Sums of unit vectors contained in a half-space












2












$begingroup$


Consider $n$ unit vectors ${v_1,...,v_n}$ with $v_iin mathbb{R}^3$. Now define



$text{H}(w):={w'inmathbb{R}^3 | (w',w)>0}, winmathbb{R}^3$



(where $(cdot,cdot)$ is the standard scalar product in $mathbb{R}^3$), i.e. the strict halfspace identified by the vector $w$. Assume $v_iintext{H}(w) forall i=1,...,n$.
For $n=2$, this implies



$v_1+v_2in text{H}(v_1)captext{H}(v_2)$.



Is it true that



$sum_i^n v_iin bigcap_{i}^ntext{H}(v_i)$



for $n>2$?
If the answer is yes, how can one prove it?






geometry convex-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $(w', w)$ do you mean the (standard) inner product of $w'$ and $w$?
    $endgroup$
    – Math1000
    Jan 14 at 21:17










  • $begingroup$
    Yes! Sorry for not specifying.
    $endgroup$
    – Tanatofobico
    Jan 14 at 21:19










  • $begingroup$
    If we relax the constraint that $v_1, v_2$ are unit vectors is our proposition true in the case $n=2,$ when $n>2$ it is possible that $v_i, v_j, v_k $ are nearly parallel to each other, and the result of their sum is nowhere near a unit vector.
    $endgroup$
    – Doug M
    Jan 14 at 21:31










  • $begingroup$
    Not even in the case $n=2$ the sum of two unit vectors is always a unit vector, but I don't see how it should be required that the sum of unit vectors is a unit vector for proving the statement. Am I missing something?
    $endgroup$
    – Tanatofobico
    Jan 14 at 22:04
















2












$begingroup$


Consider $n$ unit vectors ${v_1,...,v_n}$ with $v_iin mathbb{R}^3$. Now define



$text{H}(w):={w'inmathbb{R}^3 | (w',w)>0}, winmathbb{R}^3$



(where $(cdot,cdot)$ is the standard scalar product in $mathbb{R}^3$), i.e. the strict halfspace identified by the vector $w$. Assume $v_iintext{H}(w) forall i=1,...,n$.
For $n=2$, this implies



$v_1+v_2in text{H}(v_1)captext{H}(v_2)$.



Is it true that



$sum_i^n v_iin bigcap_{i}^ntext{H}(v_i)$



for $n>2$?
If the answer is yes, how can one prove it?






geometry convex-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    By $(w', w)$ do you mean the (standard) inner product of $w'$ and $w$?
    $endgroup$
    – Math1000
    Jan 14 at 21:17










  • $begingroup$
    Yes! Sorry for not specifying.
    $endgroup$
    – Tanatofobico
    Jan 14 at 21:19










  • $begingroup$
    If we relax the constraint that $v_1, v_2$ are unit vectors is our proposition true in the case $n=2,$ when $n>2$ it is possible that $v_i, v_j, v_k $ are nearly parallel to each other, and the result of their sum is nowhere near a unit vector.
    $endgroup$
    – Doug M
    Jan 14 at 21:31










  • $begingroup$
    Not even in the case $n=2$ the sum of two unit vectors is always a unit vector, but I don't see how it should be required that the sum of unit vectors is a unit vector for proving the statement. Am I missing something?
    $endgroup$
    – Tanatofobico
    Jan 14 at 22:04














2












2








2





$begingroup$


Consider $n$ unit vectors ${v_1,...,v_n}$ with $v_iin mathbb{R}^3$. Now define



$text{H}(w):={w'inmathbb{R}^3 | (w',w)>0}, winmathbb{R}^3$



(where $(cdot,cdot)$ is the standard scalar product in $mathbb{R}^3$), i.e. the strict halfspace identified by the vector $w$. Assume $v_iintext{H}(w) forall i=1,...,n$.
For $n=2$, this implies



$v_1+v_2in text{H}(v_1)captext{H}(v_2)$.



