JEE Main 2003Annual, Paper 1MathematicsMediumThree Dimensional Geometry249× since 2002Two systems of rectangular axes have the same origin. If a plane cuts then at distances a,b,ca,b,ca,b,c and a′,b′,c′a', b', c'a′,b′,c′ from the origin thenAOption A. 1a2+1b2+1c2−1a′2−1b′2−1c′2=0{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0a21+b21+c21−a′21−b′21−c′21=0BOption B. 1a2+1b2+1c2+1a′2+1b′2+1c′2=0\,{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} + {1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}} = 0a21+b21+c21+a′21+b′21+c′21=0COption C. 1a2+1b2−1c2+1a′2−1b′2−1c′2=0{1 \over {{a^2}}} + {1 \over {{b^2}}} - {1 \over {{c^2}}} + {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0a21+b21−c21+a′21−b′21−c′21=0DOption D. 1a2−1b2−1c2+1a′2−1b′2−1c′2=0{1 \over {{a^2}}} - {1 \over {{b^2}}} - {1 \over {{c^2}}} + {1 \over {a{'^2}}} - {1 \over {b{'^2}}} - {1 \over {c{'^2}}} = 0a21−b21−c21+a′21−b′21−c′21=0CheckSkip · show solution
