Finding perpendicular unit vector in $mathbb{R}^n$ to hyperplane












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Suppose I have $n-1$ linearly independent vectors $(v_1, ..., v_{n-1})$ in $mathbb{R}^n$ that together form a basis of a hyperplane. I'm looking to find a last vector $v$ that is normal to the hyperplane (i.e., perpendicular to all $n-1$ basis vectors), with the extra restrictions that $v$ is a unit vector and that $det([vquad v_1 quad ...quad v_{n-1}])>0$.



Is there a formula to find such a $v$ given $v_1, ..., v_{n-1}$, that is smooth in $v_1, ..., v_{n-1}$? I was considering the cross-product but I'm not sure how it generalizes to $n-1$ vectors and to $mathbb{R}^n$ and how to get the determinant right.






linear-algebra vectors orthogonality smooth-functions







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    Suppose I have $n-1$ linearly independent vectors $(v_1, ..., v_{n-1})$ in $mathbb{R}^n$ that together form a basis of a hyperplane. I'm looking to find a last vector $v$ that is normal to the hyperplane (i.e., perpendicular to all $n-1$ basis vectors), with the extra restrictions that $v$ is a unit vector and that $det([vquad v_1 quad ...quad v_{n-1}])>0$.



    Is there a formula to find such a $v$ given $v_1, ..., v_{n-1}$, that is smooth in $v_1, ..., v_{n-1}$? I was considering the cross-product but I'm not sure how it generalizes to $n-1$ vectors and to $mathbb{R}^n$ and how to get the determinant right.






    linear-algebra vectors orthogonality smooth-functions







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      1



      $begingroup$


      Suppose I have $n-1$ linearly independent vectors $(v_1, ..., v_{n-1})$ in $mathbb{R}^n$ that together form a basis of a hyperplane. I'm looking to find a last vector $v$ that is normal to the hyperplane (i.e., perpendicular to all $n-1$ basis vectors), with the extra restrictions that $v$ is a unit vector and that $det([vquad v_1 quad ...quad v_{n-1}])>0$.



      Is there a formula to find such a $v$ given $v_1, ..., v_{n-1}$, that is smooth in $v_1, ..., v_{n-1}$? I was considering the cross-product but I'm not sure how it generalizes to $n-1$ vectors and to $mathbb{R}^n$ and how to get the determinant right.






      linear-algebra vectors orthogonality smooth-functions







      share|cite|improve this question











      $endgroup$




      Suppose I have $n-1$ linearly independent vectors $(v_1, ..., v_{n-1})$ in $mathbb{R}^n$ that together form a basis of a hyperplane. I'm looking to find a last vector $v$ that is normal to the hyperplane (i.e., perpendicular to all $n-1$ basis vectors), with the extra restrictions that $v$ is a unit vector and that $det([vquad v_1 quad ...quad v_{n-1}])>0$.



      Is there a formula to find such a $v$ given $v_1, ..., v_{n-1}$, that is smooth in $v_1, ..., v_{n-1}$? I was considering the cross-product but I'm not sure how it generalizes to $n-1$ vectors and to $mathbb{R}^n$ and how to get the determinant right.






      linear-algebra vectors orthogonality smooth-functions




      linear-algebra vectors orthogonality smooth-functions






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      share|cite|improve this question













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      edited Dec 2 '18 at 16:40









      José Carlos Santos

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      154k22124227










      asked Dec 2 '18 at 15:30









      DashermanDasherman

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      1,020817






















          2 Answers
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          $begingroup$

          Your idea of generalizing the concept of cross-product is a good one; you can read about it here. And it will indeed give you a smooth function (a rational function, in fact).






