Question #b2bfd

2 Answers
Mar 15, 2017

See below.

Explanation:

x+y+z+3 =0x+y+z+3=0 is the equation of a plane in RR^3. This plane can be generated by a linear form with the structure

Pi->p = p_0+lambda_1 vec v_1+lambda_2 vec v_2

The given plane is given with the structure

Pi_0-> << p-p_0,vec v >> = 0

with

p = (x,y,z)
p_0=(-1,-1,-1) and
vec v = (1,1,1)

Here Pi and Pi_0 are equivalent if vec v = vec v_1 xx vec v_2 because then

Pi_0-> << p_0+lambda_1 vec v_1+lambda_2 vec v_2-p_0, vec v >> = 0
The affine space Pi is generated by the linear combinations of two independent vectors {vec v_1, vec v_2} in RR^3 with the property:

vec v_1 ne 0, vec v_2 ne 0, << vec v_1, vec v >> = << vec v_2, vec v >>=0

Mar 15, 2017

The linear algebra answer.

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