# Thread: Linear algebra, finding a subset which is perpendicular to another

1. ## Linear algebra, finding a subset which is perpendicular to another

M is a subset of R^n. We denote the set of vectors which are perpendicular to M as M* (ie each element of M* is perpendicular to each element of M).

(Question) In R^3, find M* where M={2i-j+2k,i-2j-2k}.

2. Hello,

Originally Posted by flawless
M is a subset of R^n. We denote the set of vectors which are perpendicular to M as M* (ie each element of M* is perpendicular to each element of M).

(Question) In R^3, find M* where M={2i-j+2k,i-2j-2k}.
Do you know things about the cross product ?

Here, you have $M=\{u,v\}$, where :

$u=\begin{pmatrix} 2 \\ {\color{red}-1} \\ {\color{blue}2} \end{pmatrix}$ and $v=\begin{pmatrix} 1 \\ {\color{red}-2} \\ {\color{blue}-2} \end{pmatrix}$

$w=u \times v$ will be perpendicular to both u and v.

$w=\begin{pmatrix} \begin{vmatrix} {\color{red}-1} & {\color{red}-2} \\ {\color{blue}2} & {\color{blue}-2} \end{vmatrix} \\ \\
\begin{vmatrix} 2 & 1 \\ {\color{blue}2} & {\color{blue}-2} \end{vmatrix} \\ \\
\begin{vmatrix} 2 & 1 \\ {\color{red}-1} & {\color{red}-2} \end{vmatrix} \end{pmatrix}$

(the first coordinate was obtained by ignoring the line composed of the first coordinates of u and v, then we take the determinant of the remainder... etc for second and third)

So, unless I've made mistakes : $w=\begin{pmatrix} 6 \\ -6 \\ -3 \end{pmatrix}$