value of $|vec atimes vec b - vec a times vec c |$












0












$begingroup$



If $vec a,vec b,vec c$ are unit vectors such that $vec a.vec b =
0 = vec a.vec c$ , and the angle between $b$ and $c$ is $pi/3$,
then the value of $|vec atimes vec b - vec a times vec c |$ is?




Attempt:



Let $vec z = vec a times vec b- vec a times vec c$



(Avoiding vec with modulus)



$implies |z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c) $ (since angle between a and b is pi/2, $sin theta = 1$)



$implies |z|^2 = 1+1 - 2[vec atimes vec b ~~~~vec a ~~~~vec c]$



Where denotes box product (scalar triple product)



from property of scalar triple product we get:



$ implies |z|^2 = 2 - 2[vec c ~~~~ vec atimes vec b ~~~~vec a] = 2-2 vec ctimes ((vec a times vec b) . vec a)$



Now $a $ is perpendicular to $atimes b$ so last term should be zero.



$implies |z|^2 = 2 implies |z| = sqrt 2$



But answer given is $1$. Please let me know my mistake.






vectors







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$endgroup$

















    0












    $begingroup$



    If $vec a,vec b,vec c$ are unit vectors such that $vec a.vec b =
    0 = vec a.vec c$     , and the angle between $b$ and $c$ is $pi/3$,
    then the value of $|vec atimes vec b - vec a times vec c |$ is?




    Attempt:



    Let $vec z = vec a times vec b- vec a times vec c$



    (Avoiding vec with modulus)



    $implies |z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c) $ (since angle between a and b is pi/2, $sin theta = 1$)



    $implies |z|^2 = 1+1 - 2[vec atimes vec b ~~~~vec a ~~~~vec c]$



    Where denotes box product (scalar triple product)



    from property of scalar triple product we get:



    $ implies |z|^2 = 2 - 2[vec c ~~~~ vec atimes vec b ~~~~vec a] = 2-2 vec ctimes ((vec a times vec b) . vec a)$



    Now $a $ is perpendicular to $atimes b$ so last term should be zero.



    $implies |z|^2 = 2 implies |z| = sqrt 2$



    But answer given is $1$. Please let me know my mistake.






    vectors







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      If $vec a,vec b,vec c$ are unit vectors such that $vec a.vec b =
      0 = vec a.vec c$       , and the angle between $b$ and $c$ is $pi/3$,
      then the value of $|vec atimes vec b - vec a times vec c |$ is?




      Attempt:



      Let $vec z = vec a times vec b- vec a times vec c$



      (Avoiding vec with modulus)



      $implies |z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c) $ (since angle between a and b is pi/2, $sin theta = 1$)



      $implies |z|^2 = 1+1 - 2[vec atimes vec b ~~~~vec a ~~~~vec c]$



      Where denotes box product (scalar triple product)



      from property of scalar triple product we get:



      $ implies |z|^2 = 2 - 2[vec c ~~~~ vec atimes vec b ~~~~vec a] = 2-2 vec ctimes ((vec a times vec b) . vec a)$



      Now $a $ is perpendicular to $atimes b$ so last term should be zero.



      $implies |z|^2 = 2 implies |z| = sqrt 2$



      But answer given is $1$. Please let me know my mistake.






      vectors







      share|cite|improve this question









      $endgroup$





      If $vec a,vec b,vec c$ are unit vectors such that $vec a.vec b =
      0 = vec a.vec c$       , and the angle between $b$ and $c$ is $pi/3$,
      then the value of $|vec atimes vec b - vec a times vec c |$ is?




      Attempt:



      Let $vec z = vec a times vec b- vec a times vec c$



      (Avoiding vec with modulus)



      $implies |z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c) $ (since angle between a and b is pi/2, $sin theta = 1$)



      $implies |z|^2 = 1+1 - 2[vec atimes vec b ~~~~vec a ~~~~vec c]$



      Where denotes box product (scalar triple product)



      from property of scalar triple product we get:



      $ implies |z|^2 = 2 - 2[vec c ~~~~ vec atimes vec b ~~~~vec a] = 2-2 vec ctimes ((vec a times vec b) . vec a)$



      Now $a $ is perpendicular to $atimes b$ so last term should be zero.



      $implies |z|^2 = 2 implies |z| = sqrt 2$



      But answer given is $1$. Please let me know my mistake.






      vectors




      vectors






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 28 '18 at 5:58









      AbcdAbcd

      3,03821235




      3,03821235






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          The error is that $[c atimes b a]$ does not equal $ctimes((atimes b)
          cdot a)$          . Inside this last expression, $(atimes b)cdot a$ is a scalar, and
          one cannot take the vector product of a vector with a scalar. In fact
          $$[c atimes b a]=ccdot((atimes b)times a).$$
          Now one can use the vector triple product formula to simplify this
          $$(atimes b)times a=(acdot a)b-(bcdot a)a$$
          etc.



