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.

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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Answer 2

Yes, it is true. If the planes (x = cy + bz), (y = cx + az), and (z = bx + ay) go through the same straight line, then it follows that (a^2 + b^2 + c^2 + 2abc = 1). This relationship arises from the condition that the three planes intersect along a common line.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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