          share|cite|improve this answer









          $endgroup$





















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            $begingroup$

            Let $mathbf{e}=(e_1, dots e_n)$ be the canonical basis of $mathbb{R}^n$. The vector you're looking for is $v=frac{sum_{i=1}^nlambda_ie_i}{sqrt{sum_{i=1}^nlambda_i^2}}$ where $lambda_i=det([e_i v_1 dots v_{n-1}])$. To see that it is indeed the vector you're looking for, let $varphi:mathbb{R}^ntomathbb{R}$ be the linear form defined by $varphi(w)=det([w v_1 dots v_{n-1}])$. Since the inner product $langle.,.rangle$ on $mathbb{R}^n$ is non-degenerate, there is one and only one vector $v'in mathbb{R}^n$ such that for every $win mathbb{R}^n$, $varphi(w)=langle w,v'rangle$. This vector is not zero because $varphi$ is not zero. Since $mathbf{e}$ is an orthonormal basis, we have $v'=sum_{i=1}^nlangle e_i,v'rangle e_i$ that is $v'=sum_{i=1}^nvarphi(e_i)e_i=sum_{i=1}^nlambda_ie_i$. Then we can see that, for all $1leqslant i leqslant n-1 $, $langle v_i,v'rangle=varphi(v_i)=det([v_i v_1 dots v_{n-1}])=0$ and $det([v' v_1 dots v_{n-1}])=varphi(v')=Vert v'Vert^2>0$. Finally, we see that $v=frac{v'}{Vert v'Vert}$ is a unit vector orthogonal to the $v_i$'s and $det([v v_1 dots v_{n-1}])=Vert v'Vert>0$.






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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3












              $begingroup$

              Your idea of generalizing the concept of cross-product is a good one; you can read about it here. And it will indeed give you a smooth function (a rational function, in fact).






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Your idea of generalizing the concept of cross-product is a good one; you can read about it here. And it will indeed give you a smooth function (a rational function, in fact).






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Your idea of generalizing the concept of cross-product is a good one; you can read about it here. And it will indeed give you a smooth function (a rational function, in fact).






                  share|cite|improve this answer









                  $endgroup$



                  Your idea of generalizing the concept of cross-product is a good one; you can read about it here. And it will indeed give you a smooth function (a rational function, in fact).







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 2 '18 at 15:35









                  José Carlos SantosJosé Carlos Santos

                  154k22124227




                  154k22124227























                      1












                      $begingroup$

                      Let $mathbf{e}=(e_1, dots e_n)$ be the canonical basis of $mathbb{R}^n$. The vector you're looking for is $v=frac{sum_{i=1}^nlambda_ie_i}{sqrt{sum_{i=1}^nlambda_i^2}}$ where $lambda_i=det([e_i v_1 dots v_{n-1}])$. To see that it is indeed the vector you're looking for, let $varphi:mathbb{R}^ntomathbb{R}$ be the linear form defined by $varphi(w)=det([w v_1 dots v_{n-1}])$. Since the inner product $langle.,.rangle$ on $mathbb{R}^n$ is non-degenerate, there is one and only one vector $v'in mathbb{R}^n$ such that for every $win mathbb{R}^n$, $varphi(w)=langle w,v'rangle$. This vector is not zero because $varphi$ is not zero. Since $mathbf{e}$ is an orthonormal basis, we have $v'=sum_{i=1}^nlangle e_i,v'rangle e_i$ that is $v'=sum_{i=1}^nvarphi(e_i)e_i=sum_{i=1}^nlambda_ie_i$. Then we can see that, for all $1leqslant i leqslant n-1 $, $langle v_i,v'rangle=varphi(v_i)=det([v_i v_1 dots v_{n-1}])=0$ and $det([v' v_1 dots v_{n-1}])=varphi(v')=Vert v'Vert^2>0$. Finally, we see that $v=frac{v'}{Vert v'Vert}$ is a unit vector orthogonal to the $v_i$'s and $det([v v_1 dots v_{n-1}])=Vert v'Vert>0$.






