Neep help justifying a vector relation given in a question












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$begingroup$


i'm trying to do a question for which I was given the following line equations:



$underline r = underline a + lambda underline u$



$underline r' = underline a' + lambda' underline u'$



They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.



$lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$



I know that



$lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $



but cant get any further.






linear-algebra geometry vectors







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    0












    $begingroup$


    i'm trying to do a question for which I was given the following line equations:



    $underline r = underline a + lambda underline u$



    $underline r' = underline a' + lambda' underline u'$



    They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.



    $lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$



    I know that



    $lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $



    but cant get any further.






    linear-algebra geometry vectors







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      i'm trying to do a question for which I was given the following line equations:



      $underline r = underline a + lambda underline u$



      $underline r' = underline a' + lambda' underline u'$



      They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.



      $lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$



      I know that



      $lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $



      but cant get any further.






      linear-algebra geometry vectors







      share|cite|improve this question









      $endgroup$




      i'm trying to do a question for which I was given the following line equations:



      $underline r = underline a + lambda underline u$



      $underline r' = underline a' + lambda' underline u'$



      They then gave me this relationship without any justification, i've been trying to get my head around it but have not had much luck.



      $lvert underline r-underline r'rvert^2lvert underline u times underline u'rvert^2=lvert (underline a - underline a') cdot (underline u times underline u')lvert^2+lvert (underline r - underline r') times (underline u times underline u')lvert^2$



      I know that



      $lvert (underline r - underline r') times (underline u times underline u')lvert = lvert underline r-underline r'rvertlvert underline u times underline u'rvertsin theta $



      but cant get any further.






      linear-algebra geometry vectors




      linear-algebra geometry vectors






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 29 at 12:52









      A. PavlenkoA. Pavlenko

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      123






















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          $begingroup$

          You should be familiar with some basic properties of the cross product. Especially the identity



          $$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$



          As can be seen, for instance from:



          $$
          begin{aligned}
          || mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
          &= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
          &= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
          &= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
          &= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
          end{aligned}
          $$



          Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and



          $$
          mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
          mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
          $$






          share|cite|improve this answer









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            1 Answer
            1






            active

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            active

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            active

            oldest

            votes









            1












            $begingroup$

            You should be familiar with some basic properties of the cross product. Especially the identity



            $$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$



            As can be seen, for instance from:



            $$
            begin{aligned}
            || mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
            &= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
            &= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
            &= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
            &= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
            end{aligned}
            $$



            Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and



            $$
            mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
            mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
            $$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              You should be familiar with some basic properties of the cross product. Especially the identity



              $$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$



              As can be seen, for instance from:



              $$
              begin{aligned}
              || mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
              &= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
              &= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
              &= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
              &= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
              end{aligned}
              $$



              Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and



              $$
              mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
              mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
              $$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                You should be familiar with some basic properties of the cross product. Especially the identity



                $$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$



                As can be seen, for instance from:



                $$
                begin{aligned}
                || mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
                &= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
                &= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
                &= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
                &= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
                end{aligned}
                $$



                Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and



                $$
                mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
                mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
                $$






                share|cite|improve this answer









                $endgroup$



                You should be familiar with some basic properties of the cross product. Especially the identity



                $$ || mathbf{a} times mathbf{b} ||^2 = ||mathbf{a}||^2 , || mathbf{b}||^2 - |mathbf{a} cdot mathbf{b}|^2 $$



                As can be seen, for instance from:



                $$
                begin{aligned}
                || mathbf{a} times mathbf{b} ||^2 & = (mathbf{a} times mathbf{b}) cdot (mathbf{a} times mathbf{b}) \
                &= mathbf{b} cdot ((mathbf{a} times mathbf{b}) times mathbf{a}) \
                &= -mathbf{b} cdot (mathbf{a} times (mathbf{a} times mathbf{b})) \
                &= -mathbf{b} cdot (mathbf{a},(mathbf{a}cdotmathbf{b}) - mathbf{b},(mathbf{a}cdotmathbf{a})) \
                &= ||mathbf{a}||^2,||mathbf{b}||^2 - |mathbf{a}cdotmathbf{b}|^2
                end{aligned}
                $$



                Now, with $mathbf{a} = mathbf{r} - mathbf{r'}$ and $mathbf{b} = mathbf{u} times mathbf{u'}$, use the bilinearity of the inner product and



                $$
                mathbf{r} = mathbf{a} + lambda ,mathbf{u}; ,, mathbf{r'} = mathbf{a'} + lambda' ,mathbf{u'} \
                mathbf{u} cdot (mathbf{u} times mathbf{u'}) = mathbf{u'} cdot (mathbf{u} times mathbf{u'}) = 0
                $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 29 at 14:06









                jgbjgb

                14813




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