*The Area of a Parallelogram*

Let $v=(v_1,v_2)$ and $w=(w_1,w_2)$ be two linearly independent vectors in $\mathbb{R}^2$. Then they span a parallelogram as shown in Figure 1.

Figure 1. Parallelogram spanned by two vector v and w.

The area $A$ of the parallelogram is

\begin{align*}

A&=||v||||w||\sin\theta\\

&=||v\times w||\\

&=\left|\begin{array}{cc}

v_1 & v_2\\

w_1 & w_2

\end{array}\right|.

\end{align*}

In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the area.

*Exercise*. Given two vectors $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ in $\mathbb{R}^3$, show that $||v||||w||\sin\theta=||v\times w||$ where $\theta$ is the angle between $v$ and $w$.

Hint. Note that $||v||^2||w||^2\sin^2\theta=||v||^2||w||^2-(v\cdot w)^2$.

*The Volume of a Parallelepiped*

Let $u=(u_1,u_2,u_3)$, $v=(v_1,v_2,v_3)$, $w=(w_1,w_2,w_3)$ be three linearly independent vectors in $\mathbb{R}^3$. Then they span a parallelepiped as shown in Figure 2.

Figure 2. Parallelepiped spanned by vectors u, v and w.

The volume $V$ of the parallelepiped is

\begin{align*}

V&=||u||\cos\theta ||v\times w||\\

&=u\cdot (v\times w)\\

&=\left|\begin{array}{ccc}

u_1 & u_2 & u_3\\

v_1 & v_2 & v_3\\

w_1 & w_2 & w_3

\end{array}\right|.

\end{align*}

In general, the resulting determinant is not necessarily positive. If it is negative, we need to take the absolute value of the determinant for the volume.

### Like this:

Like Loading...

*Related*