          But a simpler way to approach this problem is to note that $atimes b-atimes c
          =atimes (b-c)$          . Since $a$ and $b-c$ are orthogonal, $|atimes (b-c)|=|a||b-c|$
          etc.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How would we find |b-c| ?
            $endgroup$
            – Abcd
            Nov 28 '18 at 6:13










          • $begingroup$
            @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
            $endgroup$
            – Lord Shark the Unknown
            Nov 28 '18 at 6:15



















          1












          $begingroup$

          After writing



          $|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)$



          Notice that the vectors $vec c, vec b, (vec a times vec c)$, and $(vec a times vec b)$ are coplanar and that the angle between $(vec a times vec b)$ and $(vec a times vec c)$ is also $frac{pi}{3}$



          So,$(vec a times vec b). (vec a times vec c)=1×1×cos frac{pi}{3}= frac{1}{2}$



          Now substitue in the first equation and you'll get that
          $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)=1+1-2frac{1}{2}=1$$






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            Use the Gram-determinant:
            $$begin{align}
            |atimes(b-c)|^2&=|a|^2|b-c|^2-langle a,b-crangle^2\
            &=1cdot(|b|^2-2langle b,crangle+|c|^2)-(langle a,brangle-langle a,crangle)^2\
            &=1-2cos(pi/3)+1-(0-0)^2\
            &=1.
            end{align}$$






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              1












              $begingroup$

              The error is that $[c atimes b a]$ does not equal $ctimes((atimes b)
              cdot a)$              . Inside this last expression, $(atimes b)cdot a$ is a scalar, and
              one cannot take the vector product of a vector with a scalar. In fact
              $$[c atimes b a]=ccdot((atimes b)times a).$$
              Now one can use the vector triple product formula to simplify this
              $$(atimes b)times a=(acdot a)b-(bcdot a)a$$
              etc.



              But a simpler way to approach this problem is to note that $atimes b-atimes c
              =atimes (b-c)$              . Since $a$ and $b-c$ are orthogonal, $|atimes (b-c)|=|a||b-c|$
              etc.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How would we find |b-c| ?
                $endgroup$
                – Abcd
                Nov 28 '18 at 6:13










              • $begingroup$
                @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
                $endgroup$
                – Lord Shark the Unknown
                Nov 28 '18 at 6:15
















              1












              $begingroup$

              The error is that $[c atimes b a]$ does not equal $ctimes((atimes b)
              cdot a)$              . Inside this last expression, $(atimes b)cdot a$ is a scalar, and
              one cannot take the vector product of a vector with a scalar. In fact
              $$[c atimes b a]=ccdot((atimes b)times a).$$
              Now one can use the vector triple product formula to simplify this
              $$(atimes b)times a=(acdot a)b-(bcdot a)a$$
              etc.



              But a simpler way to approach this problem is to note that $atimes b-atimes c
              =atimes (b-c)$              . Since $a$ and $b-c$ are orthogonal, $|atimes (b-c)|=|a||b-c|$
              etc.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                How would we find |b-c| ?
                $endgroup$
                – Abcd
                Nov 28 '18 at 6:13










              • $begingroup$
                @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
                $endgroup$
                – Lord Shark the Unknown
                Nov 28 '18 at 6:15














              1












              1








              1





              $begingroup$

              The error is that $[c atimes b a]$ does not equal $ctimes((atimes b)
              cdot a)$              . Inside this last expression, $(atimes b)cdot a$ is a scalar, and
              one cannot take the vector product of a vector with a scalar. In fact
              $$[c atimes b a]=ccdot((atimes b)times a).$$
              Now one can use the vector triple product formula to simplify this
              $$(atimes b)times a=(acdot a)b-(bcdot a)a$$
              etc.



              But a simpler way to approach this problem is to note that $atimes b-atimes c
              =atimes (b-c)$              . Since $a$ and $b-c$ are orthogonal, $|atimes (b-c)|=|a||b-c|$
              etc.






              share|cite|improve this answer









              $endgroup$



              The error is that $[c atimes b a]$ does not equal $ctimes((atimes b)
              cdot a)$              . Inside this last expression, $(atimes b)cdot a$ is a scalar, and
              one cannot take the vector product of a vector with a scalar. In fact
              $$[c atimes b a]=ccdot((atimes b)times a).$$
              Now one can use the vector triple product formula to simplify this
              $$(atimes b)times a=(acdot a)b-(bcdot a)a$$
              etc.



