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Math Help - Linear algebra, finding a subset which is perpendicular to another

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    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}.
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    Hello,

    Quote Originally Posted by flawless View Post
    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} \\ \\<br />
\begin{vmatrix} 2 & 1 \\ {\color{blue}2} & {\color{blue}-2} \end{vmatrix} \\ \\<br />
\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}
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