Question

Let W be a solutions space of homogeneous equation 2a+3b+4c=0
Describe the cosets of W in R^3

Answer 1

$Solution: Given ~that~ "W~ be~ a~ solution~ space~ of~ homogenuous~ equation~ 2a+2b+4c=0"$

$\therefore solving ~2a+2b+4c=0 \\ \therefore a+b+2c=0 \\a=-b-2c~where ~b ~and~c~ are~ free ~variables. \\\ This ~can~be ~written~as~$

$W=\begin{bmatrix} a \\ b \\ c\\ \end{bmatrix}={\begin{bmatrix} -b-2c \\ b \\ c\\ \end{bmatrix}}=-b\begin{bmatrix} -1 \\ 1\\ 0\\ \end{bmatrix}+c\begin{bmatrix} -2\\ 0\\ 1\\ \end{bmatrix}$

$\therefore W=\begin{bmatrix} -1 \\ 1\\ 0\\ \end{bmatrix},\begin{bmatrix} -2\\ 0\\ 1\\ \end{bmatrix} ~ these ~are~ the~ two~vectors~in~R^3$

$now~ after~ observing ~theses~ vectors~we ~conclude~ that~,these ~are~equations~in~straight~lines.$ In three dimensional space the cosets of a line (plane) through the origin is the set of all lines (planes) parallel to it.

If we consider the equation $a=-b-2c~where ~b ~and~c~ are~ free ~variables.$

just replacing b and c values, keeping a constant, we get different solutions and all these solutions are gives parallel lines to each other. So all these are cosets of the given line 2a+2b+4c=0 in $R^3.$