A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.












-1












$begingroup$



A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.



The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.




I have no idea where to even start with this so some help would be appreciated.






geometry vectors 3d cross-product solid-geometry







share|cite|improve this question











$endgroup$

















    -1












    $begingroup$



    A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.



    The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.




    I have no idea where to even start with this so some help would be appreciated.






    geometry vectors 3d cross-product solid-geometry







    share|cite|improve this question











    $endgroup$















      -1












      -1








      -1


      0



      $begingroup$



      A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.



      The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.




      I have no idea where to even start with this so some help would be appreciated.






      geometry vectors 3d cross-product solid-geometry







      share|cite|improve this question











      $endgroup$





      A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.



      The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.




      I have no idea where to even start with this so some help would be appreciated.






      geometry vectors 3d cross-product solid-geometry




      geometry vectors 3d cross-product solid-geometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 16:03







      user593746

















      asked Dec 18 '18 at 14:25









      H.LinkhornH.Linkhorn

      35213




      35213






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
          $$
          v_1 = frac12 a times b,
          $$
          at least for one choice of orientation.



          (By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
          $$
          (-v_1) + v_2 + v_3 + v_4 = 0,
          $$
          from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
            $endgroup$
            – Shubham Johri
            Dec 18 '18 at 14:46










          • $begingroup$
            D'oh! Thanks -- fixed.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 14:47



















          1












          $begingroup$

          You have $$vec v_1=frac 12 vec atimes vec b$$
          You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            ..and fiddle with the signs to actually make things work out. :(
            $endgroup$
            – John Hughes
            Dec 18 '18 at 15:20










          • $begingroup$
            Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
            $endgroup$
            – Andrei
            Dec 18 '18 at 15:34










          • $begingroup$
            "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 16:46










          • $begingroup$
            @JohnHughes Thanks for pointing that out. I will edit my answer.
            $endgroup$
            – Andrei
            Dec 18 '18 at 16:48











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          2 Answers
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          2 Answers
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          active

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          active

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          1












          $begingroup$

          Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
          $$
          v_1 = frac12 a times b,
          $$
          at least for one choice of orientation.



          (By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
          $$
          (-v_1) + v_2 + v_3 + v_4 = 0,
          $$
          from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
            $endgroup$
            – Shubham Johri
            Dec 18 '18 at 14:46










          • $begingroup$
            D'oh! Thanks -- fixed.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 14:47
















          1












          $begingroup$

          Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
          $$
          v_1 = frac12 a times b,
          $$
          at least for one choice of orientation.



          (By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
          $$
          (-v_1) + v_2 + v_3 + v_4 = 0,
          $$
          from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
            $endgroup$
            – Shubham Johri
            Dec 18 '18 at 14:46










          • $begingroup$
            D'oh! Thanks -- fixed.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 14:47














          1












          1








          1





          $begingroup$

          Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
          $$
          v_1 = frac12 a times b,
          $$
          at least for one choice of orientation.



          (By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
          $$
          (-v_1) + v_2 + v_3 + v_4 = 0,
          $$
          from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)






          share|cite|improve this answer











          $endgroup$



          Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
          $$
          v_1 = frac12 a times b,
          $$
          at least for one choice of orientation.



          (By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
          $$
          (-v_1) + v_2 + v_3 + v_4 = 0,
          $$
          from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 14:47

























          answered Dec 18 '18 at 14:37









          John HughesJohn Hughes

          63.1k24090




          63.1k24090












          • $begingroup$
            Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
            $endgroup$
            – Shubham Johri
            Dec 18 '18 at 14:46










          • $begingroup$
            D'oh! Thanks -- fixed.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 14:47


















          • $begingroup$
            Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
            $endgroup$
            – Shubham Johri
            Dec 18 '18 at 14:46










          • $begingroup$
            D'oh! Thanks -- fixed.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 14:47
















          $begingroup$
          Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
          $endgroup$
          – Shubham Johri
          Dec 18 '18 at 14:46




          $begingroup$
          Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
          $endgroup$
          – Shubham Johri
          Dec 18 '18 at 14:46












