Question

A variable plane moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant $\lambda .$ It passes through a fixed point, which has coordinates

• A. $(\lambda ,\lambda ,\lambda )$
• B. $(\frac{1}{\lambda },\frac{1}{\lambda }.\frac{1}{\lambda })$
  
• C. $(-\lambda ,-\lambda ,-\lambda )$
• D. $(-\frac{1}{\lambda },-\frac{1}{\lambda },-\frac{1}{\lambda })$

Answer 1

The correct option is B

$(\frac{1}{\lambda },\frac{1}{\lambda }.\frac{1}{\lambda })$

Let equation of the variable plane be $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\left(1\right)$
The intercepts on the coordinate axes are a, b , c. The sum of reeciprocals of intercepts is constant $\lambda ,$ therefore
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\lambda or\frac{\left(\frac{1}{\lambda }\right)}{a}+\frac{\left(\frac{1}{\lambda }\right)}{b}+\frac{\left(\frac{1}{\lambda }\right)}{c}=1\therefore (\frac{1}{\lambda },\frac{1}{\lambda },\frac{1}{\lambda })$

lies on the plane (1).
Hence, the variable plane (1) always passes
through the fixed point $(\frac{1}{\lambda },\frac{1}{\lambda }.\frac{1}{\lambda }).$