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If (a-a')^2+(b-b')^2+(c-c')^2=p

and (ab'-a'b)^2+(bc'-b'c)^2+(ca'-c'a)^2=q
then the perpendicular distance of the line
ax+by+cz=1,a'x+b'y+c'z=1 is

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madan how did u get (p/q)^0.5
finally got it


The plane having the line is
ax+by+cz +k(a'x+b'y+c'z)=1+k
the distance from origin
d= \frac{|1+m|}{ \sqrt{(a+ma')^{2} +(b+mb')^{2} +(c+mc')^{2} } }
d^2= \alpha = \frac{1+2m+m^{2} }{(a'^2+b^2+c'^2)m^2+(2aa'+2bb'+2cc')m+(a^2+b^2+c^2)}
take (a'^2+b^2+c'^2)=A 
 \alpha = \frac{1+2m+m^2}{Am^2+Bm+C}
(A \alpha -1)m^2+(B\alpha -2)m+(C \alpha -1)=0
the dicriminat is greater than zero
u can see that
 \frac{4(A+C-B)}{(4AC-B^2)} = \frac{p}{q}
0 \leq  \alpha  \leq  \frac{p}{q} \\ d= \sqrt{ \frac{p}{q} }

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