# How do you find an equation for the plane that contains the line with parametric equations l=(8 - 7t, -5 - 2t , 5 - t) and is parallel to the line with parametric equations x=3 + t, y=-7 + 9t, z=8 - 6t?

##### 1 Answer
Sep 30, 2016

$\Pi \to 21 x - 43 y - 61 z - 78 = 0$

#### Explanation:

Given two lines ${L}_{1} , {L}_{2}$ determine a plane $\Pi$ such that

${L}_{1} \in \Pi$ and ${L}_{2} \notin \Pi$

Here

${L}_{1} \to p = {p}_{1} + \lambda {\vec{v}}_{1}$
${L}_{2} \to p = {p}_{2} + \lambda {\vec{v}}_{2}$

with

${p}_{1} = \left(8 , - 5 , 5\right) , {\vec{v}}_{1} = \left(- 7 , - 2 , - 1\right)$
${p}_{2} = \left(3 , - 7 , 8\right) , {\vec{v}}_{2} = \left(1 , 9 , - 6\right)$

The plane equation is

$\Pi \to p = {p}_{1} + {\lambda}_{1} {\vec{v}}_{1} + {\lambda}_{2} {\vec{v}}_{2}$ because

${L}_{1} \in \Pi$ ( make ${\lambda}_{2} = 0$ ) and is parallel to ${L}_{2}$ ( making ${\lambda}_{2} = 0$)

The nonparametric plane equation can be easily obtained remembering that for the plane

$\left\langlep - {p}_{1} , \vec{n}\right\rangle = 0$

here $p = \left(x , y , z\right)$ and $\vec{n} = {\vec{v}}_{1} \times {\vec{v}}_{2} = \left(21 , - 43 , - 61\right)$ so

$\Pi \to 21 x - 43 y - 61 z - 78 = 0$