Is it true that



$sum_i^n v_iin bigcap_{i}^ntext{H}(v_i)$



for $n>2$?
If the answer is yes, how can one prove it?






geometry convex-geometry







share|cite|improve this question











$endgroup$




Consider $n$ unit vectors ${v_1,...,v_n}$ with $v_iin mathbb{R}^3$. Now define



$text{H}(w):={w'inmathbb{R}^3 | (w',w)>0}, winmathbb{R}^3$



(where $(cdot,cdot)$ is the standard scalar product in $mathbb{R}^3$), i.e. the strict halfspace identified by the vector $w$. Assume $v_iintext{H}(w) forall i=1,...,n$.
For $n=2$, this implies



$v_1+v_2in text{H}(v_1)captext{H}(v_2)$.



Is it true that



$sum_i^n v_iin bigcap_{i}^ntext{H}(v_i)$



for $n>2$?
If the answer is yes, how can one prove it?






geometry convex-geometry




geometry convex-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 22:08







Tanatofobico

















asked Jan 14 at 21:12









TanatofobicoTanatofobico

425




425












  • $begingroup$
    By $(w', w)$ do you mean the (standard) inner product of $w'$ and $w$?
    $endgroup$
    – Math1000
    Jan 14 at 21:17










  • $begingroup$
    Yes! Sorry for not specifying.
    $endgroup$
    – Tanatofobico
    Jan 14 at 21:19










  • $begingroup$
    If we relax the constraint that $v_1, v_2$ are unit vectors is our proposition true in the case $n=2,$ when $n>2$ it is possible that $v_i, v_j, v_k $ are nearly parallel to each other, and the result of their sum is nowhere near a unit vector.
    $endgroup$
    – Doug M
    Jan 14 at 21:31










  • $begingroup$
    Not even in the case $n=2$ the sum of two unit vectors is always a unit vector, but I don't see how it should be required that the sum of unit vectors is a unit vector for proving the statement. Am I missing something?
    $endgroup$
    – Tanatofobico
    Jan 14 at 22:04


















  • $begingroup$
    By $(w', w)$ do you mean the (standard) inner product of $w'$ and $w$?
    $endgroup$
    – Math1000
    Jan 14 at 21:17










  • $begingroup$
    Yes! Sorry for not specifying.
    $endgroup$
    – Tanatofobico
    Jan 14 at 21:19










  • $begingroup$
    If we relax the constraint that $v_1, v_2$ are unit vectors is our proposition true in the case $n=2,$ when $n>2$ it is possible that $v_i, v_j, v_k $ are nearly parallel to each other, and the result of their sum is nowhere near a unit vector.
    $endgroup$
    – Doug M
    Jan 14 at 21:31










  • $begingroup$
    Not even in the case $n=2$ the sum of two unit vectors is always a unit vector, but I don't see how it should be required that the sum of unit vectors is a unit vector for proving the statement. Am I missing something?
    $endgroup$
    – Tanatofobico
    Jan 14 at 22:04
















$begingroup$
By $(w', w)$ do you mean the (standard) inner product of $w'$ and $w$?
$endgroup$
– Math1000
Jan 14 at 21:17




$begingroup$
By $(w', w)$ do you mean the (standard) inner product of $w'$ and $w$?
$endgroup$
– Math1000
Jan 14 at 21:17












$begingroup$
Yes! Sorry for not specifying.
$endgroup$
– Tanatofobico
Jan 14 at 21:19




$begingroup$
Yes! Sorry for not specifying.
$endgroup$
– Tanatofobico
Jan 14 at 21:19












$begingroup$
If we relax the constraint that $v_1, v_2$ are unit vectors is our proposition true in the case $n=2,$ when $n>2$ it is possible that $v_i, v_j, v_k $ are nearly parallel to each other, and the result of their sum is nowhere near a unit vector.
$endgroup$
– Doug M
Jan 14 at 21:31




$begingroup$
If we relax the constraint that $v_1, v_2$ are unit vectors is our proposition true in the case $n=2,$ when $n>2$ it is possible that $v_i, v_j, v_k $ are nearly parallel to each other, and the result of their sum is nowhere near a unit vector.
$endgroup$
– Doug M
Jan 14 at 21:31












$begingroup$
Not even in the case $n=2$ the sum of two unit vectors is always a unit vector, but I don't see how it should be required that the sum of unit vectors is a unit vector for proving the statement. Am I missing something?
$endgroup$
– Tanatofobico
Jan 14 at 22:04




$begingroup$
Not even in the case $n=2$ the sum of two unit vectors is always a unit vector, but I don't see how it should be required that the sum of unit vectors is a unit vector for proving the statement. Am I missing something?
$endgroup$
– Tanatofobico
Jan 14 at 22:04










