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  • Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a’, b’, c’ respectively from the origin, prove that: frac{1}{a^{2}}+frac{1}{b^{2}}+frac{1}{c^{2}}=frac{1}{a'^{2}}+frac{1}{b'^{2}}+frac{1}{c'^{2}}

Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a’, b’, c’ respectively from the origin, prove that:

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

Answers (1)

Given, we have 2 systems of rectangular axes. Both the systems have the same origin, and there is a plane that cuts both systems.

One system is cut at a distance of a, b, c.

The other system is cut at a distance of a’, b’, c’.

To prove:

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

Proof: Since a plane intersects both the systems at distances a, b, c, and   a’, b’, c’ respectively, this plane will have different equations in the two different systems.

Let us consider the equation of the plane in the system with distances a, b, c to be:

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

Let us consider the equation of the plane in the system with distances a’, b’, c’ be:

\frac{x}{a'}+\frac{y}{b'}+\frac{z}{c'}=1

According to the question, the plane cuts both the systems from the origin. We know, the perpendicular distance of a plane ax + by + cz + d =0 from the origin is given by:

\left | \frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}} \right |

(where not all a, b, and c are zero)

Therefore, the perpendicular distance from the origin of the first plane  is:

\left | \frac{-1}{\sqrt{\left (\frac{1}{a} \right )^{2}+\left (\frac{1}{b} \right )^{2}+\left (\frac{1}{c} \right )^{2}}} \right |

And, the perpendicular distance from the origin of the second plane:

\left | \frac{-1}{\sqrt{\left (\frac{1}{a'} \right )^{2}+\left (\frac{1}{b'} \right )^{2}+\left (\frac{1}{c'} \right )^{2}}} \right |

We also know, if two systems of lines have the same origin, their perpendicular distances from the origin to the plane in both systems     are equal.

Therefore,

\left | \frac{-1}{\sqrt{\left (\frac{1}{a} \right )^{2}+\left (\frac{1}{b} \right )^{2}+\left (\frac{1}{c} \right )^{2}}} \right |=\left | \frac{-1}{\sqrt{\left (\frac{1}{a'} \right )^{2}+\left (\frac{1}{b'} \right )^{2}+\left (\frac{1}{c'} \right )^{2}}} \right |

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

Cross-multiplying,

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

Squaring both sides,

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

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

 

Or

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

Hence, proved.

 

Posted by

infoexpert24

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