# Question #b2bfd

Mar 15, 2017

See below.

#### Explanation:

$x + y + z + 3 = 0$ is the equation of a plane in ${\mathbb{R}}^{3}$. This plane can be generated by a linear form with the structure

$\Pi \to p = {p}_{0} + {\lambda}_{1} {\vec{v}}_{1} + {\lambda}_{2} {\vec{v}}_{2}$

The given plane is given with the structure

${\Pi}_{0} \to \left\langlep - {p}_{0} , \vec{v}\right\rangle = 0$

with

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

Here $\Pi$ and ${\Pi}_{0}$ are equivalent if $\vec{v} = {\vec{v}}_{1} \times {\vec{v}}_{2}$ because then

${\Pi}_{0} \to \left\langle{p}_{0} + {\lambda}_{1} {\vec{v}}_{1} + {\lambda}_{2} {\vec{v}}_{2} - {p}_{0} , \vec{v}\right\rangle = 0$
The affine space $\Pi$ is generated by the linear combinations of two independent vectors $\left\{{\vec{v}}_{1} , {\vec{v}}_{2}\right\} \in {\mathbb{R}}^{3}$ with the property:

${\vec{v}}_{1} \ne 0 , {\vec{v}}_{2} \ne 0 , \left\langle{\vec{v}}_{1} , \vec{v}\right\rangle = \left\langle{\vec{v}}_{2} , \vec{v}\right\rangle = 0$

Mar 15, 2017

This matrix has 3 independent variables but is of rank 1 only , so it will have $3 - 1 = 2$ linearly independent solutions.