1 Answer
1






active

oldest

votes


















0












$begingroup$

Let $w=(0,1,0)$,



Let $theta in (0, frac{pi}2)$
$$v_1= (costheta, sin theta,0)$$



$$v_2=(-costheta, sin theta,0)$$
$$v_3= (cosfrac{theta}2, sin frac{theta}2,0)$$



We can check that $v_i in H(w)$.



$$sum_{i=1}^3 v_i = (cos frac{theta}2,2sin theta + sin frac{theta}2 ,0)$$



and begin{align}lim_{thetato 0^+}langle sum_{i=1}^3 v_i, v_2rangle&=lim_{thetato 0^+}left(-cos thetacos frac{theta}2+2sin^2theta+sin theta sin frac{ theta}2right) \
&=-1end{align}



In particular, if we let $theta=0.01$,



$$v_1 = (cos(0.01),sin(0.01),0),$$$$ v_2 = (-cos(0.01), sin(0.01), 0), $$$$v_3=(cos(0.005),sin(0.005),0).$$



We can check that $v_1+v_2+v_3 notin H(v_2).$



octave:3> w=[0,1,0]

w =



   0   1   0



octave:4> v1 = [cos(0.01), sin(0.01), 0]

v1 =



   0.99995   0.01000   0.00000



octave:5> v2 = [-cos(0.01), sin(0.01), 0]

v2 =



  -0.99995   0.01000   0.00000





octave:7> v3 = [cos(0.005), sin(0.005), 0]

v3 =



   0.99999   0.00500   0.00000



octave:8> (v1+v2+v3)*v2'

ans = -0.99969





share|cite|improve this answer











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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Let $w=(0,1,0)$,



    Let $theta in (0, frac{pi}2)$
    $$v_1= (costheta, sin theta,0)$$



    $$v_2=(-costheta, sin theta,0)$$
    $$v_3= (cosfrac{theta}2, sin frac{theta}2,0)$$



    We can check that $v_i in H(w)$.



    $$sum_{i=1}^3 v_i = (cos frac{theta}2,2sin theta + sin frac{theta}2 ,0)$$



    and begin{align}lim_{thetato 0^+}langle sum_{i=1}^3 v_i, v_2rangle&=lim_{thetato 0^+}left(-cos thetacos frac{theta}2+2sin^2theta+sin theta sin frac{ theta}2right) \
    &=-1end{align}



    In particular, if we let $theta=0.01$,



    $$v_1 = (cos(0.01),sin(0.01),0),$$$$ v_2 = (-cos(0.01), sin(0.01), 0), $$$$v_3=(cos(0.005),sin(0.005),0).$$



    We can check that $v_1+v_2+v_3 notin H(v_2).$



    octave:3> w=[0,1,0]
    
w =
    

    
   0   1   0
    

    
octave:4> v1 = [cos(0.01), sin(0.01), 0]
    
v1 =
    

    
   0.99995   0.01000   0.00000
    

    
octave:5> v2 = [-cos(0.01), sin(0.01), 0]
    
v2 =
    

    
  -0.99995   0.01000   0.00000
    

    

    
octave:7> v3 = [cos(0.005), sin(0.005), 0]
    
v3 =
    

    
   0.99999   0.00500   0.00000
    

    
octave:8> (v1+v2+v3)*v2'
    
ans = -0.99969





    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Let $w=(0,1,0)$,



      Let $theta in (0, frac{pi}2)$
      $$v_1= (costheta, sin theta,0)$$



      $$v_2=(-costheta, sin theta,0)$$
      $$v_3= (cosfrac{theta}2, sin frac{theta}2,0)$$



      We can check that $v_i in H(w)$.



      $$sum_{i=1}^3 v_i = (cos frac{theta}2,2sin theta + sin frac{theta}2 ,0)$$



      and begin{align}lim_{thetato 0^+}langle sum_{i=1}^3 v_i, v_2rangle&=lim_{thetato 0^+}left(-cos thetacos frac{theta}2+2sin^2theta+sin theta sin frac{ theta}2right) \
      &=-1end{align}



      In particular, if we let $theta=0.01$,



      $$v_1 = (cos(0.01),sin(0.01),0),$$$$ v_2 = (-cos(0.01), sin(0.01), 0), $$$$v_3=(cos(0.005),sin(0.005),0).$$



