Test Index

NCERT Class XII Mathematics Chapter - - Solutions

© examsnet.com
Question : 80 of 78
Marks: +1, -0
Prove that if a plane has the intercepts a, b, c and is at a distance of p units from the origin, then 1a2+1b2+1c2=1p2\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{p^2}
Solution:  
The equation of the plane in the intercept form is xa+yb+zc=1\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1
Distance of the plane from origin is p units.
⇒∣1a⋅0+1b⋅0+1c⋅0−1∣(1a)2+(1b)2+(1c)2=p\Rightarrow \frac{\left| \frac{1}{a} \cdot 0 + \frac{1}{b} \cdot 0 + \frac{1}{c} \cdot 0 - 1 \right|}{\sqrt{ \left(\frac{1}{a}\right)^2 + \left(\frac{1}{b}\right)^2 + \left(\frac{1}{c}\right)^2 }} = p
⇒1a2+1b2+1c2=1p2\Rightarrow \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}= \frac{1}{p^2}
© examsnet.com
Go to Question:
Prev Question
Next Question