MCQ
Two systems of rectangular axes have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$ from the origin, then
  • $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
  • B
    $\frac{1}{ a ^2}+\frac{1}{b^2}+\frac{1}{ c ^2}+\frac{1}{ a ^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{ c ^{\prime 2}}=0$
  • C
    $\frac{1}{a^2}+\frac{1}{b^2}-\frac{1}{c^2}+\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
  • D
    $\frac{1}{ a ^2}-\frac{1}{b^2}-\frac{1}{ c ^2}+\frac{1}{ a ^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{ c ^{\prime 2}}=0$

Answer

Correct option: A.
$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{a^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{c^{\prime 2}}=0$
(A)
The equations of the plane with reference to the two systems of rectangular axes are
$\frac{x}{ a }+\frac{y}{b}+\frac{ z }{ c }=1... (i)$
and $\frac{ X }{ a ^{\prime}}+\frac{ Y }{ b ^{\prime}}+\frac{ Z }{ c ^{\prime}}=1... (ii)$
Since the origin of axes is same.
∴ Length of the perpendicular from $(0,0,0)$ on plane (i)
$=$ Length of the perpendicular from $(0,0,0)$ on plane (ii)
$\Rightarrow\left|\frac{-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\left|\frac{-1}{\sqrt{\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}}}\right|$
$\Rightarrow \frac{1}{ a ^2}+\frac{1}{b^2}+\frac{1}{ c ^2}-\frac{1}{ a ^{\prime 2}}-\frac{1}{b^{\prime 2}}-\frac{1}{ c ^{\prime 2}}=0$

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