Answer
The area of the parallelogram is $\sqrt {15} $.
Work Step by Step
We have $||{\bf{v}}|| = ||{\bf{w}}|| = 2$ and ${\bf{v}}\cdot{\bf{w}} = 1$.
By Eq. (6) of Section 13.4, the area of the parallelogram spanned by vectors ${\bf{v}}$ and ${\bf{w}}$ is $||{\bf{v}} \times {\bf{w}}||$, that is, the magnitude of the vector ${\bf{v}} \times {\bf{w}}$.
By Theorem 1. (ii) of Section 13.4, ${\bf{v}} \times {\bf{w}}$ has length $||{\bf{v}}||||{\bf{w}}||\sin \theta $, where $\theta$ is the angle between ${\bf{v}}$ and ${\bf{w}}$ for $0 \le \theta \le \pi $. Thus,
$||{\bf{v}} \times {\bf{w}}|| = ||{\bf{v}}||||{\bf{w}}||\sin \theta $
Since ${\bf{v}}\cdot{\bf{w}} = 1$, so
${\bf{v}}\cdot{\bf{w}} = ||{\bf{v}}||||{\bf{w}}||\cos \theta = 1$
$2\cdot2\cos \theta = 1$, ${\ \ }$ $\cos \theta = \frac{1}{4}$
Since ${\sin ^2}\theta + {\cos ^2}\theta = 1$, we have
$\sin \theta = \sqrt {1 - {{\cos }^2}\theta } = \sqrt {1 - \frac{1}{{16}}} = \sqrt {\frac{{15}}{{16}}} $
Therefore,
$||{\bf{v}} \times {\bf{w}}|| = ||{\bf{v}}||||{\bf{w}}||\sin \theta $
$||{\bf{v}} \times {\bf{w}}|| = 2\cdot2\cdot\sqrt {\frac{{15}}{{16}}} = \sqrt {15} $
The area of the parallelogram is $\sqrt {15} $.
