Two systems of rectangular axes have the same origin. If a plane cuts them at distance a, b, c and a', b', c' from the origin, then
A
1a2+1b2+1c2+1a′2+1b′2+1c′2=0
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B
1a2+1b2−1c2+1a′2+1b′2−1c′2=0
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C
1a2−1b2−1c2+1a′2−1b′2−1c′2=0
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D
1a2+1b2+1c2−1a′2−1b′2−1c′2=0
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Solution
The correct option is D1a2+1b2+1c2−1a′2−1b′2−1c′2=0 Equation of planes be xa+yb+zc=1andxa′+yb′+zc′=1 (Perpendicular distances of planes from origin is same) ∣∣
∣
∣∣−1√1a2+1b2+1c2∣∣
∣
∣∣=∣∣
∣
∣∣−1√1a′2+1b′2+1c′2∣∣
∣
∣∣∴∑1a2−∑1a′2=0.