# Find a Cartesian Equation of the plane with contains the line (x-2)/3=(y+4)/2=(z-1)/2 and passes through the point (1,1,1)?

## The answer in the textbook is $- 10 x - 2 y + 16 z = 4$

Feb 17, 2018

See below.

#### Explanation:

Given the line

$L \to p = {p}_{0} + \lambda \vec{v}$ with

$p = \left(x , y , z\right)$
${p}_{0} = \left(2 , - 4 , 1\right)$ and
$\vec{v} = \left(3 , 2 , 2\right)$

and the point

${p}_{1} = \left(1 , 1 , 1\right)$

we have the plane equation

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

it should obey the condition

$\left\langle{p}_{0} - {p}_{1} + \lambda \vec{v} , \vec{n}\right\rangle = 0$ or

$\left\langle{p}_{0} - {p}_{1} , \vec{n}\right\rangle + \lambda \left\langle\vec{v} , \vec{n}\right\rangle = 0$

so we conclude

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

so we conclude that

$\left({p}_{0} - {p}_{1}\right) \bot \vec{n}$ and $\vec{v} \bot \vec{n}$

so we can choose

$\vec{n} = \left({p}_{0} - {p}_{1}\right) \times \vec{v}$ and then

$\Pi \to \left\langlep - {p}_{1} , \left({p}_{0} - {p}_{1}\right) \times \vec{v}\right\rangle = 0$ is the plane equation or putting values

$10 x + 2 y - 17 z + 5 = 0$