Question

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`?

Answer 1

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`