这是一份Sydney悉尼大学MATH2021/MATH2921的成功案例

The plane $5 x-3 y+z-10=0$ has normal vector $\mathbf{n}{1}=(5,-3,1)$ and the plane $2 x+4 y-z+3=0$ has normal vector $\mathbf{n}{2}=(2,4,-1)$. Since $\mathbf{n}{1}$ and $\mathbf{n}{2}$ are not scalar multiples, then the two planes are not parallel and hence will intersect. A point $(x, y, z)$ on both planes will satisfy the following system of two equations in three unknowns:
$$
\begin{aligned}
&5 x-3 y+z-10=0 \
&2 x+4 y-z+3=0
\end{aligned}
$$
Set $x=0$ (why is that a good choice?). Then the above equations are reduced to:
$$
\begin{array}{r}
-3 y+z-10=0 \
4 y-z+3=0
\end{array}
$$
The second equation gives $z=4 y+3$, substituting that into the first equation gives $y=7$. Then $z=31$, and so the point $(0,7,31)$ is on $L$. Since $\mathbf{n}{1} \times \mathbf{n}{2}=(-1,7,26)$, then $L$ is given by:
$$
\mathbf{r}+t\left(\mathbf{n}{1} \times \mathbf{n}{2}\right)=(0,7,31)+t(-1,7,26), \text { for }-\infty<t<\infty
$$
or in parametric form:
$$
x=-t, \quad y=7+7 t, \quad z=31+26 t, \text { for }-\infty<t<\infty
$$

MATH2021/MATH2921 COURSE NOTES ：

Recall that if you are given, for example, the definite integral
$$
\int_{1}^{2} x^{3} \sqrt{x^{2}-1} d x,
$$
then you would make the substitution
$$
\begin{aligned}
u &=x^{2}-1 \Rightarrow x^{2}=u+1 \
d u &=2 x d x
\end{aligned}
$$
which changes the limits of integration
$$
\begin{aligned}
&x=1 \Rightarrow u=0 \
&x=2 \Rightarrow u=3
\end{aligned}
$$