If p,q,r,x,y,z are non zero real number such that p x + q y + r z + ( p 2 + q 2 + r 2 ) ( x 2 + y 2 + z 2 ) = 0Then p y q x + q z r y + r x p z iswhat try ( p x + q y + r z ) 2 = ( p 2 + q 2 + r 2 ) ( x 2 + y 2 + z 2 ) p 2 x 2 + q 2 y 2 + r 2 z 2 + 2 p q x y + 2 q r y z + 2 p r x z = p 2 x 2 + p 2 y 2 + p 2 z 2 + q 2 x 2 + q 2 y 2 + q 2 z 2 + r 2 x 2 + r 2 y 2 + r 2 z 2 2 p q x y + 2 q r y z + 2 p r x z = p 2 y 2 + p 2 z 2 + q 2 x 2 + q 2 z 2 + r 2 x 2 + r 2 y 2 How do i solve it?

Question

If p,q,r,x,y,z are non zero real number such that  p  x  +  q  y  +  r  z  +      (          p      2        +          q      2        +          r      2        )    (          x      2        +          y      2        +          z      2        )    =  0Then                     p        y                    q        x              +                  q        z                    r        y              +                  r        x                    p        z             iswhat try  (  p  x  +  q  y  +  r  z      )    2    =  (      p    2    +      q    2    +      r    2    )  (      x    2    +      y    2    +      z    2    )      p    2        x    2    +      q    2        y    2    +      r    2        z    2    +  2  p  q  x  y  +  2  q  r  y  z  +  2  p  r  x  z  =      p    2        x    2    +      p    2        y    2    +      p    2        z    2    +      q    2        x    2    +      q    2        y    2    +      q    2        z    2    +      r    2        x    2    +      r    2        y    2    +      r    2        z    2    2  p  q  x  y  +  2  q  r  y  z  +  2  p  r  x  z  =      p    2        y    2    +      p    2        z    2    +      q    2        x    2    +      q    2        z    2    +      r    2        x    2    +      r    2        y    2  How do i solve it?

Zackary Galloway · Accepted Answer

Write   v  =  (  p  ,  q  ,  r  ) and   w  =  (  x  ,  y  ,  z  ). Then the given relation states  v  ⋅  w  +      |    v      |        |    w      |    =  0But   v  ⋅  w  =      |    v      |        |    w      |    cos  ⁡  θ where   θ is the angle between them, so      |    v      |        |    w      |    (  cos  ⁡  θ  +  1  )  =  0Since none of the scalars are zero, we get   cos  ⁡  θ  =  −  1, so v=kw for some nonzero   k  ∈      R   and            p      y              q      x        +            q      z              r      y        +            r      x              p      z        =            k      x      y              k      y      x        +            k      y      z              k      z      y        +            k      z      x              k      x      z        =  3

Kallie Fritz · Accepted Answer

By C-S  0  =  p  x  +  q  y  +  r  z  +      (          p      2        +          q      2        +          r      2        )    (          x      2        +          y      2        +          z      2        )    ≥  p  x  +  q  y  +  r  z  +      |    p  x  +  q  y  +  r  z      |  and since  p  x  +  q  y  +  r  z  +      |    p  x  +  q  y  +  r  z      |    ≥  0  ,we obtain  p  x  +  q  y  +  r  z  +      |    p  x  +  q  y  +  r  z      |    =  0  ,which gives  p  x  +  q  y  +  r  z  ≤  0.Also, the equality occurs for  (  x  ,  y  ,  z  )      |        |    (  p  ,  q  ,  r  )  ,which says that there is k&amp;lt;0, for which  (  p  ,  q  ,  r  )  =  k  (  x  ,  y  ,  z  )  .Thus,      ∑          c      y      c                  p      y              q      x        =      ∑          c      y      c                  k      x      y              k      y      x        =  3.

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