                      share|cite|improve this answer











                      $endgroup$


















                        1












                        $begingroup$

                        Let $mathbf{e}=(e_1, dots e_n)$ be the canonical basis of $mathbb{R}^n$. The vector you're looking for is $v=frac{sum_{i=1}^nlambda_ie_i}{sqrt{sum_{i=1}^nlambda_i^2}}$ where $lambda_i=det([e_i v_1 dots v_{n-1}])$. To see that it is indeed the vector you're looking for, let $varphi:mathbb{R}^ntomathbb{R}$ be the linear form defined by $varphi(w)=det([w v_1 dots v_{n-1}])$. Since the inner product $langle.,.rangle$ on $mathbb{R}^n$ is non-degenerate, there is one and only one vector $v'in mathbb{R}^n$ such that for every $win mathbb{R}^n$, $varphi(w)=langle w,v'rangle$. This vector is not zero because $varphi$ is not zero. Since $mathbf{e}$ is an orthonormal basis, we have $v'=sum_{i=1}^nlangle e_i,v'rangle e_i$ that is $v'=sum_{i=1}^nvarphi(e_i)e_i=sum_{i=1}^nlambda_ie_i$. Then we can see that, for all $1leqslant i leqslant n-1 $, $langle v_i,v'rangle=varphi(v_i)=det([v_i v_1 dots v_{n-1}])=0$ and $det([v' v_1 dots v_{n-1}])=varphi(v')=Vert v'Vert^2>0$. Finally, we see that $v=frac{v'}{Vert v'Vert}$ is a unit vector orthogonal to the $v_i$'s and $det([v v_1 dots v_{n-1}])=Vert v'Vert>0$.






                        share|cite|improve this answer











                        $endgroup$
















                          1












                          1








                          1





                          $begingroup$

                          Let $mathbf{e}=(e_1, dots e_n)$ be the canonical basis of $mathbb{R}^n$. The vector you're looking for is $v=frac{sum_{i=1}^nlambda_ie_i}{sqrt{sum_{i=1}^nlambda_i^2}}$ where $lambda_i=det([e_i v_1 dots v_{n-1}])$. To see that it is indeed the vector you're looking for, let $varphi:mathbb{R}^ntomathbb{R}$ be the linear form defined by $varphi(w)=det([w v_1 dots v_{n-1}])$. Since the inner product $langle.,.rangle$ on $mathbb{R}^n$ is non-degenerate, there is one and only one vector $v'in mathbb{R}^n$ such that for every $win mathbb{R}^n$, $varphi(w)=langle w,v'rangle$. This vector is not zero because $varphi$ is not zero. Since $mathbf{e}$ is an orthonormal basis, we have $v'=sum_{i=1}^nlangle e_i,v'rangle e_i$ that is $v'=sum_{i=1}^nvarphi(e_i)e_i=sum_{i=1}^nlambda_ie_i$. Then we can see that, for all $1leqslant i leqslant n-1 $, $langle v_i,v'rangle=varphi(v_i)=det([v_i v_1 dots v_{n-1}])=0$ and $det([v' v_1 dots v_{n-1}])=varphi(v')=Vert v'Vert^2>0$. Finally, we see that $v=frac{v'}{Vert v'Vert}$ is a unit vector orthogonal to the $v_i$'s and $det([v v_1 dots v_{n-1}])=Vert v'Vert>0$.






                          share|cite|improve this answer











                          $endgroup$



                          Let $mathbf{e}=(e_1, dots e_n)$ be the canonical basis of $mathbb{R}^n$. The vector you're looking for is $v=frac{sum_{i=1}^nlambda_ie_i}{sqrt{sum_{i=1}^nlambda_i^2}}$ where $lambda_i=det([e_i v_1 dots v_{n-1}])$. To see that it is indeed the vector you're looking for, let $varphi:mathbb{R}^ntomathbb{R}$ be the linear form defined by $varphi(w)=det([w v_1 dots v_{n-1}])$. Since the inner product $langle.,.rangle$ on $mathbb{R}^n$ is non-degenerate, there is one and only one vector $v'in mathbb{R}^n$ such that for every $win mathbb{R}^n$, $varphi(w)=langle w,v'rangle$. This vector is not zero because $varphi$ is not zero. Since $mathbf{e}$ is an orthonormal basis, we have $v'=sum_{i=1}^nlangle e_i,v'rangle e_i$ that is $v'=sum_{i=1}^nvarphi(e_i)e_i=sum_{i=1}^nlambda_ie_i$. Then we can see that, for all $1leqslant i leqslant n-1 $, $langle v_i,v'rangle=varphi(v_i)=det([v_i v_1 dots v_{n-1}])=0$ and $det([v' v_1 dots v_{n-1}])=varphi(v')=Vert v'Vert^2>0$. Finally, we see that $v=frac{v'}{Vert v'Vert}$ is a unit vector orthogonal to the $v_i$'s and $det([v v_1 dots v_{n-1}])=Vert v'Vert>0$.







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                          edited Dec 2 '18 at 22:31

























                          answered Dec 2 '18 at 22:21









                          acid_adicacid_adic

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