Find vec c=vec a×vec b and the area of the parallelogram…

Find $\vec c=\vec a\times\vec b$ and the area of the parallelogram constructed on $\vec a=2\vec i+3\vec j+0\vec k$ and $\vec b=0\vec i+3\vec j+2\vec k$.

Explanation of Method

The vector product of a and b and the area of their parallelogram.

Step 1: The vector product

\[ \vec c=\begin{vmatrix}\vec i&\vec j&\vec k\\2&3&0\\0&3&2\end{vmatrix}=\vec i(6-0)-\vec j(4-0)+\vec k(6-0). \]

\[\vec c=\vec a\times\vec b=\{6,-4,6\}\]

Check: $\vec c\cdot\vec a=12-12+0=0$ and $\vec c\cdot\vec b=0-12+12=0$ $\checkmark$

Step 2: Area

\[ S=|\vec c|=\sqrt{36+16+36}=\sqrt{88}. \]

\[S=2\sqrt{22}\approx9.38\]

Lagrange check: $|\vec a|^2|\vec b|^2-(\vec a\cdot\vec b)^2=13\cdot13-9^2=88$ $\checkmark$

$O$ $\vec a$ $\vec b$ $\tfrac12\vec c$

$\vec c=\vec a\times\vec b=\{6,-4,6\}\ \text{(half scale)}$

$\vec c$ is perpendicular to both $\vec a$ and $\vec b$ (right-angle marks); $|\vec c|=2\sqrt{22}$ is the area of the shaded parallelogram.

Result

\[\boxed{\,\vec c=\vec a\times\vec b=\{6,-4,6\}\,}\]

\[\boxed{\,S=2\sqrt{22}\approx9.38\,}\]