If the planes #x=cy+bz# , #y=cx+az# , #z=bx+ay# go through the straight line, then is it true that #a^2+b^2+c^2+2abc=1#?

1 Answer
Shwetank Mauria
Mar 9, 2017

Yes, it is true. Please see below for details.

Explanation:

Let the planes #x=cy+bz# , #y=cx+az# , #z=bx+ay# go through the straight line defined by #(p,q,r)#. The planes can also be written as

#x-cy-bz=0# , #cx-y+az=0# , #bx+ay-z=0#

As the planes pass through line #(p,q,r)#, the line is perpendicular to the normal of the plane, say #x-cy-bz=0# and hence dot product should be zero i.e.

#p-cq-br=0#

Similarly #cp-q+ar=0# and

#bp+aq-r=0#

Solving them for #p,q# and #r# from first two equations, we get

#p/(-ac-b)=(-q)/(a+bc)=r/(-1+c^2)#

which implies #p=-ac-b#, #q=-a-bc# and #r=c^2-1#

and substituting in third we get

#b(-ac-b)+a(-a-bc)-(c^2-1)=0#

or #-abc-b^2-a^2-abc-c^2+1=0#

or #a^2+b^2+c^2+2abc=1#

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