Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
0 votes

Prove that if a plane has the intercepts $a, b, c$ and is at a distance of $p$ units from the origin, then $\large\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} = \frac{1}{p^2}$.

Can you answer this question?

1 Answer

0 votes
  • The perpendicular distance of a plane from the origin is $p=\large\frac{d}{\sqrt{A^2+B^2+C^2}}$
Step 1:
Equation of a plane with intercepts $a,b,c$ is $\large\frac{x}{a}+\large\frac{y}{b}+\large\frac{z}{c}=$$1$
The perpendicular distance of a plane is $p=\large\frac{Ax_1+By_1+cz_1+d}{\sqrt{A^2+B^2+C^2}}$
From the origin $(x_1,y_1,z_1)\Rightarrow (0,0,0)$
We can written as
The perpendicular distance of a plane from the origin is $p=\large\frac{d}{\sqrt{A^2+B^2+C^2}}$
$\Rightarrow p=\large\frac{1}{\sqrt{\big(\Large\frac{1}{a}\big)^2+\big(\Large\frac{1}{b}\big)^2+\big(\Large\frac{1}{c}\big)^2}}$
$\Rightarrow p=\large\frac{1}{\sqrt{\big(\Large\frac{1}{a^2}\big)+\big(\Large\frac{1}{b^2}\big)+\big(\Large\frac{1}{c^2}\big)}}$
Step 2:
Squaring on both sides we get
$\Rightarrow p^2=\large\frac{a^2b^2c^2}{b^2c^2+a^2c^2+a^2b^2}$
Step 3:
Reciprocating on both sides and simplifying we get
Hence proved.
answered Jun 5, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App