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Two systems of rectangular axes have the same origin. If a plane cuts then at distances $$a,b,c$$ and $$a', b', c'$$ from the origin then

JEE · Math · previous-year question

  1. A.$${1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0$$correct
  2. B.$$\,{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} + {1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}} = 0$$
  3. C.$${1 \over {{a^2}}} + {1 \over {{b^2}}} - {1 \over {{c^2}}} + {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0$$
  4. D.$${1 \over {{a^2}}} - {1 \over {{b^2}}} - {1 \over {{c^2}}} + {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0$$

Answer

A. $${1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0$$

Explanation

Equation of planes be $${x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1$$ So the distance from (0, 0, 0) to both the plane is same. $$ \therefore $$ $$\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }}} \right| = \left| {{{ - 1} \over {\sqrt {{1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}}} }}} \right|$$ $$ \Rightarrow $$ $${1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {{{a'}^{2}}}} - {1 \over {{{b'}^{2}}}} - {1 \over {{{c'}^{2}}}} = 0$$

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