      We can check that $v_1+v_2+v_3 notin H(v_2).$



      octave:3> w=[0,1,0]
      
w =
      

      
   0   1   0
      

      
octave:4> v1 = [cos(0.01), sin(0.01), 0]
      
v1 =
      

      
   0.99995   0.01000   0.00000
      

      
octave:5> v2 = [-cos(0.01), sin(0.01), 0]
      
v2 =
      

      
  -0.99995   0.01000   0.00000
      

      

      
octave:7> v3 = [cos(0.005), sin(0.005), 0]
      
v3 =
      

      
   0.99999   0.00500   0.00000
      

      
octave:8> (v1+v2+v3)*v2'
      
ans = -0.99969





      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Let $w=(0,1,0)$,



        Let $theta in (0, frac{pi}2)$
        $$v_1= (costheta, sin theta,0)$$



        $$v_2=(-costheta, sin theta,0)$$
        $$v_3= (cosfrac{theta}2, sin frac{theta}2,0)$$



        We can check that $v_i in H(w)$.



        $$sum_{i=1}^3 v_i = (cos frac{theta}2,2sin theta + sin frac{theta}2 ,0)$$



        and begin{align}lim_{thetato 0^+}langle sum_{i=1}^3 v_i, v_2rangle&=lim_{thetato 0^+}left(-cos thetacos frac{theta}2+2sin^2theta+sin theta sin frac{ theta}2right) \
        &=-1end{align}



        In particular, if we let $theta=0.01$,



        $$v_1 = (cos(0.01),sin(0.01),0),$$$$ v_2 = (-cos(0.01), sin(0.01), 0), $$$$v_3=(cos(0.005),sin(0.005),0).$$



        We can check that $v_1+v_2+v_3 notin H(v_2).$



        octave:3> w=[0,1,0]
        
w =
        

        
   0   1   0
        

        
octave:4> v1 = [cos(0.01), sin(0.01), 0]
        
v1 =
        

        
   0.99995   0.01000   0.00000
        

        
octave:5> v2 = [-cos(0.01), sin(0.01), 0]
        
v2 =
        

        
  -0.99995   0.01000   0.00000
        

        

        
octave:7> v3 = [cos(0.005), sin(0.005), 0]
        
v3 =
        

        
   0.99999   0.00500   0.00000
        

        
octave:8> (v1+v2+v3)*v2'
        
ans = -0.99969





        share|cite|improve this answer











        $endgroup$



        Let $w=(0,1,0)$,



        Let $theta in (0, frac{pi}2)$
        $$v_1= (costheta, sin theta,0)$$



        $$v_2=(-costheta, sin theta,0)$$
        $$v_3= (cosfrac{theta}2, sin frac{theta}2,0)$$



        We can check that $v_i in H(w)$.



        $$sum_{i=1}^3 v_i = (cos frac{theta}2,2sin theta + sin frac{theta}2 ,0)$$



        and begin{align}lim_{thetato 0^+}langle sum_{i=1}^3 v_i, v_2rangle&=lim_{thetato 0^+}left(-cos thetacos frac{theta}2+2sin^2theta+sin theta sin frac{ theta}2right) \
        &=-1end{align}



        In particular, if we let $theta=0.01$,



        $$v_1 = (cos(0.01),sin(0.01),0),$$$$ v_2 = (-cos(0.01), sin(0.01), 0), $$$$v_3=(cos(0.005),sin(0.005),0).$$



        We can check that $v_1+v_2+v_3 notin H(v_2).$



        octave:3> w=[0,1,0]
        
w =
        

        
   0   1   0
        

        
octave:4> v1 = [cos(0.01), sin(0.01), 0]
        
v1 =
        

        
   0.99995   0.01000   0.00000
        

        
octave:5> v2 = [-cos(0.01), sin(0.01), 0]
        
v2 =
        

        
  -0.99995   0.01000   0.00000
        

        

        
octave:7> v3 = [cos(0.005), sin(0.005), 0]
        
v3 =
        

        
   0.99999   0.00500   0.00000
        

        
octave:8> (v1+v2+v3)*v2'
        
ans = -0.99969






        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 15 at 11:56

























        answered Jan 15 at 1:16









        Siong Thye GohSiong Thye Goh

        103k1468119




        103k1468119






























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            張江高科駅

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            張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2
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