              But a simpler way to approach this problem is to note that $atimes b-atimes c
              =atimes (b-c)$              . Since $a$ and $b-c$ are orthogonal, $|atimes (b-c)|=|a||b-c|$
              etc.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 28 '18 at 6:09









              Lord Shark the UnknownLord Shark the Unknown

              103k1160132




              103k1160132












              • $begingroup$
                How would we find |b-c| ?
                $endgroup$
                – Abcd
                Nov 28 '18 at 6:13










              • $begingroup$
                @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
                $endgroup$
                – Lord Shark the Unknown
                Nov 28 '18 at 6:15


















              • $begingroup$
                How would we find |b-c| ?
                $endgroup$
                – Abcd
                Nov 28 '18 at 6:13










              • $begingroup$
                @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
                $endgroup$
                – Lord Shark the Unknown
                Nov 28 '18 at 6:15
















              $begingroup$
              How would we find |b-c| ?
              $endgroup$
              – Abcd
              Nov 28 '18 at 6:13




              $begingroup$
              How would we find |b-c| ?
              $endgroup$
              – Abcd
              Nov 28 '18 at 6:13












              $begingroup$
              @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
              $endgroup$
              – Lord Shark the Unknown
              Nov 28 '18 at 6:15




              $begingroup$
              @Abcd Either by simple plane geometry, or by working out $|b-c|^2$.
              $endgroup$
              – Lord Shark the Unknown
              Nov 28 '18 at 6:15











              1












              $begingroup$

              After writing



              $|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)$



              Notice that the vectors $vec c, vec b, (vec a times vec c)$, and $(vec a times vec b)$ are coplanar and that the angle between $(vec a times vec b)$ and $(vec a times vec c)$ is also $frac{pi}{3}$



              So,$(vec a times vec b). (vec a times vec c)=1×1×cos frac{pi}{3}= frac{1}{2}$



              Now substitue in the first equation and you'll get that
              $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)=1+1-2frac{1}{2}=1$$






              share|cite|improve this answer











              $endgroup$


















                1












                $begingroup$

                After writing



                $|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)$



                Notice that the vectors $vec c, vec b, (vec a times vec c)$, and $(vec a times vec b)$ are coplanar and that the angle between $(vec a times vec b)$ and $(vec a times vec c)$ is also $frac{pi}{3}$



                So,$(vec a times vec b). (vec a times vec c)=1×1×cos frac{pi}{3}= frac{1}{2}$



                Now substitue in the first equation and you'll get that
                $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)=1+1-2frac{1}{2}=1$$






                share|cite|improve this answer











                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  After writing



                  $|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)$



                  Notice that the vectors $vec c, vec b, (vec a times vec c)$, and $(vec a times vec b)$ are coplanar and that the angle between $(vec a times vec b)$ and $(vec a times vec c)$ is also $frac{pi}{3}$



                  So,$(vec a times vec b). (vec a times vec c)=1×1×cos frac{pi}{3}= frac{1}{2}$



                  Now substitue in the first equation and you'll get that
                  $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)=1+1-2frac{1}{2}=1$$






                  share|cite|improve this answer











                  $endgroup$



                  After writing



                  $|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)$



                  Notice that the vectors $vec c, vec b, (vec a times vec c)$, and $(vec a times vec b)$ are coplanar and that the angle between $(vec a times vec b)$ and $(vec a times vec c)$ is also $frac{pi}{3}$



                  So,$(vec a times vec b). (vec a times vec c)=1×1×cos frac{pi}{3}= frac{1}{2}$



                  Now substitue in the first equation and you'll get that
                  $$|z|^2 = |a|^2|b|^2 + |a|^2|c|^2 - 2 (vec a times vec b). (vec a times vec c)=1+1-2frac{1}{2}=1$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 28 '18 at 6:38

























                  answered Nov 28 '18 at 6:32









                  Fareed AFFareed AF

                  47711




                  47711























                      0












                      $begingroup$

                      Use the Gram-determinant:
                      $$begin{align}
                      |atimes(b-c)|^2&=|a|^2|b-c|^2-langle a,b-crangle^2\
                      &=1cdot(|b|^2-2langle b,crangle+|c|^2)-(langle a,brangle-langle a,crangle)^2\
                      &=1-2cos(pi/3)+1-(0-0)^2\
                      &=1.
                      end{align}$$






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        Use the Gram-determinant:
                        $$begin{align}
                        |atimes(b-c)|^2&=|a|^2|b-c|^2-langle a,b-crangle^2\
                        &=1cdot(|b|^2-2langle b,crangle+|c|^2)-(langle a,brangle-langle a,crangle)^2\
                        &=1-2cos(pi/3)+1-(0-0)^2\
                        &=1.
                        end{align}$$






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Use the Gram-determinant:
                          $$begin{align}
                          |atimes(b-c)|^2&=|a|^2|b-c|^2-langle a,b-crangle^2\
                          &=1cdot(|b|^2-2langle b,crangle+|c|^2)-(langle a,brangle-langle a,crangle)^2\
                          &=1-2cos(pi/3)+1-(0-0)^2\
                          &=1.
                          end{align}$$






                          share|cite|improve this answer









                          $endgroup$



                          Use the Gram-determinant:
                          $$begin{align}
                          |atimes(b-c)|^2&=|a|^2|b-c|^2-langle a,b-crangle^2\
                          &=1cdot(|b|^2-2langle b,crangle+|c|^2)-(langle a,brangle-langle a,crangle)^2\
                          &=1-2cos(pi/3)+1-(0-0)^2\
                          &=1.
                          end{align}$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 28 '18 at 20:32









                          Michael HoppeMichael Hoppe

                          10.8k31834




                          10.8k31834






























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