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

  • Option 1)

    \frac{1}{a^{2}}+\frac{1}{b^{2}}-\frac{1}{c^{2}}+\frac{1}{a'^{2}}+\frac{1}{b'^{2}}-\frac{1}{c'^{2}}=0

  • Option 2)

    \frac{1}{a^{2}}-\frac{1}{b^{2}}-\frac{1}{c^{2}}+\frac{1}{a'^{2}}-\frac{1}{b'^{2}}-\frac{1}{c'^{2}}=0

  • Option 3)

    \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-\frac{1}{a'^{2}}-\frac{1}{b'^{2}}-\frac{1}{c'^{2}}=0

  • Option 4)

    \frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}=0

 

Answers (1)

As we learnt in

Distance of a point from plane (Cartesian form) -

The length of perpendicular from P(x_{1},y_{1},z_{1}) to the plane

ax+by+cz+d= 0 is given by  \frac{\left [ ax_{1}+by_{1} +cz_{1}+d\right ]}{\left | \sqrt{a^{2}+b^{2}+c^{2}} \right |}

 

-

 

 Clearly with reference to both rectangular axis, the distance of plane from origin doesn't change. 

 


Option 1)

\frac{1}{a^{2}}+\frac{1}{b^{2}}-\frac{1}{c^{2}}+\frac{1}{a'^{2}}+\frac{1}{b'^{2}}-\frac{1}{c'^{2}}=0

Incorrect option

Option 2)

\frac{1}{a^{2}}-\frac{1}{b^{2}}-\frac{1}{c^{2}}+\frac{1}{a'^{2}}-\frac{1}{b'^{2}}-\frac{1}{c'^{2}}=0

Incorrect option

Option 3)

\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}-\frac{1}{a'^{2}}-\frac{1}{b'^{2}}-\frac{1}{c'^{2}}=0

Correct option

Option 4)

\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}+\frac{1}{a'^{2}}+\frac{1}{b'^{2}}+\frac{1}{c'^{2}}=0

Incorrect option

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