# How do you find the parametric equations of the line of intersection of the planes x+2z=0 and 2x-3y+4?

Sep 3, 2016

$\vec{l} = \left(\begin{matrix}0 \\ \frac{4}{3} \\ 0\end{matrix}\right) + t \left(\begin{matrix}6 \\ 4 \\ - 3\end{matrix}\right)$ where $t$ is the parameter.

#### Explanation:

${\pi}_{1} : x + 2 z = 0$
${\pi}_{2} : 2 x - 3 y = - 4$

the line will lie on both planes, clearly. As such the equation of each plane in form ${\pi}_{i} : \left({\vec{r}}_{i} - {\vec{r}}_{o \setminus i}\right) \cdot {\vec{n}}_{i} = 0$ will also hold true for every point on the line apropos both planes.

So the line $\vec{l} = {\vec{l}}_{o} + \vec{d}$ will run in a direction $\vec{d}$ that is perpendicular to the ${\vec{n}}_{1 , 2}$ and its direction will be $\vec{d} = {\vec{n}}_{1} \times {\vec{n}}_{2}$

ie $\vec{d} = \det \left(\begin{matrix}\hat{x} & \hat{y} & \hat{z} \\ 1 & 0 & 2 \\ 2 & - 3 & 0\end{matrix}\right) {\hat{n}}_{d}$

$= 6 \hat{x} + 4 \hat{y} - 3 \hat{z} = \left(\begin{matrix}6 \\ 4 \\ - 3\end{matrix}\right)$

All we now need is a point on the line. if we set x = 0 for both ${\pi}_{1}$ and ${\pi}_{2}$ then we see that $z = 0 , y = \frac{4}{3}$ or ${\vec{l}}_{o} = \left(\begin{matrix}0 \\ \frac{4}{3} \\ 0\end{matrix}\right)$

therefore:

$\vec{l} = \left(\begin{matrix}0 \\ \frac{4}{3} \\ 0\end{matrix}\right) + t \left(\begin{matrix}6 \\ 4 \\ - 3\end{matrix}\right)$ where $t$ is the parameter.

Sep 3, 2016

$\frac{x}{6} = \frac{y - \frac{4}{3}}{4} = \frac{z}{-} 3$ is the reqd. Cartesian Eqn.,

OR,

$\vec{r} = \left(0 , \frac{4}{3} , 0\right) + \lambda \left(6 , 4 , - 3\right) , \lambda \in \mathbb{R}$

#### Explanation:

Let us denote, by ${\pi}_{1} \mathmr{and} {\pi}_{2}$ the given planes, resp.

From ${\pi}_{1} , x = - 2 z \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1\right)$

Sub.ing this $x$ in ${\pi}_{2} ,$ we get,

$- 4 z - 3 y + 4 = 0 , \mathmr{and} , 3 y - 4 = - 4 z \ldots \ldots \ldots \ldots \ldots \ldots \left(2\right)$

Rewriting $\left(1\right)$ as, $2 x = - 4 z \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left(1 '\right)$

And, Obviously, $- 4 z = - 4 z \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(3\right)$

Combining $\left(1 '\right) , \left(2\right) , \mathmr{and} \left(3\right)$, we have,

$2 x = 3 y - 4 = - 4 z , i . e . , 2 x = 3 \left(y - \frac{4}{3}\right) = - 4 z$, or,

$\frac{2 x}{12} = \frac{3 \left(y - \frac{4}{3}\right)}{12} = \frac{- 4 z}{12}$,

$\therefore \frac{x}{6} = \frac{y - \frac{4}{3}}{4} = \frac{z}{-} 3$ is the reqd. Cartesian Eqn. of

the Line $\subset {\pi}_{1} \cap {\pi}_{2}$.

Its vector eqn. can be written as

$\vec{r} = \left(0 , \frac{4}{3} , 0\right) + \lambda \left(6 , 4 , - 3\right) , \lambda \in \mathbb{R}$, in accordance with

the Answer submitted by Respected Eddie !