Prove that if a line l 1 ≠ l is sent to itself under a reflection through l, then l 1 and l intersect at right angles

Question

Prove that if a line       l    1    ≠ l is sent to itself under a reflection through l, then       l    1   and l intersect at right angles

Alisa Medina · Accepted Answer

l  e  t  l  1  b  e            (      x      −      x      1      )      ÷      b      1      =      (      y      −      y      1      )              b      1        =            (      y      −      y      1      )              b      2        =            (      z      −      z      1      )              b      3        a  n  d  l  b  e  t  h  e            (      x      −      x      2      )              a      1        =            (      y      −      y      2      )              a      2        =            (      z      −      z      2      )              a      3          are the cartesian equation of these two lines in space having direction ratios of these lines are    b1,b2,b3 and a1,a2,a3 respectively    as from the qyestion      l    1    ≠  l but it reflected l1 as a mirror image .....    so we know that sum of product of direction cosine of these lines must be equal to zero then        l    1    ×      l    2    +      m    1    ×      m    2    +      n    1    ×      n    2    =  0  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  .  (  1  )    and we know that   l  1  =  b  1      /    (  b      1    2    +  b      2    2    +  b      3    2        )          0.5        ,  m  1  =  b  2      /    (  b      1    2    +  b      2    2    +  b      3    2        )          0.5        ,  n  1  =  b  3      /    (  b      1    2    +  b      2    2    +  b      3    2        )          0.5          also   l  2  =  a  1      /    (  a      1    2    +  a      2    2    +  a      3    2        )          0.5        ,  m  2  =  a  2      /    (  a      1    2    +  a      2    2    +  a      3    2        )          0.5        ,  n  2  =  a  3      /    (  a      1    2    +  a      2    2    +  a      3    2        )          0.5          put these all value in equation (1) and we get,    a  1  b  1  +  a  2  b  2  +  a  3  b  3  =  0......................2    and we know that ......    cos  ⁡  T  h  e  t  a  =            (      a      1      b      1      +      a      2      b      2      +      a      3      b      3              (      a      12      +      a      22      +      a              32                  0.5                      ×  (  b      1    2    +  b      2    2    +  b      3    2        )          0.5          then from equation   2    cos  ⁡  T  h  e  t  a  =  0    therefore Theta  =  90  degree    which prove that both the lines l_1and l are make right angle to each other....    proved.

Prove that if a line l_1 neq l is sent to itself under a reflection through l, then l1 and l intersect at right angles

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