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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 OX, OY, OZ and ox, oy, oz be two rectangular systems
∴ 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)``
Hence, `1/"a"^2 + 1/"b"^2 + 1/"c"^2 = 1/"a'"^2 + 1/"b'"^2 + 1/"c'"^2`
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