这是一份Sydney悉尼大学MATH1023/MATH1923的成功案例

$$
\left[\begin{array}{c}
1 \
0 \
-3
\end{array}\right] \times\left[\begin{array}{l}
0 \
1 \
2
\end{array}\right]
$$
Solution: We have
$$
\begin{aligned}
(\vec{i}-3 \vec{k}) \times(\vec{j}+2 \vec{k}) &=\vec{i} \times \vec{j}+2 \vec{i} \times \vec{k}-3 \vec{k} \times \vec{j}-6 \vec{k} \times \vec{k} \
&=\vec{k}-2 \vec{j}+3 \vec{i}+6 \overrightarrow{0} \
&=3 \vec{i}-2 \vec{j}+\vec{k}
\end{aligned}
$$
Hence
$$
\left[\begin{array}{c}
1 \
0 \
-3
\end{array}\right] \times\left[\begin{array}{l}
0 \
1 \
2
\end{array}\right]=\left[\begin{array}{c}
3 \
-2 \
1
\end{array}\right]
$$

MATH1023/MATH1923 COURSE NOTES ：

The area of the triangle is
$$
\frac{\left|\overrightarrow{\mathrm{AD}^{\prime}} \times \overrightarrow{\mathrm{AM}}\right|}{2}=\frac{1}{2} \sqrt{3^{2}+1^{2}+6^{2}}=\frac{\sqrt{46}}{2} .
$$

We have $\overrightarrow{\mathbf{A C}^{\prime}}=\left[\begin{array}{c}-2 \ 3 \ 1\end{array}\right]$, and hence the line $\overleftrightarrow{\mathbf{A C}^{\prime}}$ has parametric equation
$$
\left(\begin{array}{l}
x \
y \
z
\end{array}\right)=\left(\begin{array}{l}
2 \
0 \
0
\end{array}\right)+t\left[\begin{array}{c}
-2 \
3 \
1
\end{array}\right] \Longrightarrow x=2-2 t, y=3 t, z=t .
$$

Since $P$ is on the $z$-axis, $P=\left(\begin{array}{c}0 \ 0 \ z^{\prime}\end{array}\right)$ for some real number $z^{\prime}>0$. The parametric equation of the line $\overleftrightarrow{\mathrm{MP}}$ is thus
$$
\left(\begin{array}{l}
x \
y \
z
\end{array}\right)=\left(\begin{array}{l}
1 \
3 \
0
\end{array}\right)+s\left[\begin{array}{c}
-1 \
-3 \
z^{\prime}
\end{array}\right] \Longrightarrow x=1-s \quad y=3-3 s, \quad z=s z^{\prime}
$$
Since $N$ is on both $\overleftrightarrow{\mathrm{MP}}$ and $\overleftrightarrow{\mathrm{AC}}$ ‘ we must have
$$
2-2 t=1-s, 3 t=3-3 s, t=s z^{\prime}
$$