mathbf{B}}$ $\newcommand{\bfC}{\mathbf{C}}$ $\newcommand{\bfF}{\mathbf{F}}$ $\newcommand{\bfI}{\mathbf{I}}$ $\newcommand{\bfa}{\mathbf{a}}$ $\newcommand{\bfb}{\mathbf{b}}$ $\newcommand{\bfc}{\mathbf{c}}$ $\newcommand{\bfd}{\mathbf{d}}$ $\newcommand{\bfe}{\mathbf{e}}$ $\newcommand{\bfi}{\mathbf{i}}$ $\newcommand{\bfj}{\mathbf{j}}$ $\newcommand{\bfk}{\mathbf{k}}$ $\newcommand{\bfn}{\mathbf{n}}$ $\newcommand{\bfr}{\mathbf{r}}$ $\newcommand{\bfu}{\mathbf{u}}$ $\newcommand{\bfv}{\mathbf{v}}$ $\newcommand{\bfw}{\mathbf{w}}$ $\newcommand{\bfx}{\mathbf{x}}$ $\newcommand{\bfy}{\mathbf{y}}$ $\newcommand{\bfz}{\mathbf{z}}$
Compute $\bfi \times (\bfi + \bfk)$ in two ways:
By the determinant formula
By expanding the sum and recalling the cross products of standard coordinate vectors with each other
Solution
Part (a)
Recall that:
Cross product definition
The cross product of $\mathbf{x} = \langle x_1, x_2, x_3\rangle$ and $\mathbf{y} = \langle y_1, y_2, y_3\rangle$ is \begin{align}\mathbf{x} \times \mathbf{y} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{vmatrix} \end{align}
We identify $\bfx = \bfi = \langle 1, 0, 0 \rangle$ and $\bfy = \bfi + \bfk = \langle 1, 0, 1 \rangle.$
Hence,
Determinant of a 3x3 matrix
\begin{align}\mathbf{x} \times \mathbf{y} &= \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 0 \\ 1 & 0 & 1 \end{vmatrix}\\ \ \\ &= \begin{vmatrix}0 & 0 \\ 0 & 1 \end{vmatrix} \ \mathbf{i} - \begin{vmatrix}1 & 0 \\ 1 & 1 \end{vmatrix} \ \mathbf{j} + \begin{vmatrix}1 & 0 \\ 1 & 0 \end{vmatrix} \ \mathbf{k} \end{align}
Determinant of a 2x2 matrix
Evaluating the $2 \times 2$ determinants, we get $$\bfx \times \bfy = - \bfj.$$
Part (b)
Recall that
Algebra of cross products
The cross product of a sum is the sum of cross products: $$\bfa \times (\bfb + \bfc) = \bfa \times \bfb + \bfa \times \bfc.$$
Hence, $\bfi \times (\bfi + \bfk) = \bfi \times \bfi + \bfi \times \bfk$.
Recall that
Cross product length and the angle between vectors
The cross product of any vector with itself is zero.
Cross products of i, j, and k
$\bfi \times \bfk = - \bfj$.
Hence, $\bfi \times \bfi = 0$, and $$\bfi \times (\bfi + \bfk) = -\bfj.$$
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