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Question
Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that
`1/a^2 + 1/b^2 + 1/c^2 = 1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2)`
Solution
Let’s take OX, OY, OZ and ox, oy, oz to be two rectangular systems.
And, the equations of two planes are
`"X"/a + "Y"/b + "Z"/c` = 1 ......(i)
And `x/(a"'") + y/(b"'") + z/(c"'")` = 1 .....(ii)
Length of perpendicular from origin to plane (i) is
= `|(0/a + 0/b + 0/c - 1)/sqrt(1/a^2 + 1/b^2 + 1/c^2)|`
= `1/sqrt(1/a^2 + 1/b^2 + 1/c^2)`
Length of perpendicular from origin to plane (ii)
= `|(0/(a"'") + 0/(b"'") + 0/(c"'") - 1)/sqrt(1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2))|`
= `1/sqrt(1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2))`
As per the condition of the question
`1/sqrt(1/a^2 + 1/b^2 + 1/c^2) = 1/sqrt(1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2))`
Thus, `1/a^2 + 1/b^2 + 1/c^2 = 1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2)`
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