If p,q,r,x,y,z are non zero real number such that px+qy+rz+sqrt((p^2+q^2+r^2)(x^2+y^2+z^2))=0 Then (py)/(qx)+(qz)/(ry)+(rx)/(pz) is what try (px+qy+rz)^2=(p^2+q^2+r^2)(x^2+y^2+z^2) p^2x^2+q^2y^2+r^2z^2+2pqxy+2qryz+2prxz=p^2x^2+p^2y^2+p^2z^2+q^2x^2+q^2y^2+q^2z^2+r^2x^2+r^2y^2+r^2z^2 2pqxy+2qryz+2prxz=p^2y^2+p^2z^2+q^2x^2+q^2z^2+r^2x^2+r^2y^2 How do i solve it Help me please

Krish Crosby 2022-09-14 Answered
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 ) = 0
Then p y q x + q z r y + r x p z is
what 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?
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Answers (2)

Zackary Galloway
Answered 2022-09-15 Author has 4 answers
Write v = ( p , q , r ) and w = ( x , y , z ). Then the given relation states
v w + | v | | w | = 0
But v w = | v | | w | cos θ where θ is the angle between them, so
| v | | w | ( cos θ + 1 ) = 0
Since 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
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Kallie Fritz
Answered 2022-09-16 Author has 1 answers
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<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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