          $begingroup$
          D'oh! Thanks -- fixed.
          $endgroup$
          – John Hughes
          Dec 18 '18 at 14:47




          $begingroup$
          D'oh! Thanks -- fixed.
          $endgroup$
          – John Hughes
          Dec 18 '18 at 14:47











          1












          $begingroup$

          You have $$vec v_1=frac 12 vec atimes vec b$$
          You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            ..and fiddle with the signs to actually make things work out. :(
            $endgroup$
            – John Hughes
            Dec 18 '18 at 15:20










          • $begingroup$
            Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
            $endgroup$
            – Andrei
            Dec 18 '18 at 15:34










          • $begingroup$
            "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 16:46










          • $begingroup$
            @JohnHughes Thanks for pointing that out. I will edit my answer.
            $endgroup$
            – Andrei
            Dec 18 '18 at 16:48
















          1












          $begingroup$

          You have $$vec v_1=frac 12 vec atimes vec b$$
          You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.






          share|cite|improve this answer











          $endgroup$









          • 2




            $begingroup$
            ..and fiddle with the signs to actually make things work out. :(
            $endgroup$
            – John Hughes
            Dec 18 '18 at 15:20










          • $begingroup$
            Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
            $endgroup$
            – Andrei
            Dec 18 '18 at 15:34










          • $begingroup$
            "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 16:46










          • $begingroup$
            @JohnHughes Thanks for pointing that out. I will edit my answer.
            $endgroup$
            – Andrei
            Dec 18 '18 at 16:48














          1












          1








          1





          $begingroup$

          You have $$vec v_1=frac 12 vec atimes vec b$$
          You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.






          share|cite|improve this answer











          $endgroup$



          You have $$vec v_1=frac 12 vec atimes vec b$$
          You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 16:56

























          answered Dec 18 '18 at 14:59









          AndreiAndrei

          11.7k21026




          11.7k21026








          • 2




            $begingroup$
            ..and fiddle with the signs to actually make things work out. :(
            $endgroup$
            – John Hughes
            Dec 18 '18 at 15:20










          • $begingroup$
            Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
            $endgroup$
            – Andrei
            Dec 18 '18 at 15:34










          • $begingroup$
            "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 16:46










          • $begingroup$
            @JohnHughes Thanks for pointing that out. I will edit my answer.
            $endgroup$
            – Andrei
            Dec 18 '18 at 16:48














          • 2




            $begingroup$
            ..and fiddle with the signs to actually make things work out. :(
            $endgroup$
            – John Hughes
            Dec 18 '18 at 15:20










          • $begingroup$
            Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
            $endgroup$
            – Andrei
            Dec 18 '18 at 15:34










          • $begingroup$
            "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
            $endgroup$
            – John Hughes
            Dec 18 '18 at 16:46










          • $begingroup$
            @JohnHughes Thanks for pointing that out. I will edit my answer.
            $endgroup$
            – Andrei
            Dec 18 '18 at 16:48








          2




          2




          $begingroup$
          ..and fiddle with the signs to actually make things work out. :(
          $endgroup$
          – John Hughes
          Dec 18 '18 at 15:20




          $begingroup$
          ..and fiddle with the signs to actually make things work out. :(
          $endgroup$
          – John Hughes
          Dec 18 '18 at 15:20












          $begingroup$
          Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
          $endgroup$
          – Andrei
          Dec 18 '18 at 15:34




          $begingroup$
          Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
          $endgroup$
          – Andrei
          Dec 18 '18 at 15:34












          $begingroup$
          "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
          $endgroup$
          – John Hughes
          Dec 18 '18 at 16:46




          $begingroup$
          "But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
          $endgroup$
          – John Hughes
          Dec 18 '18 at 16:46












          $begingroup$
          @JohnHughes Thanks for pointing that out. I will edit my answer.
          $endgroup$
          – Andrei
          Dec 18 '18 at 16:48




          $begingroup$
          @JohnHughes Thanks for pointing that out. I will edit my answer.
          $endgroup$
          – Andrei
          Dec 18 '18 at 16:48


















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          .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
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