Find the straight line from two plane vectors Two planes have equations $x + 3y - 2z = 4$ and $2x +y +3z = 5$. The planes intersect in the straight line $l$.
Find a vector equation for the line $l$.
How can I do this type of question?
 A: Method I:
Let $z=t$. Solve $\begin{cases} x+3y=2t+4 \\ 2x+y=5-3t \end{cases}$ for $x$ and $y$. Then you have $(x,y,z)=(at+b, ct+d,t)$ for some $a,b,c,d\in\mathbb{R}$. The vector equation is $\mathbf{r}=(a,c,1)t+(b,d,0)$.
Method II:
$(1,3,-2)\times (2,1,3)$ is a direction vector of the line. Take an arbitrary value of $z$ (say $0$) and solve for $x$, $y$ to obtain a point on the line.
Method III:
Take two arbitrary values of $z$ (say $0$ and $1$). Solve for $x$ and $y$ in each case to obtain two points on the line. The difference of the two points gives a direction vector of the line.
A: Let $\vec{l}(a,b,c)$.
Hence, $a+3b-2c=0$ and $2a+b+3c=0$, which gives $b=-\frac{7}{11}a$ and $c=-\frac{5}{11}a$.
Let $a=11$. Hence, $b=-7$ and $c=-5$ and $\vec{l}(11,-7,-5)$.
But for $z=0$ we get $x=\frac{11}{5}$ and $y=\frac{3}{5}$, which gets an answer:
$$\left\{\left(\frac{11}{5}+11t,\frac{3}{5}-7t,-5t\right)|t\in\mathbb R\right\}$$
A: Find the intersection:
$$\begin{cases}I\;\;&x+3y-2z=4\\{}\\II\;\;&2x+y+3z=5\end{cases}\implies II-2I:\;\;-5y+7z=-3\implies\color{red}{y=\frac75z+\frac35}$$
and substituting in $\;I\;$ :
$$x+\frac{21}5z+\frac95-2z=4\implies \color{red}{x=-\frac{11}5z+\frac{11}5}$$
and thus the line is
$$\left(\frac{11}5\,,\,\,\frac35\,,\,0\right)+t\left(-11,\,7,\,5\right)\;,\;\;t\in\Bbb R$$
