벡터의 연산

Note that

• the point on $W$ with $x=0$, $y=0$ obeys $-0+3(0)+3z=6$ and so has $z=2$
• the point on $W$ with $x=0$, $y=2$ obeys $-0+3(2)+3z=6$ and so has $z=0$
• the point on $W$ with $x=3$, $y=0$ obeys $-3+3(0)+3z=6$ and so has $z=3$
• the point on $W$ with $x=3$, $y=2$ obeys $-3+3(2)+3z=6$ and so has $z=1$

So the four corners of the parallelogram are $(0,0,2)$, $(0,2,0)$, $(3,0,3)$ and $(3,2,1)$. The vectors

$$\begin{aligned} \mathbf{d}_1&=\left< 0-0 \,,\, 2-0 \,,\, 0-2 \right> = \left< 0 \,,\, 2 \,,\, -2\right> \\ \mathbf{d}_2&=\left< 3-0 \,,\, 0-0 \,,\, 3-2 \right> = \left< 3 \,,\, 0 \,,\, 1\right> \end{aligned}$$

form two sides of the paralleogram. So the area of the parallelogram is

$$\begin{aligned} \big|\mathbf{d}_1\times\mathbf{d}_2\big| =\left|\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 2 & -2 \\ 3 & 0 & 1 \end{matrix}\right]\right| =\left| 2\,\hat{\boldsymbol{\imath}} - 6\,\hat{\boldsymbol{\jmath}} -6\hat{\mathbf{k}} \right| =\sqrt{76} =2\sqrt{19} \end{aligned}$$